easy solution in all languages with their runtime (ms) - guess which is the fastest(this solution)
go - 0ms\n- java - 1ms\n- c++ - 6ms\n- python - 24ms\n- python3 - 53ms\n- javasript - 74ms\n- dart - 413ms\n\n\npython3 []\nclass Solution:\n    def numberOfPai
- go - 0ms\n- java - 1ms\n- c++ - 6ms\n- python - 24ms\n- python3 - 53ms\n- javasript - 74ms\n- dart - 413ms\n\n\n```python3 []\nclass Solution:\n    def numberOfPairs(self, nums1: List[int], nums2: List[int], k: int) -> int:\n        count = 0\n        for i in nums1:\n            for j in nums2:\n                if i % (j * k) == 0:\n                    count += 1\n        return count\n```\n\n```javascript []\nvar numberOfPairs = function(nums1, nums2, k) {\n    let count = 0\n    for(let i = 0;i< nums1.length;i++){\n        for(let 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```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```go []\nfunc numberOfPairs(nums1 []int, nums2 []int, k int) int {\n    count := 0;\n    for i := 0 ; i<len(nums1) ;i++{\n        for j := 0 ;j<len(nums2);j++{\n            if nums1[i] % (nums2[j]*k)==0{\n                count ++;\n            }\n        }\n    }\n    return count;\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            if(nums1[i]%k!=0) continue;\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```\n```python []\nclass Solution(object):\n    def numberOfPairs(self, nums1, nums2, k):\n        """\n        :type nums1: List[int]\n        :type nums2: List[int]\n        :type k: int\n        :rtype: int\n        """\n        count = 0\n        for i in nums1:\n            for j in nums2:\n                if i % (j * k) == 0:\n                    count += 1\n        return count\n```\n```dart []\nclass Solution {\n  int numberOfPairs(List<int> nums1, List<int> nums2, int k) {\n    int count = 0;\n        for (int i = 0; i<nums1.length;i++){\n            if(nums1[i]%k!=0) continue;\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\n```\n\n
['Array', 'C', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'C#', 'Dart']