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# 15.2   Fractional knapsack problem
!!! question
Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown below.
![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
<p align="center"> Figure 15-3 &nbsp; Example data of the fractional knapsack problem </p>
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack.
The difference is that, in this problem, only a part of an item can be chosen. As shown in the Figure 15-4 , **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value.
2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
![Value per unit weight of the item](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png){ class="animation-figure" }
<p align="center"> Figure 15-4 &nbsp; Value per unit weight of the item </p>
### 1. &nbsp; Greedy strategy determination
Maximizing the total value of the items in the knapsack essentially means maximizing the value per unit weight. From this, the greedy strategy shown below can be deduced.
1. Sort the items by their unit value from high to low.
2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**.
3. If the remaining capacity of the knapsack is insufficient, use part of the current item to fill the knapsack.
![Greedy strategy of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png){ class="animation-figure" }
<p align="center"> Figure 15-5 &nbsp; Greedy strategy of the fractional knapsack problem </p>
### 2. &nbsp; Code implementation
We have created an `Item` class in order to sort the items by their unit value. We loop and make greedy choices until the knapsack is full, then exit and return the solution:
=== "Python"
```python title="fractional_knapsack.py"
class Item:
"""物品"""
def __init__(self, w: int, v: int):
self.w = w # 物品重量
self.v = v # 物品价值
def fractional_knapsack(wgt: list[int], val: list[int], cap: int) -> int:
"""分数背包：贪心"""
# 创建物品列表，包含两个属性：重量、价值
items = [Item(w, v) for w, v in zip(wgt, val)]
# 按照单位价值 item.v / item.w 从高到低进行排序
items.sort(key=lambda item: item.v / item.w, reverse=True)
# 循环贪心选择
res = 0
for item in items:
if item.w <= cap:
# 若剩余容量充足，则将当前物品整个装进背包
res += item.v
cap -= item.w
else:
# 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap
# 已无剩余容量，因此跳出循环
break
return res
```
=== "C++"
```cpp title="fractional_knapsack.cpp"
/* 物品 */
class Item {
public:
int w; // 物品重量
int v; // 物品价值
Item(int w, int v) : w(w), v(v) {
}
};
/* 分数背包：贪心 */
double fractionalKnapsack(vector<int> &wgt, vector<int> &val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
vector<Item> items;
for (int i = 0; i < wgt.size(); i++) {
items.push_back(Item(wgt[i], val[i]));
}
// 按照单位价值 item.v / item.w 从高到低进行排序
sort(items.begin(), items.end(), [](Item &a, Item &b) { return (double)a.v / a.w > (double)b.v / b.w; });
// 循环贪心选择
double res = 0;
for (auto &item : items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double)item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Java"
```java title="fractional_knapsack.java"
/* 物品 */
class Item {
int w; // 物品重量
int v; // 物品价值
public Item(int w, int v) {
this.w = w;
this.v = v;
}
}
/* 分数背包：贪心 */
double fractionalKnapsack(int[] wgt, int[] val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item[] items = new Item[wgt.length];
for (int i = 0; i < wgt.length; i++) {
items[i] = new Item(wgt[i], val[i]);
}
// 按照单位价值 item.v / item.w 从高到低进行排序
Arrays.sort(items, Comparator.comparingDouble(item -> -((double) item.v / item.w)));
// 循环贪心选择
double res = 0;
for (Item item : items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double) item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "C#"
```csharp title="fractional_knapsack.cs"
/* 物品 */
class Item(int w, int v) {
public int w = w; // 物品重量
public int v = v; // 物品价值
}
/* 分数背包：贪心 */
double FractionalKnapsack(int[] wgt, int[] val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item[] items = new Item[wgt.Length];
for (int i = 0; i < wgt.Length; i++) {
items[i] = new Item(wgt[i], val[i]);
}
// 按照单位价值 item.v / item.w 从高到低进行排序
Array.Sort(items, (x, y) => (y.v / y.w).CompareTo(x.v / x.w));
// 循环贪心选择
double res = 0;
foreach (Item item in items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (double)item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Go"
```go title="fractional_knapsack.go"
/* 物品 */
type Item struct {
w int // 物品重量
v int // 物品价值
}
/* 分数背包：贪心 */
func fractionalKnapsack(wgt []int, val []int, cap int) float64 {
// 创建物品列表，包含两个属性：重量、价值
items := make([]Item, len(wgt))
for i := 0; i < len(wgt); i++ {
items[i] = Item{wgt[i], val[i]}
}
// 按照单位价值 item.v / item.w 从高到低进行排序
sort.Slice(items, func(i, j int) bool {
return float64(items[i].v)/float64(items[i].w) > float64(items[j].v)/float64(items[j].w)
})
// 循环贪心选择
res := 0.0
for _, item := range items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += float64(item.v)
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += float64(item.v) / float64(item.w) * float64(cap)
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "Swift"
```swift title="fractional_knapsack.swift"
/* 物品 */
class Item {
var w: Int // 物品重量
var v: Int // 物品价值
init(w: Int, v: Int) {
self.w = w
self.v = v
}
}
/* 分数背包：贪心 */
func fractionalKnapsack(wgt: [Int], val: [Int], cap: Int) -> Double {
// 创建物品列表，包含两个属性：重量、价值
var items = zip(wgt, val).map { Item(w: $0, v: $1) }
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort { -(Double($0.v) / Double($0.w)) < -(Double($1.v) / Double($1.w)) }
// 循环贪心选择
var res = 0.0
var cap = cap
for item in items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += Double(item.v)
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += Double(item.v) / Double(item.w) * Double(cap)
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "JS"
```javascript title="fractional_knapsack.js"
/* 物品 */
class Item {
constructor(w, v) {
this.w = w; // 物品重量
this.v = v; // 物品价值
}
}
/* 分数背包：贪心 */
function fractionalKnapsack(wgt, val, cap) {
// 创建物品列表，包含两个属性：重量、价值
const items = wgt.map((w, i) => new Item(w, val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => b.v / b.w - a.v / a.w);
// 循环贪心选择
let res = 0;
for (const item of items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "TS"
```typescript title="fractional_knapsack.ts"
/* 物品 */
class Item {
w: number; // 物品重量
v: number; // 物品价值
constructor(w: number, v: number) {
this.w = w;
this.v = v;
}
}
/* 分数背包：贪心 */
function fractionalKnapsack(wgt: number[], val: number[], cap: number): number {
// 创建物品列表，包含两个属性：重量、价值
const items: Item[] = wgt.map((w, i) => new Item(w, val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => b.v / b.w - a.v / a.w);
// 循环贪心选择
let res = 0;
for (const item of items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (item.v / item.w) * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Dart"
```dart title="fractional_knapsack.dart"
/* 物品 */
class Item {
int w; // 物品重量
int v; // 物品价值
Item(this.w, this.v);
}
/* 分数背包：贪心 */
double fractionalKnapsack(List<int> wgt, List<int> val, int cap) {
// 创建物品列表，包含两个属性：重量、价值
List<Item> items = List.generate(wgt.length, (i) => Item(wgt[i], val[i]));
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort((a, b) => (b.v / b.w).compareTo(a.v / a.w));
// 循环贪心选择
double res = 0;
for (Item item in items) {
if (item.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v / item.w * cap;
// 已无剩余容量，因此跳出循环
