translation: binary search updates (#1569)

* translation: binary search updates

* fix minor gramma and expression issues

en/docs/chapter_searching/binary_search.md

@@ -1,24 +1,24 @@
 # Binary search
 
-<u>Binary search</u> is an efficient search algorithm based on the divide-and-conquer strategy. It utilizes the orderliness of data, reducing the search range by half each round until the target element is found or the search interval is empty.
+<u>Binary search</u> is an efficient search algorithm that uses a divide-and-conquer strategy. It takes advantage of the sorted order of elements in an array by reducing the search interval by half in each iteration, continuing until either the target element is found or the search interval becomes empty.
 
 !!! question
 
-    Given an array `nums` of length $n$, with elements arranged in ascending order and non-repeating. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown in the figure below.
+    Given an array `nums` of length $n$, where elements are arranged in ascending order without duplicates. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown in the figure below.
 
 ![Binary search example data](binary_search.assets/binary_search_example.png)
 
-As shown in the figure below, we first initialize pointers $i = 0$ and $j = n - 1$, pointing to the first and last elements of the array, representing the search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
+As shown in the figure below, we firstly initialize pointers with $i = 0$ and $j = n - 1$, pointing to the first and last element of the array respectively. They also represent the whole search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
 
-Next, perform the following two steps in a loop.
+And then the following two steps may be performed in a loop.
 
 1. Calculate the midpoint index $m = \lfloor {(i + j) / 2} \rfloor$, where $\lfloor \: \rfloor$ denotes the floor operation.
-2. Compare the size of `nums[m]` and `target`, divided into the following three scenarios.
+2. Based on the comparison between the value of `nums[m]` and `target`, one of the following three cases will be chosen to execute.
     1. If `nums[m] < target`, it indicates that `target` is in the interval $[m + 1, j]$, thus set $i = m + 1$.
     2. If `nums[m] > target`, it indicates that `target` is in the interval $[i, m - 1]$, thus set $j = m - 1$.
     3. If `nums[m] = target`, it indicates that `target` is found, thus return index $m$.
 
-If the array does not contain the target element, the search interval will eventually reduce to empty. In this case, return $-1$.
+If the array does not contain the target element, the search interval will eventually reduce to empty, ending up returning $-1$.
 
 === "<1>"
     ![Binary search process](binary_search.assets/binary_search_step1.png)
@@ -41,7 +41,7 @@ If the array does not contain the target element, the search interval will event
 === "<7>"
     ![binary_search_step7](binary_search.assets/binary_search_step7.png)
 
-It's worth noting that since $i$ and $j$ are both of type `int`, **$i + j$ might exceed the range of `int` type**. To avoid large number overflow, we usually use the formula $m = \lfloor {i + (j - i) / 2} \rfloor$ to calculate the midpoint.
+It's worth noting that as $i$ and $j$ are both of type `int`, **$i + j$ might exceed the range of `int` type**. To avoid large number overflow, we usually use the formula $m = \lfloor {i + (j - i) / 2} \rfloor$ to calculate the midpoint.
 
 The code is as follows:
 
@@ -49,13 +49,13 @@ The code is as follows:
 [file]{binary_search}-[class]{}-[func]{binary_search}
 ```
 
-**Time complexity is $O(\log n)$** : In the binary loop, the interval reduces by half each round, hence the number of iterations is $\log_2 n$.
+**Time complexity is $O(\log n)$** : In the binary loop, the interval decreases by half each round, hence the number of iterations is $\log_2 n$.
 
-**Space complexity is $O(1)$** : Pointers $i$ and $j$ use constant size space.
+**Space complexity is $O(1)$** : Pointers $i$ and $j$ occupies constant size of space.
 
 ## Interval representation methods
 
-Besides the aforementioned closed interval, a common interval representation is the "left-closed right-open" interval, defined as $[0, n)$, where the left boundary includes itself, and the right boundary does not include itself. In this representation, the interval $[i, j)$ is empty when $i = j$.
+Besides the above closed interval, another common interval representation is the "left-closed right-open" interval, defined as $[0, n)$, where the left boundary includes itself, and the right boundary does not. In this representation, the interval $[i, j)$ is empty when $i = j$.
 
 We can implement a binary search algorithm with the same functionality based on this representation:
 
@@ -63,9 +63,9 @@ We can implement a binary search algorithm with the same functionality based on
 [file]{binary_search}-[class]{}-[func]{binary_search_lcro}
 ```
 
