Skip to content

Commit f563c72

Browse files
yanglbmeyanglbme
authored
feat: add solutions to lc problem: No.1442 (#4267)
No.1442.Count Triplets That Can Form Two Arrays of Equal XOR
1 parent 6793e6d commit f563c72

File tree

8 files changed

+217
-149
lines changed

8 files changed

+217
-149
lines changed

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

+77-49
Original file line numberDiff line numberDiff line change
@@ -85,7 +85,13 @@ tags:
8585

8686
<!-- solution:start -->
8787

88-
### 方法一
88+
### 方法一：枚举
89+
90+
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。
91+
92+
枚举结束后，返回答案即可。
93+
94+
时间复杂度 $O(n^2)$，其中 $n$ 是数组 $\textit{arr}$ 的长度。空间复杂度 $O(1)$。
8995

9096
<!-- tabs:start -->
9197

@@ -94,17 +100,13 @@ tags:
94100
```python
95101
class Solution:
96102
def countTriplets(self, arr: List[int]) -> int:
97-
n = len(arr)
98-
pre = [0] * (n + 1)
99-
for i in range(n):
100-
pre[i + 1] = pre[i] ^ arr[i]
101-
ans = 0
102-
for i in range(n - 1):
103-
for j in range(i + 1, n):
104-
for k in range(j, n):
105-
a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
106-
if a == b:
107-
ans += 1
103+
ans, n = 0, len(arr)
104+
for i, x in enumerate(arr):
105+
s = x
106+
for k in range(i + 1, n):
107+
s ^= arr[k]
108+
if s == 0:
109+
ans += k - i
108110
return ans
109111
```
110112