break;
}
}
return res;
}
```
=== "Rust"
```rust title="fractional_knapsack.rs"
/* 物品 */
struct Item {
w: i32, // 物品重量
v: i32, // 物品价值
}
impl Item {
fn new(w: i32, v: i32) -> Self {
Self { w, v }
}
}
/* 分数背包：贪心 */
fn fractional_knapsack(wgt: &[i32], val: &[i32], mut cap: i32) -> f64 {
// 创建物品列表，包含两个属性：重量、价值
let mut items = wgt
.iter()
.zip(val.iter())
.map(|(&w, &v)| Item::new(w, v))
.collect::<Vec<Item>>();
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sort_by(|a, b| {
(b.v as f64 / b.w as f64)
.partial_cmp(&(a.v as f64 / a.w as f64))
.unwrap()
});
// 循环贪心选择
let mut res = 0.0;
for item in &items {
if item.w <= cap {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v as f64;
cap -= item.w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v as f64 / item.w as f64 * cap as f64;
// 已无剩余容量，因此跳出循环
break;
}
}
res
}
```
=== "C"
```c title="fractional_knapsack.c"
/* 物品 */
typedef struct {
int w; // 物品重量
int v; // 物品价值
} Item;
/* 分数背包：贪心 */
float fractionalKnapsack(int wgt[], int val[], int itemCount, int cap) {
// 创建物品列表，包含两个属性：重量、价值
Item *items = malloc(sizeof(Item) * itemCount);
for (int i = 0; i < itemCount; i++) {
items[i] = (Item){.w = wgt[i], .v = val[i]};
}
// 按照单位价值 item.v / item.w 从高到低进行排序
qsort(items, (size_t)itemCount, sizeof(Item), sortByValueDensity);
// 循环贪心选择
float res = 0.0;
for (int i = 0; i < itemCount; i++) {
if (items[i].w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += items[i].v;
cap -= items[i].w;
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += (float)cap / items[i].w * items[i].v;
cap = 0;
break;
}
}
free(items);
return res;
}
```
=== "Kotlin"
```kotlin title="fractional_knapsack.kt"
/* 物品 */
class Item(
val w: Int, // 物品
val v: Int // 物品价值
)
/* 分数背包：贪心 */
fun fractionalKnapsack(wgt: IntArray, _val: IntArray, c: Int): Double {
// 创建物品列表，包含两个属性：重量、价值
var cap = c
val items = arrayOfNulls<Item>(wgt.size)
for (i in wgt.indices) {
items[i] = Item(wgt[i], _val[i])
}
// 按照单位价值 item.v / item.w 从高到低进行排序
items.sortBy { item: Item? -> -(item!!.v.toDouble() / item.w) }
// 循环贪心选择
var res = 0.0
for (item in items) {
if (item!!.w <= cap) {
// 若剩余容量充足，则将当前物品整个装进背包
res += item.v
cap -= item.w
} else {
// 若剩余容量不足，则将当前物品的一部分装进背包
res += item.v.toDouble() / item.w * cap
// 已无剩余容量，因此跳出循环
break
}
}
return res
}
```
=== "Ruby"
```ruby title="fractional_knapsack.rb"
[class]{Item}-[func]{}
[class]{}-[func]{fractional_knapsack}
```
=== "Zig"
```zig title="fractional_knapsack.zig"
[class]{Item}-[func]{}
[class]{}-[func]{fractionalKnapsack}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=class%20Item%3A%0A%20%20%20%20%22%22%22%E7%89%A9%E5%93%81%22%22%22%0A%20%20%20%20def%20__init__%28self,%20w%3A%20int,%20v%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.w%20%3D%20w%20%20%23%20%E7%89%A9%E5%93%81%E9%87%8D%E9%87%8F%0A%20%20%20%20%20%20%20%20self.v%20%3D%20v%20%20%23%20%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%0A%0Adef%20fractional_knapsack%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%88%86%E6%95%B0%E8%83%8C%E5%8C%85%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9B%E5%BB%BA%E7%89%A9%E5%93%81%E5%88%97%E8%A1%A8%EF%BC%8C%E5%8C%85%E5%90%AB%E4%B8%A4%E4%B8%AA%E5%B1%9E%E6%80%A7%EF%BC%9A%E9%87%8D%E9%87%8F%E3%80%81%E4%BB%B7%E5%80%BC%0A%20%20%20%20items%20%3D%20%5BItem%28w,%20v%29%20for%20w,%20v%20in%20zip%28wgt,%20val%29%5D%0A%20%20%20%20%23%20%E6%8C%89%E7%85%A7%E5%8D%95%E4%BD%8D%E4%BB%B7%E5%80%BC%20item.v%20/%20item.w%20%E4%BB%8E%E9%AB%98%E5%88%B0%E4%BD%8E%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20items.sort%28key%3Dlambda%20item%3A%20item.v%20/%20item.w,%20reverse%3DTrue%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20for%20item%20in%20items%3A%0A%20%20%20%20%20%20%20%20if%20item.w%20%3C%3D%20cap%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E5%85%85%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E6%95%B4%E4%B8%AA%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20item.v%0A%20%20%20%20%20%20%20%20%20%20%20%20cap%20-%3D%20item.w%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E4%B8%8D%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E7%9A%84%E4%B8%80%E9%83%A8%E5%88%86%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20%28item.v%20/%20item.w%29%20*%20cap%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B7%B2%E6%97%A0%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%9B%A0%E6%AD%A4%E8%B7%B3%E5%87%BA%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20fractional_knapsack%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=8&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=class%20Item%3A%0A%20%20%20%20%22%22%22%E7%89%A9%E5%93%81%22%22%22%0A%20%20%20%20def%20__init__%28self,%20w%3A%20int,%20v%3A%20int%29%3A%0A%20%20%20%20%20%20%20%20self.w%20%3D%20w%20%20%23%20%E7%89%A9%E5%93%81%E9%87%8D%E9%87%8F%0A%20%20%20%20%20%20%20%20self.v%20%3D%20v%20%20%23%20%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%0A%0Adef%20fractional_knapsack%28wgt%3A%20list%5Bint%5D,%20val%3A%20list%5Bint%5D,%20cap%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%88%86%E6%95%B0%E8%83%8C%E5%8C%85%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9B%E5%BB%BA%E7%89%A9%E5%93%81%E5%88%97%E8%A1%A8%EF%BC%8C%E5%8C%85%E5%90%AB%E4%B8%A4%E4%B8%AA%E5%B1%9E%E6%80%A7%EF%BC%9A%E9%87%8D%E9%87%8F%E3%80%81%E4%BB%B7%E5%80%BC%0A%20%20%20%20items%20%3D%20%5BItem%28w,%20v%29%20for%20w,%20v%20in%20zip%28wgt,%20val%29%5D%0A%20%20%20%20%23%20%E6%8C%89%E7%85%A7%E5%8D%95%E4%BD%8D%E4%BB%B7%E5%80%BC%20item.v%20/%20item.w%20%E4%BB%8E%E9%AB%98%E5%88%B0%E4%BD%8E%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20items.sort%28key%3Dlambda%20item%3A%20item.v%20/%20item.w,%20reverse%3DTrue%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20for%20item%20in%20items%3A%0A%20%20%20%20%20%20%20%20if%20item.w%20%3C%3D%20cap%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E5%85%85%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E6%95%B4%E4%B8%AA%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20item.v%0A%20%20%20%20%20%20%20%20%20%20%20%20cap%20-%3D%20item.w%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%E4%B8%8D%E8%B6%B3%EF%BC%8C%E5%88%99%E5%B0%86%E5%BD%93%E5%89%8D%E7%89%A9%E5%93%81%E7%9A%84%E4%B8%80%E9%83%A8%E5%88%86%E8%A3%85%E8%BF%9B%E8%83%8C%E5%8C%85%0A%20%20%20%20%20%20%20%20%20%20%20%20res%20%2B%3D%20%28item.v%20/%20item.w%29%20*%20cap%0A%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E5%B7%B2%E6%97%A0%E5%89%A9%E4%BD%99%E5%AE%B9%E9%87%8F%EF%BC%8C%E5%9B%A0%E6%AD%A4%E8%B7%B3%E5%87%BA%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20return%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20wgt%20%3D%20%5B10,%2020,%2030,%2040,%2050%5D%0A%20%20%20%20val%20%3D%20%5B50,%20120,%20150,%20210,%20240%5D%0A%20%20%20%20cap%20%3D%2050%0A%20%20%20%20n%20%3D%20len%28wgt%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20fractional_knapsack%28wgt,%20val,%20cap%29%0A%20%20%20%20print%28f%22%E4%B8%8D%E8%B6%85%E8%BF%87%E8%83%8C%E5%8C%85%E5%AE%B9%E9%87%8F%E7%9A%84%E6%9C%80%E5%A4%A7%E7%89%A9%E5%93%81%E4%BB%B7%E5%80%BC%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=8&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
Apart from sorting, in the worst case, the entire list of items needs to be traversed, **hence the time complexity is $O(n)$**, where $n$ is the number of items.
Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
### 3. &nbsp; Correctness proof
Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$.
Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
As shown in the Figure 15-6 , if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }
<p align="center"> Figure 15-6 &nbsp; Geometric representation of the fractional knapsack problem </p>
en/docs/chapter_greedy/greedy_algorithm.md
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# 15.1 &nbsp; Greedy algorithms
<u>Greedy algorithm</u> is a common algorithm for solving optimization problems, which fundamentally involves making the seemingly best choice at each decision-making stage of the problem, i.e., greedily making locally optimal decisions in hopes of finding a globally optimal solution. Greedy algorithms are concise and efficient, and are widely used in many practical problems.
Greedy algorithms and dynamic programming are both commonly used to solve optimization problems. They share some similarities, such as relying on the property of optimal substructure, but they operate differently.
- Dynamic programming considers all previous decisions at the current decision stage and uses solutions to past subproblems to construct solutions for the current subproblem.
- Greedy algorithms do not consider past decisions; instead, they proceed with greedy choices, continually narrowing the scope of the problem until it is solved.
Let's first understand the working principle of the greedy algorithm through the example of "coin change," which has been introduced in the "Complete Knapsack Problem" chapter. I believe you are already familiar with it.
!!! question
Given $n$ types of coins, where the denomination of the $i$th type of coin is $coins[i - 1]$, and the target amount is $amt$, with each type of coin available indefinitely, what is the minimum number of coins needed to make up the target amount? If it is not possible to make up the target amount, return $-1$.
The greedy strategy adopted in this problem is shown in the following figure. Given the target amount, **we greedily choose the coin that is closest to and not greater than it**, repeatedly following this step until the target amount is met.
![Greedy strategy for coin change](greedy_algorithm.assets/coin_change_greedy_strategy.png){ class="animation-figure" }
<p align="center"> Figure 15-1 &nbsp; Greedy strategy for coin change </p>
The implementation code is as follows:
=== "Python"
```python title="coin_change_greedy.py"
def coin_change_greedy(coins: list[int], amt: int) -> int:
"""零钱兑换：贪心"""
# 假设 coins 列表有序
i = len(coins) - 1
count = 0
# 循环进行贪心选择，直到无剩余金额
while amt > 0:
# 找到小于且最接近剩余金额的硬币
while i > 0 and coins[i] > amt:
i -= 1
# 选择 coins[i]
amt -= coins[i]
count += 1
# 若未找到可行方案，则返回 -1
return count if amt == 0 else -1
```
=== "C++"
```cpp title="coin_change_greedy.cpp"
/* 零钱兑换：贪心 */
int coinChangeGreedy(vector<int> &coins, int amt) {
// 假设 coins 列表有序
int i = coins.size() - 1;
int count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1;
}
```
=== "Java"
```java title="coin_change_greedy.java"
/* 零钱兑换：贪心 */
int coinChangeGreedy(int[] coins, int amt) {
// 假设 coins 列表有序
int i = coins.length - 1;
int count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1;
}
```
=== "C#"
```csharp title="coin_change_greedy.cs"
/* 零钱兑换：贪心 */
int CoinChangeGreedy(int[] coins, int amt) {
// 假设 coins 列表有序
int i = coins.Length - 1;
int count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1;
}
```
=== "Go"
```go title="coin_change_greedy.go"
/* 零钱兑换：贪心 */
func coinChangeGreedy(coins []int, amt int) int {
// 假设 coins 列表有序
i := len(coins) - 1
count := 0
// 循环进行贪心选择，直到无剩余金额
for amt > 0 {
// 找到小于且最接近剩余金额的硬币
for i > 0 && coins[i] > amt {
i--
}
// 选择 coins[i]
amt -= coins[i]
count++
}
// 若未找到可行方案，则返回 -1
if amt != 0 {
return -1
}
return count
}
```
=== "Swift"
```swift title="coin_change_greedy.swift"
/* 零钱兑换：贪心 */
func coinChangeGreedy(coins: [Int], amt: Int) -> Int {
// 假设 coins 列表有序
var i = coins.count - 1
var count = 0
var amt = amt
// 循环进行贪心选择，直到无剩余金额
while amt > 0 {
// 找到小于且最接近剩余金额的硬币
while i > 0 && coins[i] > amt {
i -= 1
}
// 选择 coins[i]
amt -= coins[i]
count += 1
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1
}
```
=== "JS"
```javascript title="coin_change_greedy.js"
/* 零钱兑换：贪心 */
function coinChangeGreedy(coins, amt) {
// 假设 coins 数组有序
let i = coins.length - 1;
let count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt === 0 ? count : -1;
}
```
=== "TS"
```typescript title="coin_change_greedy.ts"
/* 零钱兑换：贪心 */
function coinChangeGreedy(coins: number[], amt: number): number {
// 假设 coins 数组有序
let i = coins.length - 1;
let count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt === 0 ? count : -1;
}
```
=== "Dart"
```dart title="coin_change_greedy.dart"
/* 零钱兑换：贪心 */
int coinChangeGreedy(List<int> coins, int amt) {
// 假设 coins 列表有序
int i = coins.length - 1;
int count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1;
}
```
=== "Rust"
```rust title="coin_change_greedy.rs"
/* 零钱兑换：贪心 */
fn coin_change_greedy(coins: &[i32], mut amt: i32) -> i32 {
// 假设 coins 列表有序
let mut i = coins.len() - 1;
let mut count = 0;
// 循环进行贪心选择，直到无剩余金额
while amt > 0 {
// 找到小于且最接近剩余金额的硬币
while i > 0 && coins[i] > amt {
i -= 1;
}
// 选择 coins[i]
amt -= coins[i];
count += 1;
}
// 若未找到可行方案，则返回 -1
if amt == 0 {
count
} else {
-1
}
}
```
=== "C"
```c title="coin_change_greedy.c"
/* 零钱兑换：贪心 */
int coinChangeGreedy(int *coins, int size, int amt) {
// 假设 coins 列表有序
int i = size - 1;
int count = 0;
// 循环进行贪心选择，直到无剩余金额
while (amt > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > amt) {
i--;
}
// 选择 coins[i]
amt -= coins[i];
count++;
}
// 若未找到可行方案，则返回 -1
return amt == 0 ? count : -1;
}
```
=== "Kotlin"
```kotlin title="coin_change_greedy.kt"
/* 零钱兑换：贪心 */
fun coinChangeGreedy(coins: IntArray, amt: Int): Int {
// 假设 coins 列表有序
var am = amt
var i = coins.size - 1
var count = 0
// 循环进行贪心选择，直到无剩余金额
while (am > 0) {
// 找到小于且最接近剩余金额的硬币
while (i > 0 && coins[i] > am) {
i--
}
// 选择 coins[i]
am -= coins[i]
count++
}
// 若未找到可行方案，则返回 -1
return if (am == 0) count else -1
}
```
=== "Ruby"
```ruby title="coin_change_greedy.rb"
[class]{}-[func]{coin_change_greedy}
```
=== "Zig"
```zig title="coin_change_greedy.zig"
[class]{}-[func]{coinChangeGreedy}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_greedy%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%81%87%E8%AE%BE%20coins%20%E5%88%97%E8%A1%A8%E6%9C%89%E5%BA%8F%0A%20%20%20%20i%20%3D%20len%28coins%29%20-%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%BF%9B%E8%A1%8C%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E5%88%B0%E6%97%A0%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%0A%20%20%20%20while%20amt%20%3E%200%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E5%B0%8F%E4%BA%8E%E4%B8%94%E6%9C%80%E6%8E%A5%E8%BF%91%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%E7%9A%84%E7%A1%AC%E5%B8%81%0A%20%20%20%20%20%20%20%20while%20i%20%3E%200%20and%20coins%5Bi%5D%20%3E%20amt%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20-%3D%201%0A%20%20%20%20%20%20%20%20%23%20%E9%80%89%E6%8B%A9%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20amt%20-%3D%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%23%20%E8%8B%A5%E6%9C%AA%E6%89%BE%E5%88%B0%E5%8F%AF%E8%A1%8C%E6%96%B9%E6%A1%88%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20return%20count%20if%20amt%20%3D%3D%200%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E8%83%BD%E5%A4%9F%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%205,%2010,%2020,%2050,%20100%5D%0A%20%20%20%20amt%20%3D%20186%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E6%97%A0%E6%B3%95%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%2020,%2050%5D%0A%20%20%20%20amt%20%3D%2060%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%20%20%20%20print%28f%22%E5%AE%9E%E9%99%85%E4%B8%8A%E9%9C%80%E8%A6%81%E7%9A%84%E6%9C%80%E5%B0%91%E6%95%B0%E9%87%8F%E4%B8%BA%203%20%EF%BC%8C%E5%8D%B3%2020%20%2B%2020%20%2B%2020%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_greedy%