2176. Count-Equal-And-Divisible-Pairs-In-An-Array

Very simple solution, we can brute force it. Iterate through all possible combinations and count the number of combinations that is valid.

class Solution(object):
    def countPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        res = 0
        for i in range(len(nums) - 1):
            for j in range(i + 1, len(nums)):
                if nums[i] == nums[j] and (i * j) % k == 0:
                    res += 1

        return res

Time Complexity: O(n^2)

Space Complexity: O(1)

Time Taken: 52s