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Yudong Jin
97 lines
4.9 KiB
Markdown
97 lines
4.9 KiB
Markdown
# Exercises
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## Concept Review
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### How Binary Search Narrows the Search Interval
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Search for 16 in the sorted array `[2, 5, 8, 12, 16, 23, 38]`.
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Use the closed interval `[i, j]` and calculate the midpoint as
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$m=i+(j-i)/2$, rounded down.
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For each round, write `(i, j, m)`, the middle element, and how the interval is narrowed next,
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until the target is found.
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??? success "Answer"
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The search proceeds as follows:
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| Round | `(i, j, m)` | Middle element | Next step |
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| --- | --- | --- | --- |
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| 1 | `(0, 6, 3)` | 12 | `12 < 16`, set `i = 4` |
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| 2 | `(4, 6, 5)` | 23 | `23 > 16`, set `j = 4` |
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| 3 | `(4, 4, 4)` | 16 | Target found; return index 4 |
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Because the array is sorted, when the middle value is smaller than the target, the middle and everything to its left can be excluded.
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When the middle value is greater than the target, the middle and everything to its right can be excluded.
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### Left and Right Boundaries of Duplicate Elements
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Search for the number 2 in the array `[1, 2, 2, 2, 4, 6]`.
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A student uses binary search, returns immediately after finding the target at index 2, and says,
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"Index 2 is the left boundary of the number 2."
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<!-- numbered-subquestions -->
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1. Is the student's statement correct? What are the left and right boundaries of 2? Explain why.
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2. When searching for the left boundary, if the middle element equals the target, which side should be searched next?
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3. When searching for the right boundary, which side should be searched next? State only the direction; you do not need to write the complete search process.
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??? success "Answer"
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1. The student's statement is incorrect. Returning as soon as a 2 is found guarantees only that some occurrence of 2 was found, not the leftmost or rightmost one.
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Here, the left boundary is index 1 and the right boundary is index 3.
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2. When searching for the left boundary, continue searching on the left even if the middle element equals 2.
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For example, with a closed interval, set `j = m - 1`.
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3. When searching for the right boundary, continue searching on the right after the middle element equals 2.
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For example, set `i = m + 1`.
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### Choosing a Search Method for Different Data
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For each situation below, choose an appropriate method from "linear search," "binary search," and "hash table," and explain why:
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<!-- numbered-subquestions -->
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1. Repeatedly search among $10^7$ integers that are already sorted and never change, without building any additional data structure.
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2. Repeatedly test whether a key exists in a collection with frequent insertions and deletions. The collection need not remain sorted, and no range searches are needed.
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3. Search an unsorted array for a value only once.
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??? success "Answer"
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1. Binary search: the data is sorted and static, so searching takes $O(\log n)$ time without extra space.
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2. Hash table: when the hash function distributes keys fairly evenly among the buckets, insertion, deletion, and lookup by key all take $O(1)$ time on average.
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3. Scan directly from beginning to end. When searching only once, sorting the array or building a hash table still requires processing the entire array first,
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so neither reduces the total work for this single task.
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The choice depends on whether the data is sorted, whether an additional structure is allowed, how many searches are needed, and which operations must be supported.
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## Programming Exercises
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### Binary Search in a Sorted Array
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Given an integer array `nums` in strictly increasing order and a target value `target`, use binary search to find `target`. If it exists, return its array index; otherwise, return -1.
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??? tip "Hints"
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1. The initial interval is left = 0 and right = n - 1; it is nonempty while left <= right
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2. Calculate the midpoint with mid = left + (right - left) // 2
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3. If `nums[mid]` is less than `target`, move the left boundary to `mid + 1`; if `nums[mid]` is greater than `target`, move the right boundary to `mid - 1`; if they are equal, return immediately
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[LeetCode](https://leetcode.com/problems/binary-search/){ .rounded-button .exercise-button target="_blank" rel="noopener noreferrer" }
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### Insertion Point in a Sorted Array
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You are given an integer array `nums` in strictly increasing order and a target value `target`.
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Return the index of `target` if it is already in the array. Otherwise, return the insertion point at which `target` can be inserted while keeping the array in strictly increasing order.
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The insertion point may be 0 or may equal the array's length. Use binary search.
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??? tip "Hints"
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1. The answer may be 0 or the array length n
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2. With a closed interval, if `nums[mid]` is greater than or equal to `target`, set `right = mid - 1` and continue checking farther left; otherwise, set `left = mid + 1`
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3. When the loop ends, left is the insertion point
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[LeetCode](https://leetcode.com/problems/search-insert-position/){ .rounded-button .exercise-button target="_blank" rel="noopener noreferrer" }
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