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%81%87%E8%AE%BE%20coins%20%E5%88%97%E8%A1%A8%E6%9C%89%E5%BA%8F%0A%20%20%20%20i%20%3D%20len%28coins%29%20-%201%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%BF%9B%E8%A1%8C%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E5%88%B0%E6%97%A0%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%0A%20%20%20%20while%20amt%20%3E%200%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%89%BE%E5%88%B0%E5%B0%8F%E4%BA%8E%E4%B8%94%E6%9C%80%E6%8E%A5%E8%BF%91%E5%89%A9%E4%BD%99%E9%87%91%E9%A2%9D%E7%9A%84%E7%A1%AC%E5%B8%81%0A%20%20%20%20%20%20%20%20while%20i%20%3E%200%20and%20coins%5Bi%5D%20%3E%20amt%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20-%3D%201%0A%20%20%20%20%20%20%20%20%23%20%E9%80%89%E6%8B%A9%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20amt%20-%3D%20coins%5Bi%5D%0A%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20%23%20%E8%8B%A5%E6%9C%AA%E6%89%BE%E5%88%B0%E5%8F%AF%E8%A1%8C%E6%96%B9%E6%A1%88%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20-1%0A%20%20%20%20return%20count%20if%20amt%20%3D%3D%200%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E8%83%BD%E5%A4%9F%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%205,%2010,%2020,%2050,%20100%5D%0A%20%20%20%20amt%20%3D%20186%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%EF%BC%9A%E6%97%A0%E6%B3%95%E4%BF%9D%E8%AF%81%E6%89%BE%E5%88%B0%E5%85%A8%E5%B1%80%E6%9C%80%E4%BC%98%E8%A7%A3%0A%20%20%20%20coins%20%3D%20%5B1,%2020,%2050%5D%0A%20%20%20%20amt%20%3D%2060%0A%20%20%20%20res%20%3D%20coin_change_greedy%28coins,%20amt%29%0A%20%20%20%20print%28f%22%5Cncoins%20%3D%20%7Bcoins%7D,%20amt%20%3D%20%7Bamt%7D%22%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%20%7Bamt%7D%20%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29%0A%20%20%20%20print%28f%22%E5%AE%9E%E9%99%85%E4%B8%8A%E9%9C%80%E8%A6%81%E7%9A%84%E6%9C%80%E5%B0%91%E6%95%B0%E9%87%8F%E4%B8%BA%203%20%EF%BC%8C%E5%8D%B3%2020%20%2B%2020%20%2B%2020%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
You might exclaim: So clean! The greedy algorithm solves the coin change problem in about ten lines of code.
## 15.1.1 &nbsp; Advantages and limitations of greedy algorithms
**Greedy algorithms are not only straightforward and simple to implement, but they are also usually very efficient**. In the code above, if the smallest coin denomination is $\min(coins)$, the greedy choice loops at most $amt / \min(coins)$ times, giving a time complexity of $O(amt / \min(coins))$. This is an order of magnitude smaller than the time complexity of the dynamic programming solution, which is $O(n \times amt)$.
However, **for some combinations of coin denominations, greedy algorithms cannot find the optimal solution**. The following figure provides two examples.
- **Positive example $coins = [1, 5, 10, 20, 50, 100]$**: In this coin combination, given any $amt$, the greedy algorithm can find the optimal solution.
- **Negative example $coins = [1, 20, 50]$**: Suppose $amt = 60$, the greedy algorithm can only find the combination $50 + 1 \times 10$, totaling 11 coins, but dynamic programming can find the optimal solution of $20 + 20 + 20$, needing only 3 coins.
- **Negative example $coins = [1, 49, 50]$**: Suppose $amt = 98$, the greedy algorithm can only find the combination $50 + 1 \times 48$, totaling 49 coins, but dynamic programming can find the optimal solution of $49 + 49$, needing only 2 coins.
![Examples where greedy algorithms do not find the optimal solution](greedy_algorithm.assets/coin_change_greedy_vs_dp.png){ class="animation-figure" }
<p align="center"> Figure 15-2 &nbsp; Examples where greedy algorithms do not find the optimal solution </p>
This means that for the coin change problem, greedy algorithms cannot guarantee finding the globally optimal solution, and they might find a very poor solution. They are better suited for dynamic programming.
Generally, the suitability of greedy algorithms falls into two categories.
1. **Guaranteed to find the optimal solution**: In these cases, greedy algorithms are often the best choice, as they tend to be more efficient than backtracking or dynamic programming.
2. **Can find a near-optimal solution**: Greedy algorithms are also applicable here. For many complex problems, finding the global optimal solution is very challenging, and being able to find a high-efficiency suboptimal solution is also very commendable.
## 15.1.2 &nbsp; Characteristics of greedy algorithms
So, what kind of problems are suitable for solving with greedy algorithms? Or rather, under what conditions can greedy algorithms guarantee to find the optimal solution?
Compared to dynamic programming, greedy algorithms have stricter usage conditions, focusing mainly on two properties of the problem.
- **Greedy choice property**: Only when the locally optimal choice can always lead to a globally optimal solution can greedy algorithms guarantee to obtain the optimal solution.
- **Optimal substructure**: The optimal solution to the original problem contains the optimal solutions to its subproblems.
Optimal substructure has already been introduced in the "Dynamic Programming" chapter, so it is not discussed further here. It's important to note that some problems do not have an obvious optimal substructure, but can still be solved using greedy algorithms.
We mainly explore the method for determining the greedy choice property. Although its description seems simple, **in practice, proving the greedy choice property for many problems is not easy**.
For example, in the coin change problem, although we can easily cite counterexamples to disprove the greedy choice property, proving it is much more challenging. If asked, **what conditions must a coin combination meet to be solvable using a greedy algorithm**? We often have to rely on intuition or examples to provide an ambiguous answer, as it is difficult to provide a rigorous mathematical proof.
!!! quote
A paper presents an algorithm with a time complexity of $O(n^3)$ for determining whether a coin combination can use a greedy algorithm to find the optimal solution for any amount.
Pearson, D. A polynomial-time algorithm for the change-making problem[J]. Operations Research Letters, 2005, 33(3): 231-234.
## 15.1.3 &nbsp; Steps for solving problems with greedy algorithms
The problem-solving process for greedy problems can generally be divided into the following three steps.
1. **Problem analysis**: Sort out and understand the characteristics of the problem, including state definition, optimization objectives, and constraints, etc. This step is also involved in backtracking and dynamic programming.
2. **Determine the greedy strategy**: Determine how to make a greedy choice at each step. This strategy can reduce the scale of the problem at each step and eventually solve the entire problem.
3. **Proof of correctness**: It is usually necessary to prove that the problem has both a greedy choice property and optimal substructure. This step may require mathematical proofs, such as induction or reductio ad absurdum.
Determining the greedy strategy is the core step in solving the problem, but it may not be easy to implement, mainly for the following reasons.
- **Greedy strategies vary greatly between different problems**. For many problems, the greedy strategy is fairly straightforward, and we can come up with it through some general thinking and attempts. However, for some complex problems, the greedy strategy may be very elusive, which is a real test of individual problem-solving experience and algorithmic capability.
- **Some greedy strategies are quite misleading**. When we confidently design a greedy strategy, write the code, and submit it for testing, it is quite possible that some test cases will not pass. This is because the designed greedy strategy is only "partially correct," as described above with the coin change example.
To ensure accuracy, we should provide rigorous mathematical proofs for the greedy strategy, **usually involving reductio ad absurdum or mathematical induction**.
