---
description: >-
  Author: @wingkwong |
  https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array/
---

# 2176 - Count Equal and Divisible Pairs in an Array (Easy)

## Problem Link

https://leetcode.com/problems/count-equal-and-divisible-pairs-in-an-array/

## Problem Statement

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`

## Approach 1: Brute Force

Just do what it says.

<SolutionAuthor name="@wingkwong"/>

```cpp
class Solution {
public:
    int countPairs(vector<int>& nums, int k) {
        int n = nums.size(), ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                ans += nums[i] == nums[j] && (i * j) % k == 0;
            }
        }
        return ans;
    }
};
```