@@ -113,20 +115,13 @@ class Solution:
113115
```java
114116
class Solution {
115117
public int countTriplets(int[] arr) {
116-
int n = arr.length;
117-
int[] pre = new int[n + 1];
118+
int ans = 0, n = arr.length;
118119
for (int i = 0; i < n; ++i) {
119-
pre[i + 1] = pre[i] ^ arr[i];
120-
}
121-
int ans = 0;
122-
for (int i = 0; i < n - 1; ++i) {
123-
for (int j = i + 1; j < n; ++j) {
124-
for (int k = j; k < n; ++k) {
125-
int a = pre[j] ^ pre[i];
126-
int b = pre[k + 1] ^ pre[j];
127-
if (a == b) {
128-
++ans;
129-
}
120+
int s = arr[i];
121+
for (int k = i + 1; k < n; ++k) {
122+
s ^= arr[k];
123+
if (s == 0) {
124+
ans += k - i;
130125
}
131126
}
132127
}
@@ -141,15 +136,13 @@ class Solution {
141136
class Solution {
142137
public:
143138
int countTriplets(vector<int>& arr) {
144-
int n = arr.size();
145-
vector<int> pre(n + 1);
146-
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
147-
int ans = 0;
148-
for (int i = 0; i < n - 1; ++i) {
149-
for (int j = i + 1; j < n; ++j) {
150-
for (int k = j; k < n; ++k) {
151-
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
152-
if (a == b) ++ans;
139+
int ans = 0, n = arr.size();
140+
for (int i = 0; i < n; ++i) {
141+
int s = arr[i];
142+
for (int k = i + 1; k < n; ++k) {
143+
s ^= arr[k];
144+
if (s == 0) {
145+
ans += k - i;
153146
}
154147
}
155148
}
@@ -161,24 +154,59 @@ public:
161154
#### Go
162155
163156
```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-
}
157+
func countTriplets(arr []int) (ans int) {
158+
for i, x := range arr {
159+
s := x
160+
for k := i + 1; k < len(arr); k++ {
161+
s ^= arr[k]
162+
if s == 0 {
163+
ans += k - i
178164
}
179165
}
180166
}
181-
return ans
167+
return
168+
}
169+
```
170+
171+
#### TypeScript
172+
173+
```ts
174+
function countTriplets(arr: number[]): number {
175+
const n = arr.length;
176+
let ans = 0;
177+
for (let i = 0; i < n; ++i) {
178+
let s = arr[i];
179+
for (let k = i + 1; k < n; ++k) {
180+
s ^= arr[k];
181+
if (s === 0) {
182+
ans += k - i;
183+
}
184+
}
185+
}
186+
return ans;
187+
}
188+
```
189+
190+
#### Rust
191+
192+
```rust
193+
impl Solution {
194+
pub fn count_triplets(arr: Vec<i32>) -> i32 {
195+
let mut ans = 0;
196+
let n = arr.len();
197+
198+
for i in 0..n {
199+
let mut s = arr[i];
200+
for k in (i + 1)..n {
201+
s ^= arr[k];
202+
if s == 0 {
203+
ans += (k - i) as i32;
204+
}
205+
}
206+
}
207+
208+
ans
209+
}
182210
}
183211
```
184212

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

+77-49
Original file line numberDiff line numberDiff line change
@@ -67,7 +67,13 @@ tags:
6767

6868
<!-- solution:start -->
6969

70-
### Solution 1
70+
### Solution 1: Enumeration
71+
72+
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.
73+
74+
After the enumeration is complete, we return the answer.
75+
76+
The time complexity is $O(n^2)$, where $n$ is the length of the array $\textit{arr}$. The space complexity is $O(1)$.
7177

7278
<!-- tabs:start -->
7379

@@ -76,17 +82,13 @@ tags:
7682
```python
7783
class Solution:
7884
def countTriplets(self, arr: List[int]) -> int:
79-
n = len(arr)
80-
pre = [0] * (n + 1)
81-
for i in range(n):
82-
pre[i + 1] = pre[i] ^ arr[i]
83-
ans = 0
84-
for i in range(n - 1):
85-
for j in range(i + 1, n):
86-
for k in range(j, n):
87-
a, b = pre[j] ^ pre[i], pre[k + 1] ^ pre[j]
88-
if a == b:
89-
ans += 1
85+
ans, n = 0, len(arr)
86+
for i, x in enumerate(arr):
87+
s = x
88+
for k in range(i + 1, n):
89+
s ^= arr[k]
90+
if s == 0:
91+
ans += k - i
9092
return ans
9193
```
9294

@@ -95,20 +97,13 @@ class Solution:
9597
```java
9698
class Solution {
9799
public int countTriplets(int[] arr) {
98-
int n = arr.length;
99-
int[] pre = new int[n + 1];
100+
int ans = 0, n = arr.length;
100101
for (int i = 0; i < n; ++i) {
101-
pre[i + 1] = pre[i] ^ arr[i];
102-
}
103-
int ans = 0;
104-
for (int i = 0; i < n - 1; ++i) {
105-
for (int j = i + 1; j < n; ++j) {
106-
for (int k = j; k < n; ++k) {
107-
int a = pre[j] ^ pre[i];
108-
int b = pre[k + 1] ^ pre[j];
109-
if (a == b) {
110-
++ans;
111-
}
102+
int s = arr[i];
103+
for (int k = i + 1; k < n; ++k) {
104+
s ^= arr[k];
105+
if (s == 0) {
106+
ans += k - i;
112107
}
113108
}
114109
}
@@ -123,15 +118,13 @@ class Solution {
123118
class Solution {
124119
public:
125120
int countTriplets(vector<int>& arr) {
126-
int n = arr.size();
127-
vector<int> pre(n + 1);
128-
for (int i = 0; i < n; ++i) pre[i + 1] = pre[i] ^ arr[i];
129-
int ans = 0;
130-
for (int i = 0; i < n - 1; ++i) {
131-
for (int j = i + 1; j < n; ++j) {
132-
for (int k = j; k < n; ++k) {
133-
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
134-
if (a == b) ++ans;
121+
int ans = 0, n = arr.size();
122+
for (int i = 0; i < n; ++i) {
123+
int s = arr[i];
124+
for (int k = i + 1; k < n; ++k) {
125+
s ^= arr[k];
126+
if (s == 0) {
127+
ans += k - i;
135128
}
136129
}
137130
}
@@ -143,24 +136,59 @@ public:
143136
#### Go
144137
145138
```go
146-
func countTriplets(arr []int) int {
147-
n := len(arr)
148-
pre := make([]int, n+1)
149-
for i := 0; i < n; i++ {
150-
pre[i+1] = pre[i] ^ arr[i]
151-
}
152-
ans := 0
153-
for i := 0; i < n-1; i++ {
154-
for j := i + 1; j < n; j++ {
155-
for k := j; k < n; k++ {
156-
a, b := pre[j]^pre[i], pre[k+1]^pre[j]
157-
if a == b {
158-
ans++
159-
}
139+
func countTriplets(arr []int) (ans int) {
140+
for i, x := range arr {
141+
s := x
142+
for k := i + 1; k < len(arr); k++ {
143+
s ^= arr[k]
144+
if s == 0 {
145+
ans += k - i
160146
}
161147
}
162148
}
163-
return ans
149+
return
150+
}
151+
```
152+
153+
#### TypeScript
154+
155+
```ts
156+
function countTriplets(arr: number[]): number {
157+
const n = arr.length;
158+
let ans = 0;
159+
for (let i = 0; i < n; ++i) {
160+
let s = arr[i];
161+
for (let k = i + 1; k < n; ++k) {
162+
s ^= arr[k];
163+
if (s === 0) {
164+
ans += k - i;
165+
}
166+
}
167+
}
168+
return ans;
169+
}
170+
```
171+
172+
#### Rust
173+
174+
```rust
175+
impl Solution {
176+
pub fn count_triplets(arr: Vec<i32>) -> i32 {
177+
let mut ans = 0;
178+
let n = arr.len();
179+
180+
for i in 0..n {
181+
let mut s = arr[i];
182+
for k in (i + 1)..n {
183+
s ^= arr[k];
184+
if s == 0 {
185+
ans += (k - i) as i32;
186+
}
187+
}
188+
}
189+
190+
ans
191+
}
164192
}
165193
```
166194

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

+8-10
Original file line numberDiff line numberDiff line change
@@ -1,18 +1,16 @@
11
class Solution {
22
public:
33
int countTriplets(vector<int>& arr) {
4-
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-
for (int j = i + 1; j < n; ++j) {
10-
for (int k = j; k < n; ++k) {
11-
int a = pre[j] ^ pre[i], b = pre[k + 1] ^ pre[j];
12-
if (a == b) ++ans;
4+
int ans = 0, n = arr.size();
5+
for (int i = 0; i < n; ++i) {
6+
int s = arr[i];
7+
for (int k = i + 1; k < n; ++k) {
8+
s ^= arr[k];
9+
if (s == 0) {
10+
ans += k - i;
1311
}
1412
}
1513
}
1614
return ans;
1715
}
18-
};
16+
};

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

+9-16
Original file line numberDiff line numberDiff line change
@@ -1,19 +1,12 @@
1-
func countTriplets(arr []int) int {
2-
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-
}
1+
func countTriplets(arr []int) (ans int) {
2+
for i, x := range arr {
3+
s := x
4+
for k := i + 1; k < len(arr); k++ {
5+
s ^= arr[k]
6+
if s == 0 {
7+
ans += k - i
158
}
169
}
1710
}
18-
return ans
19-
}
11+
return
12+
}

0 commit comments

Comments
 (0)