However, proving correctness may not be an easy task. If we are at a loss, we usually choose to debug the code based on test cases, modifying and verifying the greedy strategy step by step.
## 15.1.4 &nbsp; Typical problems solved by greedy algorithms
Greedy algorithms are often applied to optimization problems that satisfy the properties of greedy choice and optimal substructure. Below are some typical greedy algorithm problems.
- **Coin change problem**: In some coin combinations, the greedy algorithm always provides the optimal solution.
- **Interval scheduling problem**: Suppose you have several tasks, each of which takes place over a period of time. Your goal is to complete as many tasks as possible. If you always choose the task that ends the earliest, then the greedy algorithm can achieve the optimal solution.
- **Fractional knapsack problem**: Given a set of items and a carrying capacity, your goal is to select a set of items such that the total weight does not exceed the carrying capacity and the total value is maximized. If you always choose the item with the highest value-to-weight ratio (value / weight), the greedy algorithm can achieve the optimal solution in some cases.
- **Stock trading problem**: Given a set of historical stock prices, you can make multiple trades, but you cannot buy again until after you have sold if you already own stocks. The goal is to achieve the maximum profit.
- **Huffman coding**: Huffman coding is a greedy algorithm used for lossless data compression. By constructing a Huffman tree, it always merges the two nodes with the lowest frequency, resulting in a Huffman tree with the minimum weighted path length (coding length).
- **Dijkstra's algorithm**: It is a greedy algorithm for solving the shortest path problem from a given source vertex to all other vertices.
en/docs/chapter_greedy/index.md
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---
comments: true
icon: material/head-heart-outline
---
# Chapter 15. &nbsp; Greedy
![Greedy](../assets/covers/chapter_greedy.jpg){ class="cover-image" }
!!! abstract
Sunflowers turn towards the sun, always seeking the greatest possible growth for themselves.
Greedy strategy guides to the best answer step by step through rounds of simple choices.
## Chapter contents
- [15.1 &nbsp; Greedy algorithms](greedy_algorithm.md)
- [15.2 &nbsp; Fractional knapsack problem](fractional_knapsack_problem.md)
- [15.3 &nbsp; Maximum capacity problem](max_capacity_problem.md)
- [15.4 &nbsp; Maximum product cutting problem](max_product_cutting_problem.md)
- [15.5 &nbsp; Summary](summary.md)
en/docs/chapter_greedy/max_capacity_problem.md
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---
comments: true
---
# 15.3 &nbsp; Maximum capacity problem
!!! question
Input an array $ht$, where each element represents the height of a vertical partition. Any two partitions in the array, along with the space between them, can form a container.
The capacity of the container is the product of the height and the width (area), where the height is determined by the shorter partition, and the width is the difference in array indices between the two partitions.
Please select two partitions in the array that maximize the container's capacity and return this maximum capacity. An example is shown in the following figure.
![Example data for the maximum capacity problem](max_capacity_problem.assets/max_capacity_example.png){ class="animation-figure" }
<p align="center"> Figure 15-7 &nbsp; Example data for the maximum capacity problem </p>
The container is formed by any two partitions, **therefore the state of this problem is represented by the indices of the two partitions, denoted as $[i, j]$**.
According to the problem statement, the capacity equals the product of height and width, where the height is determined by the shorter partition, and the width is the difference in array indices between the two partitions. The formula for capacity $cap[i, j]$ is:
$$
cap[i, j] = \min(ht[i], ht[j]) \times (j - i)
$$
Assuming the length of the array is $n$, the number of combinations of two partitions (total number of states) is $C_n^2 = \frac{n(n - 1)}{2}$. The most straightforward approach is to **enumerate all possible states**, resulting in a time complexity of $O(n^2)$.
### 1. &nbsp; Determination of a greedy strategy
There is a more efficient solution to this problem. As shown in the following figure, we select a state $[i, j]$ where the indices $i < j$ and the height $ht[i] < ht[j]$, meaning $i$ is the shorter partition, and $j$ is the taller one.
![Initial state](max_capacity_problem.assets/max_capacity_initial_state.png){ class="animation-figure" }
<p align="center"> Figure 15-8 &nbsp; Initial state </p>
As shown in the following figure, **if we move the taller partition $j$ closer to the shorter partition $i$, the capacity will definitely decrease**.
This is because when moving the taller partition $j$, the width $j-i$ definitely decreases; and since the height is determined by the shorter partition, the height can only remain the same (if $i$ remains the shorter partition) or decrease (if the moved $j$ becomes the shorter partition).
![State after moving the taller partition inward](max_capacity_problem.assets/max_capacity_moving_long_board.png){ class="animation-figure" }
<p align="center"> Figure 15-9 &nbsp; State after moving the taller partition inward </p>
Conversely, **we can only possibly increase the capacity by moving the shorter partition $i$ inward**. Although the width will definitely decrease, **the height may increase** (if the moved shorter partition $i$ becomes taller). For example, in the Figure 15-10 , the area increases after moving the shorter partition.
![State after moving the shorter partition inward](max_capacity_problem.assets/max_capacity_moving_short_board.png){ class="animation-figure" }
<p align="center"> Figure 15-10 &nbsp; State after moving the shorter partition inward </p>
This leads us to the greedy strategy for this problem: initialize two pointers at the ends of the container, and in each round, move the pointer corresponding to the shorter partition inward until the two pointers meet.
The following figures illustrate the execution of the greedy strategy.
1. Initially, the pointers $i$ and $j$ are positioned at the ends of the array.
2. Calculate the current state's capacity $cap[i, j]$ and update the maximum capacity.
3. Compare the heights of partitions $i$ and $j$, and move the shorter partition inward by one step.
4. Repeat steps `2.` and `3.` until $i$ and $j$ meet.
=== "<1>"
![The greedy process for maximum capacity problem](max_capacity_problem.assets/max_capacity_greedy_step1.png){ class="animation-figure" }
=== "<2>"
![max_capacity_greedy_step2](max_capacity_problem.assets/max_capacity_greedy_step2.png){ class="animation-figure" }
=== "<3>"
![max_capacity_greedy_step3](max_capacity_problem.assets/max_capacity_greedy_step3.png){ class="animation-figure" }
=== "<4>"
![max_capacity_greedy_step4](max_capacity_problem.assets/max_capacity_greedy_step4.png){ class="animation-figure" }
=== "<5>"
![max_capacity_greedy_step5](max_capacity_problem.assets/max_capacity_greedy_step5.png){ class="animation-figure" }
=== "<6>"
![max_capacity_greedy_step6](max_capacity_problem.assets/max_capacity_greedy_step6.png){ class="animation-figure" }
=== "<7>"
![max_capacity_greedy_step7](max_capacity_problem.assets/max_capacity_greedy_step7.png){ class="animation-figure" }
=== "<8>"
![max_capacity_greedy_step8](max_capacity_problem.assets/max_capacity_greedy_step8.png){ class="animation-figure" }
=== "<9>"
![max_capacity_greedy_step9](max_capacity_problem.assets/max_capacity_greedy_step9.png){ class="animation-figure" }
<p align="center"> Figure 15-11 &nbsp; The greedy process for maximum capacity problem </p>
### 2. &nbsp; Implementation
The code loops at most $n$ times, **thus the time complexity is $O(n)$**.
The variables $i$, $j$, and $res$ use a constant amount of extra space, **thus the space complexity is $O(1)$**.
