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Commit e84c011

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feat: add solutions to lc problem: No.1442
No.1442.Count Triplets That Can Form Two Arrays of Equal XOR
1 parent 5b0cd14 commit e84c011

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8 files changed

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

Lines changed: 87 additions & 2 deletions
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@@ -63,27 +63,112 @@
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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
127+
class Solution {
128+
public:
129+
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>
@@ -123,13 +89,97 @@
12389
### **Python3**
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```python
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class Solution:
93+
def countTriplets(self, arr: List[int]) -> int:
94+
n = len(arr)
95+
pre = [0] * (n + 1)
96+
for i in range(n):
97+
pre[i + 1] = pre[i] ^ arr[i]
98+
ans = 0
99+
for i in range(n - 1):
100+
for j in range(i + 1, n):
101+
for k in range(j, n):
102+
a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
103+
if a == b:
104+
ans += 1
105+
return ans
127106
```
128107

129108
### **Java**
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131110
```java
111+
class Solution {
112+
public int countTriplets(int[] arr) {
113+
int n = arr.length;
114+
int[] pre = new int[n + 1];
115+
for (int i = 0; i < n; ++i) {
116+
pre[i + 1] = pre[i] ^ arr[i];
117+
}
118+
int ans = 0;
119+
for (int i = 0; i < n - 1; ++i) {
120+
for (int j = i + 1; j < n; ++j) {
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for (int k = j; k < n; ++k) {
122+
int a = pre[j] ^ pre[i];
123+
int b = pre[k + 1] ^ pre[j];
124+
if (a == b) {
125+
++ans;
126+
}
127+
}
128+
}
129+
}
130+
return ans;
131+
}
132+
}
133+
```
134+
135+
### **C++**
136+
137+
```cpp
138+
class Solution {
139+
public:
140+
int countTriplets(vector<int>& arr) {
141+
int n = arr.size();
142+
vector<int> pre(n + 1);
143+
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
144+
int ans = 0;
145+
for (int i = 0; i < n - 1; ++i)
146+
{
147+
for (int j = i + 1; j < n; ++j)
148+
{
149+
for (int k = j; k < n; ++k)
150+
{
151+
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
152+
if (a == b) ++ans;
153+
}
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}
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}
156+
return ans;
157+
}
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};
159+
```
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### **Go**
162+
163+
```go
164+
func countTriplets(arr []int) int {
165+
n := len(arr)
166+
pre := make([]int, n+1)
167+
for i := 0; i < n; i++ {
168+
pre[i+1] = pre[i] ^ arr[i]
169+
}
170+
ans := 0
171+
for i := 0; i < n-1; i++ {
172+
for j := i + 1; j < n; j++ {
173+
for k := j; k < n; k++ {
174+
a, b := pre[j]^pre[i], pre[k+1]^pre[j]
175+
if a == b {
176+
ans++
177+
}
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}
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}
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}
181+
return ans
182+
}
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```
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### **...**
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class Solution {
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public:
3+
int countTriplets(vector<int>& arr) {
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int n = arr.size();
5+
vector<int> pre(n + 1);
6+
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
7+
int ans = 0;
8+
for (int i = 0; i < n - 1; ++i)
9+
{
10+
for (int j = i + 1; j < n; ++j)
11+
{
12+
for (int k = j; k < n; ++k)
13+
{
14+
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
15+
if (a == b) ++ans;
16+
}
17+
}
18+
}
19+
return ans;
20+
}
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};
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Original file line numberDiff line numberDiff line change
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func countTriplets(arr []int) int {
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n := len(arr)
3+
pre := make([]int, n+1)
4+
for i := 0; i < n; i++ {
5+
pre[i+1] = pre[i] ^ arr[i]
6+
}
7+
ans := 0
8+
for i := 0; i < n-1; i++ {
9+
for j := i + 1; j < n; j++ {
10+
for k := j; k < n; k++ {
11+
a, b := pre[j]^pre[i], pre[k+1]^pre[j]
12+
if a == b {
13+
ans++
14+
}
15+
}
16+
}
17+
}
18+
return ans
19+
}
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Original file line numberDiff line numberDiff line change
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1+
class Solution {
2+
public int countTriplets(int[] arr) {
3+
int n = arr.length;
4+
int[] pre = new int[n + 1];
5+
for (int i = 0; i < n; ++i) {
6+
pre[i + 1] = pre[i] ^ arr[i];
7+
}
8+
int ans = 0;
9+
for (int i = 0; i < n - 1; ++i) {
10+
for (int j = i + 1; j < n; ++j) {
11+
for (int k = j; k < n; ++k) {
12+
int a = pre[j] ^ pre[i];
13+
int b = pre[k + 1] ^ pre[j];
14+
if (a == b) {
15+
++ans;
16+
}
17+
}
18+
}
19+
}
20+
return ans;
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}
22+
}

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