-As shown in the figure below, in the two types of interval representations, the initialization of the binary search algorithm, the loop condition, and the narrowing interval operation are different.
+As shown in the figure below, under the two types of interval representations, the initialization, loop condition, and narrowing interval operation of the binary search algorithm differ.
 
-Since both boundaries in the "closed interval" representation are defined as closed, the operations to narrow the interval through pointers $i$ and $j$ are also symmetrical. This makes it less prone to errors, **therefore, it is generally recommended to use the "closed interval" approach**.
+Since both boundaries in the "closed interval" representation are inclusive, the operations to narrow the interval through pointers $i$ and $j$ are also symmetrical. This makes it less prone to errors, **therefore, it is generally recommended to use the "closed interval" approach**.
 
 ![Two types of interval definitions](binary_search.assets/binary_search_ranges.png)
 
@@ -73,11 +73,11 @@ Since both boundaries in the "closed interval" representation are defined as clo
 
 Binary search performs well in both time and space aspects.
 
-- Binary search is time-efficient. With large data volumes, the logarithmic time complexity has a significant advantage. For instance, when the data size $n = 2^{20}$, linear search requires $2^{20} = 1048576$ iterations, while binary search only requires $\log_2 2^{20} = 20$ iterations.
+- Binary search is time-efficient. With large dataset, the logarithmic time complexity offers a major advantage. For instance, given a dataset with size $n = 2^{20}$, linear search requires $2^{20} = 1048576$ iterations, while binary search only demands $\log_2 2^{20} = 20$ loops.
-- Binary search does not require extra space. Compared to search algorithms that rely on additional space (like hash search), binary search is more space-efficient.
+- Binary search does not need extra space. Compared to search algorithms that rely on additional space (like hash search), binary search is more space-efficient.
 
-However, binary search is not suitable for all situations, mainly for the following reasons.
+However, binary search may not be suitable for all scenarios due to the following concerns.
 
-- Binary search is only applicable to ordered data. If the input data is unordered, it is not worth sorting it just to use binary search, as sorting algorithms typically have a time complexity of $O(n \log n)$, which is higher than both linear and binary search. For scenarios with frequent element insertion to maintain array order, inserting elements into specific positions has a time complexity of $O(n)$, which is also quite costly.
+- Binary search can only be applied to sorted data. Unsorted data must be sorted before applying binary search, which may not be worthwhile as sorting algorithm typically has a time complexity of $O(n \log n)$. Such cost is even higher than linear search, not to mention binary search itself. For scenarios with frequent insertion, the cost of remaining the array in order is pretty high as the time complexity of inserting new elements into specific positions is $O(n)$.
-- Binary search is only applicable to arrays. Binary search requires non-continuous (jumping) element access, which is inefficient in linked lists, thus not suitable for use in linked lists or data structures based on linked lists.
+- Binary search may use array only. Binary search requires non-continuous (jumping) element access, which is inefficient in linked list. As a result, linked list or data structures based on linked list may not be suitable for this algorithm.
-- With small data volumes, linear search performs better. In linear search, each round only requires 1 decision operation; whereas in binary search, it involves 1 addition, 1 division, 1 to 3 decision operations, 1 addition (subtraction), totaling 4 to 6 operations; therefore, when data volume $n$ is small, linear search can be faster than binary search.
+- Linear search performs better on small dataset. In linear search, only 1 decision operation is required for each iteration; whereas in binary search, it involves 1 addition, 1 division, 1 to 3 decision operations, 1 addition (subtraction), totaling 4 to 6 operations. Therefore, if data size $n$ is small, linear search is faster than binary search.