2176. Count Equal and Divisible Pairs in an Array || using nested loops
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int countPairs(vector<int>& nums, int k) {\n        int n = nums.size();\n        int count = 0 ;\n        for(int i = 0 ; i<n-1 ; i++)\n        {\n            for(int j = i+1 ; j<n ; j++)\n            {\n                if(nums[i]==nums[j] && (i*j)%k==0)\n                {\n                    count++;\n                }\n            }\n        }\n        return count;\n    }\n};\n```
['C++']