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feat: update solutions to lc problem: No.2176 #3526

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68 changes: 30 additions & 38 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,11 @@ tags:

<!-- solution:start -->

### 方法一：暴力枚举
### 方法一：枚举

我们先在 $[0, n)$ 的范围内枚举下标 $j$，然后在 $[0, j)$ 的范围内枚举下标 $i$，统计满足 $\textit{nums}[i] = \textit{nums}[j]$ 且 $(i \times j) \bmod k = 0$ 的数对个数。

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

<!-- tabs:start -->

Expand All @@ -65,26 +69,22 @@ tags:
```python
class Solution:
def countPairs(self, nums: List[int], k: int) -> int:
n = len(nums)
return sum(
nums[i] == nums[j] and (i * j) % k == 0
for i in range(n)
for j in range(i + 1, n)
)
ans = 0
for j, y in enumerate(nums):
for i, x in enumerate(nums[:j]):
ans += int(x == y and i * j % k == 0)
return ans
```

#### Java

```java
class Solution {
public int countPairs(int[] nums, int k) {
int n = nums.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) {
++ans;
}
for (int j = 1; j < nums.length; ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0 ? 1 : 0;
}
}
return ans;
Expand All @@ -98,11 +98,10 @@ class Solution {
class Solution {
public:
int countPairs(vector<int>& nums, int k) {
int n = nums.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) ++ans;
for (int j = 1; j < nums.size(); ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0;
}
}
return ans;
Expand All @@ -113,30 +112,27 @@ public:
#### Go

```go
func countPairs(nums []int, k int) int {
n := len(nums)
ans := 0
for i, v := range nums {
for j := i + 1; j < n; j++ {
if v == nums[j] && (i*j)%k == 0 {
func countPairs(nums []int, k int) (ans int) {
for j, y := range nums {
for i, x := range nums[:j] {
if x == y && (i*j%k) == 0 {
ans++
}
}
}
return ans
return
}
```

#### TypeScript

```ts
function countPairs(nums: number[], k: number): number {
const n = nums.length;
let ans = 0;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
for (let j = 1; j < nums.length; ++j) {
for (let i = 0; i < j; ++i) {
if (nums[i] === nums[j] && (i * j) % k === 0) {
ans++;
++ans;
}
}
}
Expand All @@ -149,12 +145,10 @@ function countPairs(nums: number[], k: number): number {
```rust
impl Solution {
pub fn count_pairs(nums: Vec<i32>, k: i32) -> i32 {
let k = k as usize;
let n = nums.len();
let mut ans = 0;
for i in 0..n - 1 {
for j in i + 1..n {
if nums[i] == nums[j] && (i * j) % k == 0 {
for j in 1..nums.len() {
for (i, &x) in nums[..j].iter().enumerate() {
if x == nums[j] && (i * j) as i32 % k == 0 {
ans += 1;
}
}
Expand All @@ -169,11 +163,9 @@ impl Solution {
```c
int countPairs(int* nums, int numsSize, int k) {
int ans = 0;
for (int i = 0; i < numsSize - 1; i++) {
for (int j = i + 1; j < numsSize; j++) {
if (nums[i] == nums[j] && i * j % k == 0) {
ans++;
}
for (int j = 1; j < numsSize; ++j) {
for (int i = 0; i < j; ++i) {
ans += (nums[i] == nums[j] && (i * j % k) == 0);
}
}
return ans;
Expand Down
68 changes: 30 additions & 38 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,11 @@ There are 4 pairs that meet all the requirements:

<!-- solution:start -->

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

We first enumerate the index $j$ in the range $[0, n)$, and then enumerate the index $i$ in the range $[0, j)$. We count the number of pairs that satisfy $\textit{nums}[i] = \textit{nums}[j]$ and $(i \times j) \bmod k = 0$.

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

<!-- tabs:start -->

Expand All @@ -65,26 +69,22 @@ There are 4 pairs that meet all the requirements:
```python
class Solution:
def countPairs(self, nums: List[int], k: int) -> int:
n = len(nums)
return sum(
nums[i] == nums[j] and (i * j) % k == 0
for i in range(n)
for j in range(i + 1, n)
)
ans = 0
for j, y in enumerate(nums):
for i, x in enumerate(nums[:j]):
ans += int(x == y and i * j % k == 0)
return ans
```

#### Java

```java
class Solution {
public int countPairs(int[] nums, int k) {
int n = nums.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) {
++ans;
}
for (int j = 1; j < nums.length; ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0 ? 1 : 0;
}
}
return ans;
Expand All @@ -98,11 +98,10 @@ class Solution {
class Solution {
public:
int countPairs(vector<int>& nums, int k) {
int n = nums.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) ++ans;
for (int j = 1; j < nums.size(); ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0;
}
}
return ans;
Expand All @@ -113,30 +112,27 @@ public:
#### Go

```go
func countPairs(nums []int, k int) int {
n := len(nums)
ans := 0
for i, v := range nums {
for j := i + 1; j < n; j++ {
if v == nums[j] && (i*j)%k == 0 {
func countPairs(nums []int, k int) (ans int) {
for j, y := range nums {
for i, x := range nums[:j] {
if x == y && (i*j%k) == 0 {
ans++
}
}
}
return ans
return
}
```

#### TypeScript

