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feat: add solutions to lc problem: No.1442
No.1442.Count Triplets That Can Form Two Arrays of Equal XOR
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‎solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/README.md

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<li><code>1 &lt;= arr[i] &lt;= 10^8</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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前缀异或，然后暴力枚举即可。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countTriplets(self, arr: List[int]) -> int:
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n = len(arr)
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pre = [0] * (n + 1)
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for i in range(n):
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pre[i + 1] = pre[i] ^ arr[i]
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ans = 0
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for i in range(n - 1):
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for j in range(i + 1, n):
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for k in range(j, n):
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a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
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if a == b:
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ans += 1
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countTriplets(int[] arr) {
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int n = arr.length;
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int[] pre = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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pre[i + 1] = pre[i] ^ arr[i];
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}
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int ans = 0;
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for (int i = 0; i < n - 1; ++i) {
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for (int j = i + 1; j < n; ++j) {
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for (int k = j; k < n; ++k) {
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int a = pre[j] ^ pre[i];
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int b = pre[k + 1] ^ pre[j];
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if (a == b) {
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++ans;
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}
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTriplets(vector<int>& arr) {
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int n = arr.size();
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vector<int> pre(n + 1);
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for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
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int ans = 0;
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for (int i = 0; i < n - 1; ++i)
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{
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for (int j = i + 1; j < n; ++j)
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{
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for (int k = j; k < n; ++k)
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{
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int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
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if (a == b) ++ans;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countTriplets(arr []int) int {
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n := len(arr)
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pre := make([]int, n+1)
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for i := 0; i < n; i++ {
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pre[i+1] = pre[i] ^ arr[i]
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}
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ans := 0
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for i := 0; i < n-1; i++ {
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for j := i + 1; j < n; j++ {
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for k := j; k < n; k++ {
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a, b := pre[j]^pre[i], pre[k+1]^pre[j]
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if a == b {
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ans++
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}
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}
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}
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}
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return ans
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}
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```
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### **...**

‎solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/README_EN.md

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<p>Given an array of&nbsp;integers <code>arr</code>.</p>
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<p>We want to select three indices <code>i</code>, <code>j</code> and <code>k</code> where <code>(0 &lt;= i &lt; j &lt;= k &lt; arr.length)</code>.</p>
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<p>Let&#39;s define <code>a</code> and <code>b</code> as follows:</p>
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<ul>
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<li><code>a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]</code></li>
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<li><code>b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]</code></li>
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</ul>
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<p>Note that <strong>^</strong> denotes the <strong>bitwise-xor</strong> operation.</p>
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<p>Return <em>the number of triplets</em> (<code>i</code>, <code>j</code> and <code>k</code>) Where <code>a == b</code>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [2,3,1,6,7]
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [1,1,1,1,1]
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [2,3]
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [1,3,5,7,9]
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</pre>
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<p><strong>Example 5:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [7,11,12,9,5,2,7,17,22]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= arr.length &lt;= 300</code></li>
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<li><code>1 &lt;= arr[i] &lt;= 10^8</code></li>
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### **Python3**
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```python
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class Solution:
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def countTriplets(self, arr: List[int]) -> int:
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n = len(arr)
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pre = [0] * (n + 1)
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for i in range(n):
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pre[i + 1] = pre[i] ^ arr[i]
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ans = 0
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for i in range(n - 1):
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for j in range(i + 1, n):
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for k in range(j, n):
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a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
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if a == b:
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ans += 1
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int countTriplets(int[] arr) {
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int n = arr.length;
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int[] pre = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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pre[i + 1] = pre[i] ^ arr[i];
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}
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int ans = 0;
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for (int i = 0; i < n - 1; ++i) {
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for (int j = i + 1; j < n; ++j) {
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for (int k = j; k < n; ++k) {
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int a = pre[j] ^ pre[i];
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int b = pre[k + 1] ^ pre[j];
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if (a == b) {
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++ans;
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}
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTriplets(vector<int>& arr) {
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int n = arr.size();
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vector<int> pre(n + 1);
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for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
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int ans = 0;
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for (int i = 0; i < n - 1; ++i)
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{
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for (int j = i + 1; j < n; ++j)
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{
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for (int k = j; k < n; ++k)
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{
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int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
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if (a == b) ++ans;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countTriplets(arr []int) int {
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n := len(arr)
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pre := make([]int, n+1)
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for i := 0; i < n; i++ {
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pre[i+1] = pre[i] ^ arr[i]
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}
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ans := 0
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for i := 0; i < n-1; i++ {
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for j := i + 1; j < n; j++ {
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for k := j; k < n; k++ {
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a, b := pre[j]^pre[i], pre[k+1]^pre[j]
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if a == b {
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ans++
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}
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}
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}
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}
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return ans
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}
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```
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### **...**

‎solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/Solution.cpp

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class Solution {
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public:
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int countTriplets(vector<int>& arr) {
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int n = arr.size();
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vector<int> pre(n + 1);
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for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
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int ans = 0;
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for (int i = 0; i < n - 1; ++i)
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{
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for (int j = i + 1; j < n; ++j)
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{
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for (int k = j; k < n; ++k)
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{
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int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
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if (a == b) ++ans;
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}
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}
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}
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return ans;
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}
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};

‎solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/Solution.go

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func countTriplets(arr []int) int {
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n := len(arr)
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pre := make([]int, n+1)
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for i := 0; i < n; i++ {
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pre[i+1] = pre[i] ^ arr[i]
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}
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ans := 0
8+
for i := 0; i < n-1; i++ {
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for j := i + 1; j < n; j++ {
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for k := j; k < n; k++ {
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a, b := pre[j]^pre[i], pre[k+1]^pre[j]
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if a == b {
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ans++
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}
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}
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}
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}
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return ans
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}

‎solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/Solution.java

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class Solution {
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public int countTriplets(int[] arr) {
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int n = arr.length;
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int[] pre = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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pre[i + 1] = pre[i] ^ arr[i];
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}
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int ans = 0;
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for (int i = 0; i < n - 1; ++i) {
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for (int j = i + 1; j < n; ++j) {
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for (int k = j; k < n; ++k) {
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int a = pre[j] ^ pre[i];
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int b = pre[k + 1] ^ pre[j];
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if (a == b) {
15+
++ans;
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}
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}
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}
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}
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return ans;
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}
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}

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