C++ Multiple approaches
To address the problem of counting pairs of integers (nums1[i], nums2[j]) where nums1[i] is divisible by nums2[j]*k, we explore two distinct approaches: a brute
To address the problem of counting pairs of integers `(nums1[i], nums2[j])` where `nums1[i]` is divisible by `nums2[j]*k`, we explore two distinct approaches: a brute force method and an optimized strategy employing sorting and two pointers.\n\n# Brute Force Approach\n\n## Algorithm:\n1. **Initialize Counter**: Set `count` to 0.\n2. **Nested Loops**: Iterate through each element of `nums1` and `nums2` using nested loops.\n3. **Pair Validation**: For each pair `(nums1[i], nums2[j])`, increment `count` if `nums1[i]` is divisible by `nums2[j]*k`.\n4. **Return Count**: After all iterations, return `count`.\n\n## Complexity Analysis:\n- **Time Complexity**: O(n * m), where n and m are the sizes of `nums1` and `nums2` respectively, due to nested loops.\n- **Space Complexity**: O(1), minimal extra space apart from a few variables.\n\n## Code Implementation:\n```cpp\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) ++count;\n        return count;\n    }\n};\n```\n\n# Optimized Approach\n\n## Algorithm:\n1. **Sort Arrays**: Sort `nums1` and `nums2`.\n2. **Two Pointers**: Initialize pointers `i` and `j` to iterate over `nums1` and `nums2`.\n3. **Pair Validation**: While `i` and `j` are within bounds, compare `nums1[i]` with `nums2[j]*k`:\n   - If equal, increment `count` and move both pointers forward.\n   - If `nums1[i]` is less than `nums2[j]*k`, increment `i`.\n   - If greater, increment `j`.\n4. **Return Count**: Return `count`.\n\n## Complexity Analysis:\n- **Time Complexity**: O(n*log(n) + m*log(m) + n + m) due to sorting and traversal with two pointers.\n- **Space Complexity**: O(1), as sorting is in-place and minimal extra space is used.\n\n## Code Implementation:\n```cpp\n#include <vector>\n#include <algorithm>\n\nclass Solution {\npublic:\n    int numberOfPairs(vector<int>& nums1, vector<int>& nums2, int k) {\n        sort(nums1.begin(), nums1.end());\n        sort(nums2.begin(), nums2.end());\n        \n        int count = 0, i = 0, j = 0;\n        while (i < nums1.size() && j < nums2.size()) {\n            long long product = (long long)nums2[j] * k;\n            if (nums1[i] == product) {\n                ++count; ++i; ++j;\n            } else if (nums1[i] < product) ++i;\n            else ++j;\n        }\n        return count;\n    }\n};\n```\n\n# Conclusion\n\nWhile the brute force approach offers a direct solution, its quadratic time complexity makes it less efficient for larger inputs. Conversely, the optimized method with sorting and two pointers provides a more efficient solution, significantly reducing time complexity. By understanding problem constraints and employing appropriate algorithms and data structures, we can effectively solve such counting problems.\n
['Divide and Conquer', 'Ordered Map', 'C++']