```ts
function countPairs(nums: number[], k: number): number {
const n = nums.length;
let ans = 0;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
for (let j = 1; j < nums.length; ++j) {
for (let i = 0; i < j; ++i) {
if (nums[i] === nums[j] && (i * j) % k === 0) {
ans++;
++ans;
}
}
}
Expand All @@ -149,12 +145,10 @@ function countPairs(nums: number[], k: number): number {
```rust
impl Solution {
pub fn count_pairs(nums: Vec<i32>, k: i32) -> i32 {
let k = k as usize;
let n = nums.len();
let mut ans = 0;
for i in 0..n - 1 {
for j in i + 1..n {
if nums[i] == nums[j] && (i * j) % k == 0 {
for j in 1..nums.len() {
for (i, &x) in nums[..j].iter().enumerate() {
if x == nums[j] && (i * j) as i32 % k == 0 {
ans += 1;
}
}
Expand All @@ -169,11 +163,9 @@ impl Solution {
```c
int countPairs(int* nums, int numsSize, int k) {
int ans = 0;
for (int i = 0; i < numsSize - 1; i++) {
for (int j = i + 1; j < numsSize; j++) {
if (nums[i] == nums[j] && i * j % k == 0) {
ans++;
}
for (int j = 1; j < numsSize; ++j) {
for (int i = 0; i < j; ++i) {
ans += (nums[i] == nums[j] && (i * j % k) == 0);
}
}
return ans;
Expand Down
10 changes: 4 additions & 6 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.c
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Original file line number Diff line number Diff line change
@@ -1,11 +1,9 @@
int countPairs(int* nums, int numsSize, int k) {
int ans = 0;
for (int i = 0; i < numsSize - 1; i++) {
for (int j = i + 1; j < numsSize; j++) {
if (nums[i] == nums[j] && i * j % k == 0) {
ans++;
}
for (int j = 1; j < numsSize; ++j) {
for (int i = 0; i < j; ++i) {
ans += (nums[i] == nums[j] && (i * j % k) == 0);
}
}
return ans;
}
}
9 changes: 4 additions & 5 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,13 +1,12 @@
class Solution {
public:
int countPairs(vector<int>& nums, int k) {
int n = nums.size();
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) ++ans;
for (int j = 1; j < nums.size(); ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0;
}
}
return ans;
}
};
};
14 changes: 6 additions & 8 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,12 +1,10 @@
func countPairs(nums []int, k int) int {
n := len(nums)
ans := 0
for i, v := range nums {
for j := i + 1; j < n; j++ {
if v == nums[j] && (i*j)%k == 0 {
func countPairs(nums []int, k int) (ans int) {
for j, y := range nums {
for i, x := range nums[:j] {
if x == y && (i*j%k) == 0 {
ans++
}
}
}
return ans
}
return
}
11 changes: 4 additions & 7 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,14 +1,11 @@
class Solution {
public int countPairs(int[] nums, int k) {
int n = nums.length;
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (nums[i] == nums[j] && (i * j) % k == 0) {
++ans;
}
for (int j = 1; j < nums.length; ++j) {
for (int i = 0; i < j; ++i) {
ans += nums[i] == nums[j] && (i * j % k) == 0 ? 1 : 0;
}
}
return ans;
}
}
}
11 changes: 5 additions & 6 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,8 +1,7 @@
class Solution:
def countPairs(self, nums: List[int], k: int) -> int:
n = len(nums)
return sum(
nums[i] == nums[j] and (i * j) % k == 0
for i in range(n)
for j in range(i + 1, n)
)
ans = 0
for j in range(1, len(nums)):
for i, x in enumerate(nums[:j]):
ans += int(x == nums[j] and i * j % k == 0)
return ans
8 changes: 3 additions & 5 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,11 +1,9 @@
impl Solution {
pub fn count_pairs(nums: Vec<i32>, k: i32) -> i32 {
let k = k as usize;
let n = nums.len();
let mut ans = 0;
for i in 0..n - 1 {
for j in i + 1..n {
if nums[i] == nums[j] && (i * j) % k == 0 {
for j in 1..nums.len() {
for (i, &x) in nums[..j].iter().enumerate() {
if x == nums[j] && (i * j) as i32 % k == 0 {
ans += 1;
}
}
Expand Down
7 changes: 3 additions & 4 deletions solution/2100-2199/2176.Count Equal and Divisible Pairs in an Array/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,10 +1,9 @@
function countPairs(nums: number[], k: number): number {
const n = nums.length;
let ans = 0;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
for (let j = 1; j < nums.length; ++j) {
for (let i = 0; i < j; ++i) {
if (nums[i] === nums[j] && (i * j) % k === 0) {
ans++;
++ans;
}
}
}
Expand Down
6 changes: 5 additions & 1 deletion ...99/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md
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Expand Up @@ -52,7 +52,11 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Mathematics

Assume the three consecutive integers are $x-1$, $x$, and $x+1$. Their sum is $3x$, so $num$ must be a multiple of $3$. If $num$ is not a multiple of $3$, it cannot be expressed as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{num}{3}$, then $x-1$, $x$, and $x+1$ are the three consecutive integers whose sum is $num$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down
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