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feat: add solutions to lc problem: No.1442 #4267

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126 changes: 77 additions & 49 deletions ...n/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -85,7 +85,13 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：枚举

根据题目描述，要找到满足 $a = b$ 的三元组 $(i, j, k)$，即满足 $s = a \oplus b = 0$，我们只需要枚举左端点 $i$，然后计算以 $k$ 为右端点的区间 $[i, k]$ 的前缀异或和 $s$，如果 $s = 0$，那么对于任意 $j \in [i + 1, k]$，都满足 $a = b$，即 $(i, j, k)$ 是一个满足条件的三元组，一共有 $k - i$ 个，我们将其累加到答案中即可。

枚举结束后，返回答案即可。

时间复杂度 $O(n^2)$，其中 $n$ 是数组 $\textit{arr}$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -94,17 +100,13 @@ tags:
```python
class Solution:
def countTriplets(self, arr: List[int]) -> int:
n = len(arr)
pre = [0] * (n + 1)
for i in range(n):
pre[i + 1] = pre[i] ^ arr[i]
ans = 0
for i in range(n - 1):
for j in range(i + 1, n):
for k in range(j, n):
a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
if a == b:
ans += 1
ans, n = 0, len(arr)
for i, x in enumerate(arr):
s = x
for k in range(i + 1, n):
s ^= arr[k]
if s == 0:
ans += k - i
return ans
```

Expand All @@ -113,20 +115,13 @@ class Solution:
```java
class Solution {
public int countTriplets(int[] arr) {
int n = arr.length;
int[] pre = new int[n + 1];
int ans = 0, n = arr.length;
for (int i = 0; i < n; ++i) {
pre[i + 1] = pre[i] ^ arr[i];
}
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int k = j; k < n; ++k) {
int a = pre[j] ^ pre[i];
int b = pre[k + 1] ^ pre[j];
if (a == b) {
++ans;
}
int s = arr[i];
for (int k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s == 0) {
ans += k - i;
}
}
}
Expand All @@ -141,15 +136,13 @@ class Solution {
class Solution {
public:
int countTriplets(vector<int>& arr) {
int n = arr.size();
vector<int> pre(n + 1);
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int k = j; k < n; ++k) {
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
if (a == b) ++ans;
int ans = 0, n = arr.size();
for (int i = 0; i < n; ++i) {
int s = arr[i];
for (int k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s == 0) {
ans += k - i;
}
}
}
Expand All @@ -161,24 +154,59 @@ public:
#### Go

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

#### TypeScript

```ts
function countTriplets(arr: number[]): number {
const n = arr.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
let s = arr[i];
for (let k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s === 0) {
ans += k - i;
}
}
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn count_triplets(arr: Vec<i32>) -> i32 {
let mut ans = 0;
let n = arr.len();

for i in 0..n {
let mut s = arr[i];
for k in (i + 1)..n {
s ^= arr[k];
if s == 0 {
ans += (k - i) as i32;
}
}
}

ans
}
}
```

Expand Down
126 changes: 77 additions & 49 deletions ...400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -67,7 +67,13 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration

According to the problem description, to find triplets $(i, j, k)$ that satisfy $a = b$, which means $s = a \oplus b = 0$, we only need to enumerate the left endpoint $i$, and then calculate the prefix XOR sum $s$ of the interval $[i, k]$ with $k$ as the right endpoint. If $s = 0$, then for any $j \in [i + 1, k]$, the condition $a = b$ is satisfied, meaning $(i, j, k)$ is a valid triplet. There are $k - i$ such triplets, which we can add to our answer.

After the enumeration is complete, we return the answer.

The time complexity is $O(n^2)$, where $n$ is the length of the array $\textit{arr}$. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -76,17 +82,13 @@ tags:
```python
class Solution:
def countTriplets(self, arr: List[int]) -> int:
n = len(arr)
pre = [0] * (n + 1)
for i in range(n):
pre[i + 1] = pre[i] ^ arr[i]
ans = 0
for i in range(n - 1):
for j in range(i + 1, n):
for k in range(j, n):
a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
if a == b:
ans += 1
ans, n = 0, len(arr)
for i, x in enumerate(arr):
s = x
for k in range(i + 1, n):
s ^= arr[k]
if s == 0:
ans += k - i
return ans
```

Expand All @@ -95,20 +97,13 @@ class Solution:
```java
class Solution {
public int countTriplets(int[] arr) {
int n = arr.length;
int[] pre = new int[n + 1];
int ans = 0, n = arr.length;
for (int i = 0; i < n; ++i) {
pre[i + 1] = pre[i] ^ arr[i];
}
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int k = j; k < n; ++k) {
int a = pre[j] ^ pre[i];
int b = pre[k + 1] ^ pre[j];
if (a == b) {
++ans;
}
int s = arr[i];
for (int k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s == 0) {
ans += k - i;
}
}
}
Expand All @@ -123,15 +118,13 @@ class Solution {
class Solution {
public:
int countTriplets(vector<int>& arr) {
int n = arr.size();
vector<int> pre(n + 1);
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int k = j; k < n; ++k) {
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
if (a == b) ++ans;
int ans = 0, n = arr.size();
for (int i = 0; i < n; ++i) {
int s = arr[i];
for (int k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s == 0) {
ans += k - i;
}
}
}
Expand All @@ -143,24 +136,59 @@ public:
#### Go

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

#### TypeScript

```ts
function countTriplets(arr: number[]): number {
const n = arr.length;
let ans = 0;
for (let i = 0; i < n; ++i) {
let s = arr[i];
for (let k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s === 0) {
ans += k - i;
}
}
}
return ans;
}
```

#### Rust

```rust
impl Solution {
pub fn count_triplets(arr: Vec<i32>) -> i32 {
let mut ans = 0;
let n = arr.len();

for i in 0..n {
let mut s = arr[i];
for k in (i + 1)..n {
s ^= arr[k];
if s == 0 {
ans += (k - i) as i32;
}
}
}

ans
}
}
```

Expand Down
18 changes: 8 additions & 10 deletions solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,16 @@
class Solution {
public:
int countTriplets(vector<int>& arr) {
int n = arr.size();
vector<int> pre(n + 1);
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
for (int k = j; k < n; ++k) {
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
if (a == b) ++ans;
int ans = 0, n = arr.size();
for (int i = 0; i < n; ++i) {
int s = arr[i];
for (int k = i + 1; k < n; ++k) {
s ^= arr[k];
if (s == 0) {
ans += k - i;
}
}
}
return ans;
}
};
};
25 changes: 9 additions & 16 deletions solution/1400-1499/1442.Count Triplets That Can Form Two Arrays of Equal XOR/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,12 @@
func countTriplets(arr []int) int {
n := len(arr)
pre := make([]int, n+1)
for i := 0; i < n; i++ {
pre[i+1] = pre[i] ^ arr[i]
}
ans := 0
for i := 0; i < n-1; i++ {
for j := i + 1; j < n; j++ {
for k := j; k < n; k++ {
a, b := pre[j]^pre[i], pre[k+1]^pre[j]
if a == b {
ans++
}
func countTriplets(arr []int) (ans int) {
for i, x := range arr {
s := x
for k := i + 1; k < len(arr); k++ {
s ^= arr[k]
if s == 0 {
ans += k - i
}
}
}
return ans
}
return
}
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