=== "Python"
```python title="max_capacity.py"
def max_capacity(ht: list[int]) -> int:
"""最大容量：贪心"""
# 初始化 i, j，使其分列数组两端
i, j = 0, len(ht) - 1
# 初始最大容量为 0
res = 0
# 循环贪心选择，直至两板相遇
while i < j:
# 更新最大容量
cap = min(ht[i], ht[j]) * (j - i)
res = max(res, cap)
# 向内移动短板
if ht[i] < ht[j]:
i += 1
else:
j -= 1
return res
```
=== "C++"
```cpp title="max_capacity.cpp"
/* 最大容量：贪心 */
int maxCapacity(vector<int> &ht) {
// 初始化 i, j，使其分列数组两端
int i = 0, j = ht.size() - 1;
// 初始最大容量为 0
int res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
int cap = min(ht[i], ht[j]) * (j - i);
res = max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "Java"
```java title="max_capacity.java"
/* 最大容量：贪心 */
int maxCapacity(int[] ht) {
// 初始化 i, j，使其分列数组两端
int i = 0, j = ht.length - 1;
// 初始最大容量为 0
int res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
int cap = Math.min(ht[i], ht[j]) * (j - i);
res = Math.max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "C#"
```csharp title="max_capacity.cs"
/* 最大容量：贪心 */
int MaxCapacity(int[] ht) {
// 初始化 i, j，使其分列数组两端
int i = 0, j = ht.Length - 1;
// 初始最大容量为 0
int res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
int cap = Math.Min(ht[i], ht[j]) * (j - i);
res = Math.Max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "Go"
```go title="max_capacity.go"
/* 最大容量：贪心 */
func maxCapacity(ht []int) int {
// 初始化 i, j，使其分列数组两端
i, j := 0, len(ht)-1
// 初始最大容量为 0
res := 0
// 循环贪心选择，直至两板相遇
for i < j {
// 更新最大容量
capacity := int(math.Min(float64(ht[i]), float64(ht[j]))) * (j - i)
res = int(math.Max(float64(res), float64(capacity)))
// 向内移动短板
if ht[i] < ht[j] {
i++
} else {
j--
}
}
return res
}
```
=== "Swift"
```swift title="max_capacity.swift"
/* 最大容量：贪心 */
func maxCapacity(ht: [Int]) -> Int {
// 初始化 i, j，使其分列数组两端
var i = ht.startIndex, j = ht.endIndex - 1
// 初始最大容量为 0
var res = 0
// 循环贪心选择，直至两板相遇
while i < j {
// 更新最大容量
let cap = min(ht[i], ht[j]) * (j - i)
res = max(res, cap)
// 向内移动短板
if ht[i] < ht[j] {
i += 1
} else {
j -= 1
}
}
return res
}
```
=== "JS"
```javascript title="max_capacity.js"
/* 最大容量：贪心 */
function maxCapacity(ht) {
// 初始化 i, j，使其分列数组两端
let i = 0,
j = ht.length - 1;
// 初始最大容量为 0
let res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
const cap = Math.min(ht[i], ht[j]) * (j - i);
res = Math.max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i += 1;
} else {
j -= 1;
}
}
return res;
}
```
=== "TS"
```typescript title="max_capacity.ts"
/* 最大容量：贪心 */
function maxCapacity(ht: number[]): number {
// 初始化 i, j，使其分列数组两端
let i = 0,
j = ht.length - 1;
// 初始最大容量为 0
let res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
const cap: number = Math.min(ht[i], ht[j]) * (j - i);
res = Math.max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i += 1;
} else {
j -= 1;
}
}
return res;
}
```
=== "Dart"
```dart title="max_capacity.dart"
/* 最大容量：贪心 */
int maxCapacity(List<int> ht) {
// 初始化 i, j，使其分列数组两端
int i = 0, j = ht.length - 1;
// 初始最大容量为 0
int res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
int cap = min(ht[i], ht[j]) * (j - i);
res = max(res, cap);
// 向内移动短板
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "Rust"
```rust title="max_capacity.rs"
/* 最大容量：贪心 */
fn max_capacity(ht: &[i32]) -> i32 {
// 初始化 i, j，使其分列数组两端
let mut i = 0;
let mut j = ht.len() - 1;
// 初始最大容量为 0
let mut res = 0;
// 循环贪心选择，直至两板相遇
while i < j {
// 更新最大容量
let cap = std::cmp::min(ht[i], ht[j]) * (j - i) as i32;
res = std::cmp::max(res, cap);
// 向内移动短板
if ht[i] < ht[j] {
i += 1;
} else {
j -= 1;
}
}
res
}
```
=== "C"
```c title="max_capacity.c"
/* 最大容量：贪心 */
int maxCapacity(int ht[], int htLength) {
// 初始化 i, j，使其分列数组两端
int i = 0;
int j = htLength - 1;
// 初始最大容量为 0
int res = 0;
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
int capacity = myMin(ht[i], ht[j]) * (j - i);
res = myMax(res, capacity);
// 向内移动短板
if (ht[i] < ht[j]) {
i++;
} else {
j--;
}
}
return res;
}
```
=== "Kotlin"
```kotlin title="max_capacity.kt"
/* 最大容量：贪心 */
fun maxCapacity(ht: IntArray): Int {
// 初始化 i, j，使其分列数组两端
var i = 0
var j = ht.size - 1
// 初始最大容量为 0
var res = 0
// 循环贪心选择，直至两板相遇
while (i < j) {
// 更新最大容量
val cap = min(ht[i], ht[j]) * (j - i)
res = max(res, cap)
// 向内移动短板
if (ht[i] < ht[j]) {
i++
} else {
j--
}
}
return res
}
```
=== "Ruby"
```ruby title="max_capacity.rb"
[class]{}-[func]{max_capacity}
```
=== "Zig"
```zig title="max_capacity.zig"
[class]{}-[func]{maxCapacity}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20max_capacity%28ht%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20i,%20j%EF%BC%8C%E4%BD%BF%E5%85%B6%E5%88%86%E5%88%97%E6%95%B0%E7%BB%84%E4%B8%A4%E7%AB%AF%0A%20%20%20%20i,%20j%20%3D%200,%20len%28ht%29%20-%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%200%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E8%87%B3%E4%B8%A4%E6%9D%BF%E7%9B%B8%E9%81%87%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%0A%20%20%20%20%20%20%20%20cap%20%3D%20min%28ht%5Bi%5D,%20ht%5Bj%5D%29%20*%20%28j%20-%20i%29%0A%20%20%20%20%20%20%20%20res%20%3D%20max%28res,%20cap%29%0A%20%20%20%20%20%20%20%20%23%20%E5%90%91%E5%86%85%E7%A7%BB%E5%8A%A8%E7%9F%AD%E6%9D%BF%0A%20%20%20%20%20%20%20%20if%20ht%5Bi%5D%20%3C%20ht%5Bj%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20ht%20%3D%20%5B3,%208,%205,%202,%207,%207,%203,%204%5D%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_capacity%28ht%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20max_capacity%28ht%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20i,%20j%EF%BC%8C%E4%BD%BF%E5%85%B6%E5%88%86%E5%88%97%E6%95%B0%E7%BB%84%E4%B8%A4%E7%AB%AF%0A%20%20%20%20i,%20j%20%3D%200,%20len%28ht%29%20-%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%200%0A%20%20%20%20res%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E8%B4%AA%E5%BF%83%E9%80%89%E6%8B%A9%EF%BC%8C%E7%9B%B4%E8%87%B3%E4%B8%A4%E6%9D%BF%E7%9B%B8%E9%81%87%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20%23%20%E6%9B%B4%E6%96%B0%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%0A%20%20%20%20%20%20%20%20cap%20%3D%20min%28ht%5Bi%5D,%20ht%5Bj%5D%29%20*%20%28j%20-%20i%29%0A%20%20%20%20%20%20%20%20res%20%3D%20max%28res,%20cap%29%0A%20%20%20%20%20%20%20%20%23%20%E5%90%91%E5%86%85%E7%A7%BB%E5%8A%A8%E7%9F%AD%E6%9D%BF%0A%20%20%20%20%20%20%20%20if%20ht%5Bi%5D%20%3C%20ht%5Bj%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%2B%3D%201%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20-%3D%201%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20ht%20%3D%20%5B3,%208,%205,%202,%207,%207,%203,%204%5D%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_capacity%28ht%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%AE%B9%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
### 3. &nbsp; Proof of correctness
The reason why the greedy method is faster than enumeration is that each round of greedy selection "skips" some states.
For example, under the state $cap[i, j]$ where $i$ is the shorter partition and $j$ is the taller partition, greedily moving the shorter partition $i$ inward by one step leads to the "skipped" states shown below. **This means that these states' capacities cannot be verified later**.
$$
cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]
$$
![States skipped by moving the shorter partition](max_capacity_problem.assets/max_capacity_skipped_states.png){ class="animation-figure" }
<p align="center"> Figure 15-12 &nbsp; States skipped by moving the shorter partition </p>
It is observed that **these skipped states are actually all states where the taller partition $j$ is moved inward**. We have already proven that moving the taller partition inward will definitely decrease the capacity. Therefore, the skipped states cannot possibly be the optimal solution, **and skipping them does not lead to missing the optimal solution**.
