From 5f77ed363f2a7a462a3387e8ba1461b6de70e449 Mon Sep 17 00:00:00 2001 From: yanedie Date: Tue, 12 Nov 2024 02:03:18 +0800 Subject: [PATCH] translation: Update binary_search_insertion.md --- .../binary_search_insertion.md | 36 +++++++++---------- 1 file changed, 18 insertions(+), 18 deletions(-) diff --git a/en/docs/chapter_searching/binary_search_insertion.md b/en/docs/chapter_searching/binary_search_insertion.md index 8eb4a06b0..5a6eab973 100644 --- a/en/docs/chapter_searching/binary_search_insertion.md +++ b/en/docs/chapter_searching/binary_search_insertion.md @@ -6,21 +6,21 @@ Binary search is not only used to search for target elements but also to solve m !!! question - Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in the figure below. + Given a sorted array `nums` of length $n$ with unique elements and an element `target`. Now insert `target` into `nums` while maintaining its sorted order. If `target` already exists in the array, insert it to the left of the existing element. Please return the index of `target` in the array after insertion. See the example shown in the figure below. ![Example data for binary search insertion point](binary_search_insertion.assets/binary_search_insertion_example.png) If you want to reuse the binary search code from the previous section, you need to answer the following two questions. -**Question one**: When the array contains `target`, is the insertion point index the index of that element? +**Question one**: If the array already contains `target`, would the insertion point be the index of existing element? -The requirement to insert `target` to the left of equal elements means that the newly inserted `target` replaces the original `target` position. Thus, **when the array contains `target`, the insertion point index is the index of that `target`**. +The requirement to insert `target` to the left of equal elements means that the newly inserted `target` will replace the original `target` position. In other words, **when the array contains `target`, the insertion point is indeed the index of that `target`**. -**Question two**: When the array does not contain `target`, what is the index of the insertion point? +**Question two**: When the array does not contain `target`, at which index would it be inserted? -Further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`. +Let's further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`. -Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion index is $i$**. The code is as follows: +Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion point is $i$**. The code is as follows: ```src [file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion_simple} @@ -32,21 +32,21 @@ Therefore, at the end of the binary, it is certain that: $i$ points to the first Based on the previous question, assume the array may contain duplicate elements, all else remains the same. -Suppose there are multiple `target`s in the array, ordinary binary search can only return the index of one of the `target`s, **and it cannot determine how many `target`s are to the left and right of that element**. +Suppose there are multiple `target`s in the array, a regular binary search can only return the index of one `target`, **and it cannot determine how many `target`s are to the left and right of that it**. -The task requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in the figure below. +The problem requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in the figure below. -1. Perform a binary search, get an arbitrary index of `target`, denoted as $k$. -2. Start from index $k$, and perform a linear search to the left until the leftmost `target` is found and return. +1. Perform a binary search to find any `target`'s index, say $k$. +2. Starting from index $k$, perform a linear search to the left until the leftmost `target` is found and return. ![Linear search for the insertion point of duplicate elements](binary_search_insertion.assets/binary_search_insertion_naive.png) Although this method is feasible, it includes linear search, so its time complexity is $O(n)$. This method is inefficient when the array contains many duplicate `target`s. -Now consider extending the binary search code. As shown in the figure below, the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases. +Now consider extending the binary search code. As shown in the figure below, the overall process remains the same. In each round, we first calculate the middle index $m$, then compare the value of `target` and `nums[m]`, which results in the following cases. -- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search interval reduction operation, **thus making pointers $i$ and $j$ approach `target`**. -- When `nums[m] == target`, it indicates that the elements less than `target` are in the interval $[i, m - 1]$, therefore use $j = m - 1$ to narrow the interval, **thus making pointer $j$ approach elements less than `target`**. +- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search to narrow the search range, **bring the pointers $i$ and $j$ closer to `target`**. +- When `nums[m] == target`, it indicates that the elements less than `target` are in the range $[i, m - 1]$, therefore use $j = m - 1$ to narrow the range, **thus making pointer $j$ closer to the elements less than `target`**. After the loop, $i$ points to the leftmost `target`, and $j$ points to the first element less than `target`, **therefore index $i$ is the insertion point**. @@ -74,9 +74,9 @@ After the loop, $i$ points to the leftmost `target`, and $j$ points to the first === "<8>" ![binary_search_insertion_step8](binary_search_insertion.assets/binary_search_insertion_step8.png) -Observe the code, the operations of the branch `nums[m] > target` and `nums[m] == target` are the same, so the two can be combined. +Observe the following code. The operations in the branches `nums[m] > target` and `nums[m] == target` are the same, so these two branches can be merged. -Even so, we can still keep the conditions expanded, as their logic is clearer and more readable. +Even so, we can still keep the conditions expanded, as it makes the logic clearer and improves readability. ```src [file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion} @@ -84,8 +84,8 @@ Even so, we can still keep the conditions expanded, as their logic is clearer an !!! tip - The code in this section uses "closed intervals". Readers interested can implement the "left-closed right-open" method themselves. + The code in this section uses "closed interval". If you are interested in "left-closed,right-open", try to implement the code on your own. -In summary, binary search is merely about setting search targets for pointers $i$ and $j$, which might be a specific element (like `target`) or a range of elements (like elements less than `target`). +In summary, binary search essentially involves setting search targets for pointers $i$ and $j$, which might be a specific element (like `target`) or a range of elements (like elements less than `target`). -In the continuous loop of binary search, pointers $i$ and $j$ gradually approach the predefined target. Ultimately, they either find the answer or stop after crossing the boundary. +In the continuous loop of binary search, pointers $i$ and $j$ gradually approach the predefined target. In the end, they either find the answer or stop after crossing the boundary.