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Revisit the English version (#1835)

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en/docs/chapter_dynamic_programming/edit_distance_problem.md
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# Edit distance problem
Edit distance, also known as Levenshtein distance, refers to the minimum number of modifications required to transform one string into another, commonly used in information retrieval and natural language processing to measure the similarity between two sequences.
Edit distance, also known as Levenshtein distance, refers to the minimum number of edits required to transform one string into another, commonly used in information retrieval and natural language processing to measure the similarity between two sequences.
!!! question
Given two strings $s$ and $t$, return the minimum number of edits required to transform $s$ into $t$.
You can perform three types of edits on a string: insert a character, delete a character, or replace a character with any other character.
You can perform three types of edit operations on a string: insert a character, delete a character, or replace a character with any other character.
As shown in the figure below, transforming `kitten` into `sitting` requires 3 edits, including 2 replacements and 1 insertion; transforming `hello` into `algo` requires 3 steps, including 2 replacements and 1 deletion.
![Example data of edit distance](edit_distance_problem.assets/edit_distance_example.png)
![Example data for edit distance](edit_distance_problem.assets/edit_distance_example.png)
**The edit distance problem can naturally be explained with a decision tree model**. Strings correspond to tree nodes, and a round of decision (an edit operation) corresponds to an edge of the tree.
**The edit distance problem can be naturally explained using the decision tree model**. Strings correspond to tree nodes, and a round of decision (one edit operation) corresponds to an edge of the tree.
As shown in the figure below, with unrestricted operations, each node can derive many edges, each corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
As shown in the figure below, without restricting operations, each node can branch into many edges, with each edge corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
From the perspective of the decision tree, the goal of this problem is to find the shortest path between the node `hello` and the node `algo`.
From the perspective of the decision tree, the goal of this problem is to find the shortest path between node `hello` and node `algo`.
![Edit distance problem represented based on decision tree model](edit_distance_problem.assets/edit_distance_decision_tree.png)
![Representing edit distance problem based on decision tree model](edit_distance_problem.assets/edit_distance_decision_tree.png)
### Dynamic programming approach
**Step one: Think about each round of decision, define the state, thus obtaining the $dp$ table**
**Step 1: Think about the decisions in each round, define the state, and thus obtain the $dp$ table**
Each round of decision involves performing one edit operation on string $s$.
We aim to gradually reduce the problem size during the edit process, which enables us to construct subproblems. Let the lengths of strings $s$ and $t$ be $n$ and $m$, respectively. We first consider the tail characters of both strings $s[n-1]$ and $t[m-1]$.
We want the problem scale to gradually decrease during the editing process, which allows us to construct subproblems. Let the lengths of strings $s$ and $t$ be $n$ and $m$ respectively. We first consider the tail characters of the two strings, $s[n-1]$ and $t[m-1]$.
- If $s[n-1]$ and $t[m-1]$ are the same, we can skip them and directly consider $s[n-2]$ and $t[m-2]$.
- If $s[n-1]$ and $t[m-1]$ are different, we need to perform one edit on $s$ (insert, delete, replace) so that the tail characters of the two strings match, allowing us to skip them and consider a smaller-scale problem.
- If $s[n-1]$ and $t[m-1]$ are different, we need to perform one edit on $s$ (insert, delete, or replace) to make the tail characters of the two strings the same, allowing us to skip them and consider a smaller-scale problem.
Thus, each round of decision (edit operation) in string $s$ changes the remaining characters in $s$ and $t$ to be matched. Therefore, the state is the $i$-th and $j$-th characters currently considered in $s$ and $t$, denoted as $[i, j]$.
In other words, each round of decision (edit operation) we make on string $s$ will change the remaining characters to be matched in $s$ and $t$. Therefore, the state is the $i$-th and $j$-th characters currently being considered in $s$ and $t$, denoted as $[i, j]$.
State $[i, j]$ corresponds to the subproblem: **The minimum number of edits required to change the first $i$ characters of $s$ into the first $j$ characters of $t$**.
State $[i, j]$ corresponds to the subproblem: **the minimum number of edits required to change the first $i$ characters of $s$ into the first $j$ characters of $t$**.
From this, we obtain a two-dimensional $dp$ table of size $(i+1) \times (j+1)$.
**Step two: Identify the optimal substructure and then derive the state transition equation**
**Step 2: Identify the optimal substructure, and then derive the state transition equation**
Consider the subproblem $dp[i, j]$, whose corresponding tail characters of the two strings are $s[i-1]$ and $t[j-1]$, which can be divided into three scenarios as shown in the figure below.
Consider subproblem $dp[i, j]$, where the tail characters of the corresponding two strings are $s[i-1]$ and $t[j-1]$, which can be divided into the three cases shown in the figure below based on different edit operations.
