# Solving Leetcode Interviews in Seconds with AI: Count Equal and Divisible Pairs in an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2176" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate and Check:** The core approach involves iterating through all possible pairs `(i, j)` in the `nums` array where `i < j`.
*   **Condition Verification:** For each pair, we check if `nums[i] == nums[j]` and if `(i * j)` is divisible by `k`.
*   **Count Valid Pairs:** We increment a counter for each pair that satisfies both conditions.

*   **Runtime Complexity:** O(n^2), **Storage Complexity:** O(1)

	
	# Code
	```python
	def count_pairs(nums, k):
    """
    Counts the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

    Args:
        nums: A list of integers.
        k: An integer.

    Returns:
        The number of pairs that meet the requirements.
    """
    count = 0
    n = len(nums)
    for i in range(n):
        for j in range(i + 1, n):
            if nums[i] == nums[j] and (i * j) % k == 0:
                count += 1
    return count
	```