✔️✔️✔️very easy - brute force - PYTHON || C++ || JAVA⚡⚡⚡beats 99%⚡⚡⚡
Code\nPython []\nclass Solution:\n    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n        count = 0\n        for i in range(len
# Code\n```Python []\nclass Solution:\n    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n        count = 0\n        for i in range(len(nums1)):\n            for j in range(len(nums2)):\n                if nums1[i] % (nums2[j] * k) == 0:\n                    count += 1\n        return count\n```\n```C++ []\nclass Solution {\npublic:\n    int numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {\n        int count = 0;\n        for (int i = 0; i < nums1.size(); i++){\n            for (int j = 0; j < nums2.size(); j++){\n                if (nums1[i] % (nums2[j] * k) == 0){\n                    count++;\n                }\n            }\n        }\n        return count;\n    }\n};\n```\n```Java []\nclass Solution {\n    public int numberOfPairs(int[] nums1, int[] nums2, int k) {\n        int count = 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                    count++;\n                }\n            }\n        }\n        return count;\n    }\n}\n```
['Array', 'C++', 'Java', 'Python3']