1. Add $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
1. Insert $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
2. Delete $s[i-1]$, then the remaining subproblem is $dp[i-1, j]$.
3. Replace $s[i-1]$ with $t[j-1]$, then the remaining subproblem is $dp[i-1, j-1]$.
![State transition of edit distance](edit_distance_problem.assets/edit_distance_state_transfer.png)
![State transition for edit distance](edit_distance_problem.assets/edit_distance_state_transfer.png)
Based on the analysis above, we can determine the optimal substructure: The minimum number of edits for $dp[i, j]$ is the minimum among $dp[i, j-1]$, $dp[i-1, j]$, and $dp[i-1, j-1]$, plus the edit step $1$. The corresponding state transition equation is:
Based on the above analysis, the optimal substructure can be obtained: the minimum number of edits for $dp[i, j]$ equals the minimum among the minimum edit steps of $dp[i, j-1]$, $dp[i-1, j]$, and $dp[i-1, j-1]$, plus the edit step $1$ for this time. The corresponding state transition equation is:
$$
dp[i, j] = \min(dp[i, j-1], dp[i-1, j], dp[i-1, j-1]) + 1
$$
Please note, **when $s[i-1]$ and $t[j-1]$ are the same, no edit is required for the current character**, in which case the state transition equation is:
Please note that **when $s[i-1]$ and $t[j-1]$ are the same, no edit is required for the current character**, in which case the state transition equation is:
$$
dp[i, j] = dp[i-1, j-1]
$$
**Step three: Determine the boundary conditions and the order of state transitions**
**Step 3: Determine boundary conditions and state transition order**
When both strings are empty, the number of edits is $0$, i.e., $dp[0, 0] = 0$. When $s$ is empty but $t$ is not, the minimum number of edits equals the length of $t$, that is, the first row $dp[0, j] = j$. When $s$ is not empty but $t$ is, the minimum number of edits equals the length of $s$, that is, the first column $dp[i, 0] = i$.
When both strings are empty, the number of edit steps is $0$, i.e., $dp[0, 0] = 0$. When $s$ is empty but $t$ is not, the minimum number of edit steps equals the length of $t$, i.e., the first row $dp[0, j] = j$. When $s$ is not empty but $t$ is empty, the minimum number of edit steps equals the length of $s$, i.e., the first column $dp[i, 0] = i$.
Observing the state transition equation, solving $dp[i, j]$ depends on the solutions to the left, above, and upper left, so a double loop can be used to traverse the entire $dp$ table in the correct order.
Observing the state transition equation, the solution $dp[i, j]$ depends on solutions to the left, above, and upper-left, so the entire $dp$ table can be traversed in order through two nested loops.
### Code implementation
@@ -71,10 +71,10 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
[file]{edit_distance}-[class]{}-[func]{edit_distance_dp}
```
As shown in the figure below, the process of state transition in the edit distance problem is very similar to that in the knapsack problem, which can be seen as filling a two-dimensional grid.
As shown in the figure below, the state transition process for the edit distance problem is very similar to the knapsack problem and can both be viewed as the process of filling a two-dimensional grid.
=== "<1>"
![Dynamic programming process of edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png)
![Dynamic programming process for edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png)
=== "<2>"
![edit_distance_dp_step2](edit_distance_problem.assets/edit_distance_dp_step2.png)
@@ -120,9 +120,9 @@ As shown in the figure below, the process of state transition in the edit distan
### Space optimization
Since $dp[i, j]$ is derived from the solutions above $dp[i-1, j]$, to the left $dp[i, j-1]$, and to the upper left $dp[i-1, j-1]$, and direct traversal will lose the upper left solution $dp[i-1, j-1]$, and reverse traversal cannot build $dp[i, j-1]$ in advance, therefore, both traversal orders are not feasible.
Since $dp[i, j]$ is transferred from the solutions above $dp[i-1, j]$, to the left $dp[i, j-1]$, and to the upper-left $dp[i-1, j-1]$, forward traversal will lose the upper-left solution $dp[i-1, j-1]$, and reverse traversal cannot build $dp[i, j-1]$ in advance, so neither traversal order is feasible.
For this reason, we can use a variable `leftup` to temporarily store the solution from the upper left $dp[i-1, j-1]$, thus only needing to consider the solutions to the left and above. This situation is similar to the unbounded knapsack problem, allowing for direct traversal. The code is as follows:
For this reason, we can use a variable `leftup` to temporarily store the upper-left solution $dp[i-1, j-1]$, so we only need to consider the solutions to the left and above. This situation is the same as the unbounded knapsack problem, allowing for forward traversal. The code is as follows:
```src
[file]{edit_distance}-[class]{}-[func]{edit_distance_dp_comp}