The analysis shows that the operation of moving the shorter partition is "safe", and the greedy strategy is effective.
en/docs/chapter_greedy/max_product_cutting_problem.md
+405
View File
@@ -0,0 +1,405 @@
---
comments: true
---
# 15.4 &nbsp; Maximum product cutting problem
!!! question
Given a positive integer $n$, split it into at least two positive integers that sum up to $n$, and find the maximum product of these integers, as illustrated below.
![Definition of the maximum product cutting problem](max_product_cutting_problem.assets/max_product_cutting_definition.png){ class="animation-figure" }
<p align="center"> Figure 15-13 &nbsp; Definition of the maximum product cutting problem </p>
Assume we split $n$ into $m$ integer factors, where the $i$-th factor is denoted as $n_i$, that is,
$$
n = \sum_{i=1}^{m}n_i
$$
The goal of this problem is to find the maximum product of all integer factors, namely,
$$
\max(\prod_{i=1}^{m}n_i)
$$
We need to consider: How large should the number of splits $m$ be, and what should each $n_i$ be?
### 1. &nbsp; Greedy strategy determination
Experience suggests that the product of two integers is often greater than their sum. Suppose we split a factor of $2$ from $n$, then their product is $2(n-2)$. Compare this product with $n$:
$$
\begin{aligned}
2(n-2) & \geq n \newline
2n - n - 4 & \geq 0 \newline
n & \geq 4
\end{aligned}
$$
As shown below, when $n \geq 4$, splitting out a $2$ increases the product, **which indicates that integers greater than or equal to $4$ should be split**.
**Greedy strategy one**: If the splitting scheme includes factors $\geq 4$, they should be further split. The final split should only include factors $1$, $2$, and $3$.
![Product increase due to splitting](max_product_cutting_problem.assets/max_product_cutting_greedy_infer1.png){ class="animation-figure" }
<p align="center"> Figure 15-14 &nbsp; Product increase due to splitting </p>
Next, consider which factor is optimal. Among the factors $1$, $2$, and $3$, clearly $1$ is the worst, as $1 \times (n-1) < n$ always holds, meaning splitting out $1$ actually decreases the product.
As shown below, when $n = 6$, $3 \times 3 > 2 \times 2 \times 2$. **This means splitting out $3$ is better than splitting out $2$**.
**Greedy strategy two**: In the splitting scheme, there should be at most two $2$s. Because three $2$s can always be replaced by two $3$s to obtain a higher product.
![Optimal splitting factors](max_product_cutting_problem.assets/max_product_cutting_greedy_infer2.png){ class="animation-figure" }
<p align="center"> Figure 15-15 &nbsp; Optimal splitting factors </p>
From the above, the following greedy strategies can be derived.
1. Input integer $n$, continually split out factor $3$ until the remainder is $0$, $1$, or $2$.
2. When the remainder is $0$, it means $n$ is a multiple of $3$, so no further action is taken.
3. When the remainder is $2$, do not continue to split, keep it.
4. When the remainder is $1$, since $2 \times 2 > 1 \times 3$, the last $3$ should be replaced with $2$.
### 2. &nbsp; Code implementation
As shown below, we do not need to use loops to split the integer but can use the floor division operation to get the number of $3$s, $a$, and the modulo operation to get the remainder, $b$, thus:
$$
n = 3a + b
$$
Please note, for the boundary case where $n \leq 3$, a $1$ must be split out, with a product of $1 \times (n - 1)$.
=== "Python"
```python title="max_product_cutting.py"
def max_product_cutting(n: int) -> int:
"""最大切分乘积：贪心"""
# 当 n <= 3 时，必须切分出一个 1
if n <= 3:
return 1 * (n - 1)
# 贪心地切分出 3 ，a 为 3 的个数，b 为余数
a, b = n // 3, n % 3
if b == 1:
# 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return int(math.pow(3, a - 1)) * 2 * 2
if b == 2:
# 当余数为 2 时，不做处理
return int(math.pow(3, a)) * 2
# 当余数为 0 时，不做处理
return int(math.pow(3, a))
```
=== "C++"
```cpp title="max_product_cutting.cpp"
/* 最大切分乘积：贪心 */
int maxProductCutting(int n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
int a = n / 3;
int b = n % 3;
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return (int)pow(3, a - 1) * 2 * 2;
}
if (b == 2) {
// 当余数为 2 时，不做处理
return (int)pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return (int)pow(3, a);
}
```
=== "Java"
```java title="max_product_cutting.java"
/* 最大切分乘积：贪心 */
int maxProductCutting(int n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
int a = n / 3;
int b = n % 3;
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return (int) Math.pow(3, a - 1) * 2 * 2;
}
if (b == 2) {
// 当余数为 2 时，不做处理
return (int) Math.pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return (int) Math.pow(3, a);
}
```
=== "C#"
```csharp title="max_product_cutting.cs"
/* 最大切分乘积：贪心 */
int MaxProductCutting(int n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
int a = n / 3;
int b = n % 3;
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return (int)Math.Pow(3, a - 1) * 2 * 2;
}
if (b == 2) {
// 当余数为 2 时，不做处理
return (int)Math.Pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return (int)Math.Pow(3, a);
}
```
=== "Go"
```go title="max_product_cutting.go"
/* 最大切分乘积：贪心 */
func maxProductCutting(n int) int {
// 当 n <= 3 时，必须切分出一个 1
if n <= 3 {
return 1 * (n - 1)
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
a := n / 3
b := n % 3
if b == 1 {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return int(math.Pow(3, float64(a-1))) * 2 * 2
}
if b == 2 {
// 当余数为 2 时，不做处理
return int(math.Pow(3, float64(a))) * 2
}
// 当余数为 0 时，不做处理
return int(math.Pow(3, float64(a)))
}
```
=== "Swift"
```swift title="max_product_cutting.swift"
/* 最大切分乘积：贪心 */
func maxProductCutting(n: Int) -> Int {
// 当 n <= 3 时，必须切分出一个 1
if n <= 3 {
return 1 * (n - 1)
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
let a = n / 3
let b = n % 3
if b == 1 {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return pow(3, a - 1) * 2 * 2
}
if b == 2 {
// 当余数为 2 时，不做处理
return pow(3, a) * 2
}
// 当余数为 0 时，不做处理
return pow(3, a)
}
```
=== "JS"
```javascript title="max_product_cutting.js"
/* 最大切分乘积：贪心 */
function maxProductCutting(n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
let a = Math.floor(n / 3);
let b = n % 3;
if (b === 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return Math.pow(3, a - 1) * 2 * 2;
}
if (b === 2) {
// 当余数为 2 时，不做处理
return Math.pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return Math.pow(3, a);
}
```
=== "TS"
```typescript title="max_product_cutting.ts"
/* 最大切分乘积：贪心 */
function maxProductCutting(n: number): number {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
let a: number = Math.floor(n / 3);
let b: number = n % 3;
if (b === 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return Math.pow(3, a - 1) * 2 * 2;
}
if (b === 2) {
// 当余数为 2 时，不做处理
return Math.pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return Math.pow(3, a);
}
```
=== "Dart"
```dart title="max_product_cutting.dart"
/* 最大切分乘积：贪心 */
int maxProductCutting(int n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
int a = n ~/ 3;
int b = n % 3;
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return (pow(3, a - 1) * 2 * 2).toInt();
}
if (b == 2) {
// 当余数为 2 时，不做处理
return (pow(3, a) * 2).toInt();
}
// 当余数为 0 时，不做处理
return pow(3, a).toInt();
}
```
=== "Rust"
```rust title="max_product_cutting.rs"
/* 最大切分乘积：贪心 */
fn max_product_cutting(n: i32) -> i32 {
// 当 n <= 3 时，必须切分出一个 1
if n <= 3 {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
let a = n / 3;
let b = n % 3;
if b == 1 {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
3_i32.pow(a as u32 - 1) * 2 * 2
} else if b == 2 {
// 当余数为 2 时，不做处理
3_i32.pow(a as u32) * 2
} else {
// 当余数为 0 时，不做处理
3_i32.pow(a as u32)
}
}
```
=== "C"
```c title="max_product_cutting.c"
/* 最大切分乘积：贪心 */
