Simplest Brute Force Solution Java || Count Number of Pairs
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 {\n    public int numberOfPairs(int[] nums1, int[] nums2, int k) {\n     int goodPairs = 0;\n        for (int i = 0; i < nums1.length; i++) {\n            for (int j = 0; j < nums2.length; j++) {\n                if (nums1[i] % (nums2[j] * k) == 0) {\n                    goodPairs++;\n                }\n            }\n        }\n        return goodPairs;   \n    }\n}\n```
['Java']