Merge ed17b6f918 into dad0a3fd95

en/docs/chapter_searching/binary_search_edge.md

@@ -6,7 +6,7 @@
 
     Given a sorted array `nums` of length $n$, which may contain duplicate elements, return the index of the leftmost element `target`. If the element is not present in the array, return $-1$.
 
-Look back on the method of binary search for an insertion point, after the search is completed, the index $i$ will point to the leftmost occurrence of `target`. Therefore, **searching for the insertion point is essentially the same as finding the index of the leftmost `target`**.
+Recall the method of binary search for an insertion point, after the search is completed, the index $i$ will point to the leftmost occurrence of `target`. Therefore, **searching for the insertion point is essentially the same as finding the index of the leftmost `target`**.
 
 We can use the function for finding an insertion point to find the left boundary of `target`. Note that the array might not contain `target`, which could lead to the following two results:
 
@@ -29,7 +29,7 @@ Below we are going to introduce two more ingenious methods.
 
 To find the rightmost occurrence of `target`, we can reuse the logic for finding the leftmost occurrence of `target`. Specifically, we can find the leftmost `target`, and then adjust the result to point to the rightmost `target` by simply adding 1 to the index of the leftmost `target`.
 
-As shown in the figure below, after the search is complete, pointer $i$ will point to the the position just after the leftmost `target` (i.e., `target + 1`), and pointer $j$ will point to the rightmost `target`. Therefore, returning $j$ will give us the right boundary.
+As shown in the figure below, after the search is complete, pointer $i$ will point to the position just after the leftmost `target` (i.e., `target + 1`), and pointer $j$ will point to the rightmost `target`. Therefore, returning $j$ will give us the right boundary.
 
 ![Transforming the search for the right boundary into the search for the left boundary](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png)
 
@@ -52,5 +52,5 @@ Thus, as shown in the figure below, we can construct an element that does not ex
 
 The code is omitted here, but here are two important points to note about this approach.
 
-- Since the given array does not contain decimals, this means we do not need to worry about handling equal cases.
-- Because this method introduces decimals, the variable `target` in the function needs to be changed to a floating point type (no change needed in Python).
+- The given array `nums` does not contain decimal, so handling equal cases is not a concern.
+- However, the introduction of decimal in this approach requires modifying the `target` variable to a floating-point type (no change needed in Python).

en/docs/chapter_searching/binary_search_insertion.md

@@ -36,17 +36,17 @@ When there are multiple occurrences of `target` in the array, a regular binary s
 
 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 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.
+1. Perform a binary search to find any index of `target`, say $k$.
+2. Starting from index $k$, conduct a linear traversal 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)
 
 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. 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.
+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` with `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 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`**.
+- 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 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 bring 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**.
 
@@ -88,4 +88,4 @@ Even so, we can still keep the conditions expanded, as it makes the logic cleare
 
 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. In the end, 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. Ultimately, they either find the answer or stop after crossing the boundary.