feat: add solutions to lc problem: No.1311

solution/1300-1399/1310.XOR Queries of a Subarray/README.md

@@ -32,17 +32,17 @@ tags:
 
 <pre>
 <strong>输入：</strong>arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
-<strong>输出：</strong>[2,7,14,8] 
+<strong>输出：</strong>[2,7,14,8]
 <strong>解释：</strong>
 数组中元素的二进制表示形式是：
-1 = 0001 
-3 = 0011 
-4 = 0100 
-8 = 1000 
+1 = 0001
+3 = 0011
+4 = 0100
+8 = 1000
 查询的 XOR 值为：
-[0,1] = 1 xor 3 = 2 
-[1,2] = 3 xor 4 = 7 
-[0,3] = 1 xor 3 xor 4 xor 8 = 14 
+[0,1] = 1 xor 3 = 2
+[1,2] = 3 xor 4 = 7
+[0,3] = 1 xor 3 xor 4 xor 8 = 14
 [3,3] = 8
 </pre>
 
@@ -73,18 +73,18 @@ tags:
 
 ### 方法一：前缀异或
 
-我们可以用一个长度为 $n+1$ 的前缀异或数组 $s$ 来存储数组 $arr$ 的前缀异或结果，其中 $s[i] = s[i-1] \oplus arr[i-1]$，即 $s[i]$ 表示 $arr$ 中下标 $[0,i-1]$ 的元素的异或结果。
+我们可以用一个长度为 $n+1$ 的前缀异或数组 $s$ 来存储数组 $\textit{arr}$ 的前缀异或结果，其中 $s[i] = s[i-1] \oplus \textit{arr}[i-1]$，即 $s[i]$ 表示 $\textit{arr}$ 的前 $i$ 个元素的异或结果。
 
 那么对于一个查询 $[l,r]$，我们可以得到：
 
 $$
 \begin{aligned}
-arr[l] \oplus arr[l+1] \oplus \cdots \oplus arr[r] &= (arr[0] \oplus arr[1] \oplus \cdots \oplus arr[l-1]) \oplus (arr[0] \oplus arr[1] \oplus \cdots \oplus arr[r]) \\
+\textit{arr}[l] \oplus \textit{arr}[l+1] \oplus \cdots \oplus \textit{arr}[r] &= (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[l-1]) \oplus (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[r]) \\
 &= s[l] \oplus s[r+1]
 \end{aligned}
 $$
 
-时间复杂度 $O(n+m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 $arr$ 和查询数组 $queries$ 的长度。
+时间复杂度 $O(n+m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 $\textit{arr}$ 的长度和查询数组 $\textit{queries}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -162,15 +162,11 @@ func xorQueries(arr []int, queries [][]int) (ans []int) {
 ```ts
 function xorQueries(arr: number[], queries: number[][]): number[] {
     const n = arr.length;
-    const s: number[] = new Array(n + 1).fill(0);
+    const s: number[] = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans: number[] = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 }
 ```
 
@@ -184,15 +180,11 @@ function xorQueries(arr: number[], queries: number[][]): number[] {
  */
 var xorQueries = function (arr, queries) {
     const n = arr.length;
-    const s = new Array(n + 1).fill(0);
+    const s = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 };
 ```
 

solution/1300-1399/1310.XOR Queries of a Subarray/README_EN.md

@@ -31,17 +31,17 @@ tags:
 
 <pre>
 <strong>Input:</strong> arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
-<strong>Output:</strong> [2,7,14,8] 
-<strong>Explanation:</strong> 
+<strong>Output:</strong> [2,7,14,8]
+<strong>Explanation:</strong>
 The binary representation of the elements in the array are:
-1 = 0001 
-3 = 0011 
-4 = 0100 
-8 = 1000 
+1 = 0001
+3 = 0011
+4 = 0100
+8 = 1000
 The XOR values for queries are:
-[0,1] = 1 xor 3 = 2 
-[1,2] = 3 xor 4 = 7 
-[0,3] = 1 xor 3 xor 4 xor 8 = 14 
+[0,1] = 1 xor 3 = 2
+[1,2] = 3 xor 4 = 7
+[0,3] = 1 xor 3 xor 4 xor 8 = 14
 [3,3] = 8
 </pre>
 
@@ -68,7 +68,20 @@ The XOR values for queries are:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Prefix XOR
+
+We can use a prefix XOR array $s$ of length $n+1$ to store the prefix XOR results of the array $\textit{arr}$, where $s[i] = s[i-1] \oplus \textit{arr}[i-1]$. That is, $s[i]$ represents the XOR result of the first $i$ elements of $\textit{arr}$.
+
+For a query $[l, r]$, we can obtain:
+
+$$
+\begin{aligned}
+\textit{arr}[l] \oplus \textit{arr}[l+1] \oplus \cdots \oplus \textit{arr}[r] &= (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[l-1]) \oplus (\textit{arr}[0] \oplus \textit{arr}[1] \oplus \cdots \oplus \textit{arr}[r]) \\
+&= s[l] \oplus s[r+1]
+\end{aligned}
+$$
+
+Time complexity is $O(n+m)$, and space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the array $\textit{arr}$ and the query array $\textit{queries}$, respectively.
 
 <!-- tabs:start -->
 
@@ -146,15 +159,11 @@ func xorQueries(arr []int, queries [][]int) (ans []int) {
 ```ts
 function xorQueries(arr: number[], queries: number[][]): number[] {
     const n = arr.length;
-    const s: number[] = new Array(n + 1).fill(0);
+    const s: number[] = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans: number[] = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 }
 ```
 
@@ -168,15 +177,11 @@ function xorQueries(arr: number[], queries: number[][]): number[] {
  */
 var xorQueries = function (arr, queries) {
     const n = arr.length;
-    const s = new Array(n + 1).fill(0);
+    const s = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 };
 ```
 

solution/1300-1399/1310.XOR Queries of a Subarray/Solution.js

@@ -5,13 +5,9 @@
  */
 var xorQueries = function (arr, queries) {
     const n = arr.length;
-    const s = new Array(n + 1).fill(0);
+    const s = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 };

solution/1300-1399/1310.XOR Queries of a Subarray/Solution.ts

@@ -1,12 +1,8 @@
 function xorQueries(arr: number[], queries: number[][]): number[] {
     const n = arr.length;
-    const s: number[] = new Array(n + 1).fill(0);
+    const s: number[] = Array(n + 1).fill(0);
     for (let i = 0; i < n; ++i) {
         s[i + 1] = s[i] ^ arr[i];
     }
-    const ans: number[] = [];
-    for (const [l, r] of queries) {
-        ans.push(s[r + 1] ^ s[l]);
-    }
-    return ans;
+    return queries.map(([l, r]) => s[r + 1] ^ s[l]);
 }