int maxProductCutting(int n) {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1);
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
int a = n / 3;
int b = n % 3;
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return pow(3, a - 1) * 2 * 2;
}
if (b == 2) {
// 当余数为 2 时，不做处理
return pow(3, a) * 2;
}
// 当余数为 0 时，不做处理
return pow(3, a);
}
```
=== "Kotlin"
```kotlin title="max_product_cutting.kt"
/* 最大切分乘积：贪心 */
fun maxProductCutting(n: Int): Int {
// 当 n <= 3 时，必须切分出一个 1
if (n <= 3) {
return 1 * (n - 1)
}
// 贪心地切分出 3 ，a 为 3 的个数，b 为余数
val a = n / 3
val b = n % 3
if (b == 1) {
// 当余数为 1 时，将一对 1 * 3 转化为 2 * 2
return 3.0.pow((a - 1)).toInt() * 2 * 2
}
if (b == 2) {
// 当余数为 2 时，不做处理
return 3.0.pow(a).toInt() * 2 * 2
}
// 当余数为 0 时，不做处理
return 3.0.pow(a).toInt()
}
```
=== "Ruby"
```ruby title="max_product_cutting.rb"
[class]{}-[func]{max_product_cutting}
```
=== "Zig"
```zig title="max_product_cutting.zig"
[class]{}-[func]{maxProductCutting}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20math%0A%0Adef%20max_product_cutting%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%20n%20%3C%3D%203%20%E6%97%B6%EF%BC%8C%E5%BF%85%E9%A1%BB%E5%88%87%E5%88%86%E5%87%BA%E4%B8%80%E4%B8%AA%201%0A%20%20%20%20if%20n%20%3C%3D%203%3A%0A%20%20%20%20%20%20%20%20return%201%20*%20%28n%20-%201%29%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E5%9C%B0%E5%88%87%E5%88%86%E5%87%BA%203%20%EF%BC%8Ca%20%E4%B8%BA%203%20%E7%9A%84%E4%B8%AA%E6%95%B0%EF%BC%8Cb%20%E4%B8%BA%E4%BD%99%E6%95%B0%0A%20%20%20%20a,%20b%20%3D%20n%20//%203,%20n%20%25%203%0A%20%20%20%20if%20b%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%201%20%E6%97%B6%EF%BC%8C%E5%B0%86%E4%B8%80%E5%AF%B9%201%20*%203%20%E8%BD%AC%E5%8C%96%E4%B8%BA%202%20*%202%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%20-%201%29%29%20*%202%20*%202%0A%20%20%20%20if%20b%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%202%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%29%29%20*%202%0A%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%200%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20return%20int%28math.pow%283,%20a%29%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2058%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_product_cutting%28n%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=import%20math%0A%0Adef%20max_product_cutting%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%EF%BC%9A%E8%B4%AA%E5%BF%83%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%20n%20%3C%3D%203%20%E6%97%B6%EF%BC%8C%E5%BF%85%E9%A1%BB%E5%88%87%E5%88%86%E5%87%BA%E4%B8%80%E4%B8%AA%201%0A%20%20%20%20if%20n%20%3C%3D%203%3A%0A%20%20%20%20%20%20%20%20return%201%20*%20%28n%20-%201%29%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E5%9C%B0%E5%88%87%E5%88%86%E5%87%BA%203%20%EF%BC%8Ca%20%E4%B8%BA%203%20%E7%9A%84%E4%B8%AA%E6%95%B0%EF%BC%8Cb%20%E4%B8%BA%E4%BD%99%E6%95%B0%0A%20%20%20%20a,%20b%20%3D%20n%20//%203,%20n%20%25%203%0A%20%20%20%20if%20b%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%201%20%E6%97%B6%EF%BC%8C%E5%B0%86%E4%B8%80%E5%AF%B9%201%20*%203%20%E8%BD%AC%E5%8C%96%E4%B8%BA%202%20*%202%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%20-%201%29%29%20*%202%20*%202%0A%20%20%20%20if%20b%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%202%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20%20%20%20%20return%20int%28math.pow%283,%20a%29%29%20*%202%0A%20%20%20%20%23%20%E5%BD%93%E4%BD%99%E6%95%B0%E4%B8%BA%200%20%E6%97%B6%EF%BC%8C%E4%B8%8D%E5%81%9A%E5%A4%84%E7%90%86%0A%20%20%20%20return%20int%28math.pow%283,%20a%29%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%2058%0A%0A%20%20%20%20%23%20%E8%B4%AA%E5%BF%83%E7%AE%97%E6%B3%95%0A%20%20%20%20res%20%3D%20max_product_cutting%28n%29%0A%20%20%20%20print%28f%22%E6%9C%80%E5%A4%A7%E5%88%87%E5%88%86%E4%B9%98%E7%A7%AF%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
![Calculation method of the maximum product after cutting](max_product_cutting_problem.assets/max_product_cutting_greedy_calculation.png){ class="animation-figure" }
<p align="center"> Figure 15-16 &nbsp; Calculation method of the maximum product after cutting </p>
**Time complexity depends on the implementation of the power operation in the programming language**. For Python, the commonly used power calculation functions are three types:
- Both the operator `**` and the function `pow()` have a time complexity of $O(\log⁡ a)$.
- The `math.pow()` function internally calls the C language library's `pow()` function, performing floating-point exponentiation, with a time complexity of $O(1)$.
Variables $a$ and $b$ use constant size of extra space, **hence the space complexity is $O(1)$**.
### 3. &nbsp; Correctness proof
Using the proof by contradiction, only analyze cases where $n \geq 3$.
1. **All factors $\leq 3$**: Assume the optimal splitting scheme includes a factor $x \geq 4$, then it can definitely be further split into $2(x-2)$, obtaining a larger product. This contradicts the assumption.
2. **The splitting scheme does not contain $1$**: Assume the optimal splitting scheme includes a factor of $1$, then it can definitely be merged into another factor to obtain a larger product. This contradicts the assumption.
3. **The splitting scheme contains at most two $2$s**: Assume the optimal splitting scheme includes three $2$s, then they can definitely be replaced by two $3$s, achieving a higher product. This contradicts the assumption.
en/docs/chapter_greedy/summary.md
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# 15.5 &nbsp; Summary
- Greedy algorithms are often used to solve optimization problems, where the principle is to make locally optimal decisions at each decision stage in order to achieve a globally optimal solution.
- Greedy algorithms iteratively make one greedy choice after another, transforming the problem into a smaller sub-problem with each round, until the problem is resolved.
- Greedy algorithms are not only simple to implement but also have high problem-solving efficiency. Compared to dynamic programming, greedy algorithms generally have a lower time complexity.
- In the problem of coin change, greedy algorithms can guarantee the optimal solution for certain combinations of coins; for others, however, the greedy algorithm might find a very poor solution.
- Problems suitable for greedy algorithm solutions possess two main properties: greedy-choice property and optimal substructure. The greedy-choice property represents the effectiveness of the greedy strategy.
- For some complex problems, proving the greedy-choice property is not straightforward. Contrarily, proving the invalidity is often easier, such as with the coin change problem.
- Solving greedy problems mainly consists of three steps: problem analysis, determining the greedy strategy, and proving correctness. Among these, determining the greedy strategy is the key step, while proving correctness often poses the challenge.
- The fractional knapsack problem builds on the 0-1 knapsack problem by allowing the selection of a part of the items, hence it can be solved using a greedy algorithm. The correctness of the greedy strategy can be proved by contradiction.
- The maximum capacity problem can be solved using the exhaustive method, with a time complexity of $O(n^2)$. By designing a greedy strategy, each round moves inwardly shortening the board, optimizing the time complexity to $O(n)$.
- In the problem of maximum product after cutting, we deduce two greedy strategies: integers $\geq 4$ should continue to be cut, with the optimal cutting factor being $3$. The code includes power operations, and the time complexity depends on the method of implementing power operations, generally being $O(1)$ or $O(\log n)$.