build

en/docs/chapter_searching/binary_search_edge.md

@@ -10,9 +10,9 @@ comments: true
 
     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 $-1$.
 
-Recall the method of binary search for an insertion point, after the search is completed, $i$ points to the leftmost `target`, **thus searching for the insertion point is essentially searching for the index of the leftmost `target`**.
+Recalling 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`**.
 
-Consider implementing the search for the left boundary using the function for finding an insertion point. Note that the array might not contain `target`, which could lead to the following two results:
+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:
 
 - The index $i$ of the insertion point is out of bounds.
 - The element `nums[i]` is not equal to `target`.
@@ -133,21 +133,21 @@ In these cases, simply return $-1$. The code is as follows:
 
 ## 10.3.2   Find the right boundary
 
-So how do we find the rightmost `target`? The most straightforward way is to modify the code, replacing the pointer contraction operation in the case of `nums[m] == target`. The code is omitted here, but interested readers can implement it on their own.
+How do we find the rightmost occurrence of `target`? The most straightforward way is to modify the traditional binary search logic by changing how we adjust the search boundaries in the case of `nums[m] == target`. The code is omitted here. If you are interested, try to implement the code on your own.
 
-Below we introduce two more cunning methods.
+Below we are going to introduce two more ingenious methods.
 
-### 1.   Reusing the search for the left boundary
+### 1.   Reuse the left boundary search
 
-In fact, we can use the function for finding the leftmost element to find the rightmost element, specifically by **transforming the search for the rightmost `target` into a search for the leftmost `target + 1`**.
+To find the rightmost occurrence of `target`, we can reuse the function used for locating the leftmost `target`. Specifically, we transform the search for the rightmost target into a search for the leftmost target + 1.
 
-As shown in Figure 10-7, after the search is completed, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **thus returning $j$ is sufficient**.
+As shown in Figure 10-7, after the search is complete, pointer $i$ will point to the leftmost `target + 1` (if exists), while pointer $j$ will point to the rightmost occurrence of `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){ class="animation-figure" }
 
 <p align="center"> Figure 10-7 &nbsp; Transforming the search for the right boundary into the search for the left boundary </p>
 
-Please note, the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
+Note that the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
 
 === "Python"
 
@@ -267,9 +267,9 @@ Please note, the insertion point returned is $i$, therefore, it should be subtra
     [class]{}-[func]{binarySearchRightEdge}
     ```
 
-### 2. &nbsp; Transforming into an element search
+### 2. &nbsp; Transform into an element search
 
-We know that when the array does not contain `target`, $i$ and $j$ will eventually point to the first element greater and smaller than `target` respectively.
+When the array does not contain `target`, $i$ and $j$ will eventually point to the first element greater and smaller than `target` respectively.
 
 Thus, as shown in Figure 10-8, we can construct an element that does not exist in the array, to search for the left and right boundaries.
 
@@ -280,7 +280,7 @@ Thus, as shown in Figure 10-8, we can construct an element that does not exist i
 
 <p align="center"> Figure 10-8 &nbsp; Transforming the search for boundaries into the search for an element </p>
 
-The code is omitted here, but two points are worth noting.
+The code is omitted here, but here are two important points to note about this approach.
 
-- The given array does not contain decimals, meaning we do not need to worry about how to handle equal situations.
-- Since 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, introducing decimals in this approach requires modifying the `target` variable to a floating-point type (no change needed in Python).