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en/docs/chapter_searching/binary_search_insertion.md

@@ -10,7 +10,7 @@ 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 Figure 10-4.
+    Given a sorted array `nums` of length $n$ with unique elements and an element `target`, 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. Return the index of `target` in the array after insertion. See the example shown in Figure 10-4.
 
 ![Example data for binary search insertion point](binary_search_insertion.assets/binary_search_insertion_example.png){ class="animation-figure" }
 
@@ -18,15 +18,15 @@ Binary search is not only used to search for target elements but also to solve m
 
 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:
 
 === "Python"
 
@@ -160,12 +160,12 @@ 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**.
+When there are multiple occurrences of `target` in the array, a regular binary search can only return the index of one occurrence of `target`, **and it cannot determine how many occurrences of `target` are to the left and right of that position**.
 
-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 Figure 10-5.
+The problem requires inserting the target element at the leftmost position, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in Figure 10-5.
 
-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 index of `target`, say $k$.
+2. Starting from index $k$, conduct a linear search to the left until the leftmost occurrence of `target` is found, then return this index.
 
 ![Linear search for the insertion point of duplicate elements](binary_search_insertion.assets/binary_search_insertion_naive.png){ class="animation-figure" }
 
@@ -173,10 +173,10 @@ The task requires inserting the target element to the very left, **so we need to
 
 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 Figure 10-6, 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 Figure 10-6, the overall process remains the same. In each round, we first calculate the middle index $m$, then compare the value of `target` with `nums[m]`, leading to 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, **bringing 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 bringing 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**.
 
@@ -206,9 +206,9 @@ After the loop, $i$ points to the leftmost `target`, and $j$ points to the first
 
 <p align="center"> Figure 10-6 &nbsp; Steps for binary search insertion point of duplicate elements </p>
 
-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.
 
 === "Python"
 
@@ -338,8 +338,8 @@ 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 $j$. These targets could be a specific element (like `target`) or a range of elements (such as those smaller 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.