combination of binary search, prefix sums, and the inclusion-exclusion principle by coderkb
Intuition\nThe goal of this problem is to answer a series of queries, each asking for the largest GCD pair sum that is less than or equal to a given value. We n
# Intuition\nThe goal of this problem is to answer a series of queries, each asking for the largest GCD pair sum that is less than or equal to a given value. We need to first compute the possible GCD values for pairs of integers in the input array. Then, by accumulating these values, we can efficiently answer each query using binary search over the prefix sums.\n\n---\n\n# Approach\n1. **Frequency Count**: First, we calculate the frequency of each number in the `nums` array.\n  \n2. **GCD Calculation**: For each possible GCD, we use the inclusion-exclusion principle to count the number of pairs whose GCD is equal to that value. \n   - For each integer `i`, we find all pairs whose GCD is `i` by counting the frequency of multiples of `i` and subtracting pairs that would have a greater GCD.\n  \n3. **Prefix Sum Array**: We build a prefix sum array of GCD pairs so that we can answer each query quickly. This prefix sum keeps track of the cumulative number of GCD pairs for all possible values of the GCD.\n\n4. **Binary Search**: For each query, we use binary search to find the largest GCD that has a pair count less than or equal to the query value.\n\n---\n\n# Complexity\n- **Time Complexity**:\n  - **GCD Calculation**: $$O(n \\log n)$$, where `n` is the size of `nums`. The outer loop runs for each number from `1` to `mx` (maximum number in `nums`), and the inner loop runs for multiples of `i`.\n  - **Binary Search for Queries**: Each query is processed in $$O(\\log n)$$, and there are `q` queries, resulting in $$O(q \\log n)$$.\n  \n  **Overall Time Complexity**: $$O(n \\log n + q \\log n)$$\n\n- **Space Complexity**: \n  - We use space for the `freq`, `gcd`, and `prefixSum` arrays, each of size `mx + 1`. Therefore, the space complexity is $$O(n + mx)$$.\n\n---\n\n# Code with Comments\n\n```cpp\nclass Solution {\n    // Helper function to perform binary search on prefixSum array to find largest GCD.\n    int binarySearch(vector<long long>& arr, long long val, int mx) {\n        int low = 0, high = mx;\n        while(low < high) {\n            int mid = (low + high) / 2;\n            // If the value at mid is less than or equal to the query value, move right.\n            if(arr[mid] <= val) {\n                low = mid + 1;\n            } else {\n                high = mid;\n            }\n        }\n        return low;\n    }\n    \npublic:\n    vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {\n        int mx = 0;\n        // Find the maximum number in the nums array.\n        for(int num: nums) {\n            mx = max(mx, num);\n        }\n\n        // Create a frequency array to count occurrences of each number.\n        vector<int> freq(mx + 1, 0);\n        for(int num: nums) {\n            freq[num]++;\n        }\n\n        // Array to store the count of GCD pairs.\n        vector<long long> gcd(mx + 1);\n        for(int i = mx; i > 0; --i) {\n            long long curr = 0;\n            // Count how many pairs have GCD equal to \'i\'.\n            for(int j = i; j <= mx; j += i) {\n                curr += freq[j];\n            }\n            \n            // Calculate total pairs for current GCD.\n            gcd[i] = (curr * (curr - 1)) / 2;\n            \n            // Subtract pairs that would have a larger GCD.\n            for(int j = i + i; j <= mx; j += i) {\n                gcd[i] -= gcd[j];\n            }\n        }\n\n        // Create a prefix sum array to store cumulative GCD counts.\n        vector<long long> prefixSum(mx + 1);\n        for(int i = 1; i <= mx; ++i) {\n            prefixSum[i] = prefixSum[i - 1] + gcd[i];\n        }\n\n        // For each query, perform binary search to find the largest GCD.\n        vector<int> ans;\n        for(auto q: queries) {\n            ans.push_back(binarySearch(prefixSum, q, mx));\n        }\n        return ans;\n    }\n};\n```\n\nThis solution efficiently handles large inputs by leveraging mathematical principles like the inclusion-exclusion principle and binary search, resulting in an optimal solution.
['Math', 'Binary Search', 'Prefix Sum', 'C++']