# 2176. Count Equal and Divisible Pairs in an Array

## Description

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i \* j) is divisible by k.

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.


Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i], k <= 100

## Solutions

• 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;
}
}
}
return ans;
}
}

• 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;
}
}
return ans;
}
};

• 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)
)


• 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 {
ans++
}
}
}
return ans
}

• 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++) {
if (nums[i] === nums[j] && (i * j) % k === 0) {
ans++;
}
}
}
return ans;
}


• 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 {
ans += 1;
}
}
}
ans
}
}