[C++] Binary Search over frequency and validate each window of the sorted values.
Intuition\nWhen first encountering this problem, my instinct was to try sorting and binary search. Given a sorted array, it\'s often efficient to use a binary s
## Intuition\nWhen first encountering this problem, my instinct was to try sorting and binary search. Given a sorted array, it\'s often efficient to use a binary search to quickly hone in on a specific condition or target. In this case, I wanted to find the maximum window length within which the sum of the elements could be made equal by incrementing them, subject to a maximum total increment (`k`). The challenge was to do this in an optimized way.\n\n## Approach\nThe solution hinges on sorting the array and then applying a binary search to find the maximum length of a subarray (or window) that satisfies our condition. To efficiently calculate the cost of making all elements in a window equal, I used a prefix sum array. This array allows for quick calculations of the sum of elements in any given range, which is crucial for determining whether a specific window satisfies our cost constraint.\n\nHere\'s the step-by-step breakdown:\n1. **Sort the Array:** Start by sorting the array to arrange the elements in non-decreasing order. This is crucial for the sliding window technique to work effectively.\n2. **Create a Prefix Sum Table:** A prefix sum table is generated to quickly compute the sum of elements in any subarray. This table significantly reduces the time complexity of sum calculations within the binary search loop.\n3. **Binary Search for Maximum Window Length:** Implement a binary search to find the longest window such that the cost to make all its elements equal does not exceed `k`. For each potential window length, the code iteratively checks different windows of that length to see if any satisfy the cost constraint.\n4. **Cost Calculation Using Median:** For each window, the code calculates the cost of making all elements equal to the median of the window. This involves summing the differences between the median and each element in the window, which is efficiently done using the prefix sum table.\n\n## Complexity\n- **Time Complexity:** The time complexity of this solution is dominated by the sorting step and the binary search. Sorting takes $$O(n \\log n)$$ time, and the binary search takes $$O(n \\log n)$$ time, as it involves checking each potential window length against all possible windows of that length. Thus, the overall time complexity is $$O(n \\log n)$$.\n- **Space Complexity:** The space complexity is primarily due to the prefix sum table, which has a size proportional to the input array, resulting in $$O(n)$$ space complexity.\n\nThis approach effectively balances the need for quick calculations (through the prefix sum array) with the efficiency of binary search, making it a powerful solution for problems involving optimization of window-based calculations in sorted arrays.\n\n# Code\n```\nclass Solution {\n    // Using a custom type for integers to handle larger values\n    using Integer = long long;\n\npublic:\n    // Function to calculate the maximum frequency score of a vector of integers\n    int maxFrequencyScore(vector<int>& nums, long long k) {\n        // Sort the input array to enable sliding window technique\n        std::sort(nums.begin(), nums.end());\n\n        // Create a prefix sum table for efficient sum calculations in a range\n        std::vector<Integer> tbl_prefix_sum(nums.size() + 1, 0);\n        for (int i = 0; i < nums.size(); ++i) {\n            tbl_prefix_sum[i + 1] = nums[i] + tbl_prefix_sum[i];\n        }\n\n        // Initialize variables for the maximum allowed cost and binary search\n        const Integer max_cost = k;\n        int lower = 1, upper = nums.size() + 1;\n\n        // Perform binary search to find the maximum window length\n        while (lower < upper) {\n            const int window_len = lower + (upper - lower) / 2;\n            bool sat = false;\n\n            // Check if any window of \'window_len\' satisfies the condition\n            for (int i = 0, j = window_len - 1; j < nums.size(); ++i, ++j) {\n                const int med = (i + j) / 2;\n\n                // Calculate the sum of elements less than and greater than the median\n                const auto lesser_sum = tbl_prefix_sum[med + 1] - tbl_prefix_sum[i];\n                const auto greater_sum = tbl_prefix_sum[j + 1] - tbl_prefix_sum[med + 1];\n                Integer med_val = nums[med];\n\n                // Calculate the cost to make all elements equal to the median\n                const auto cost = 0\n                    + (med + 1 - i) * med_val \n                    - lesser_sum\n                    + greater_sum \n                    - (j + 1 - med - 1) * med_val;\n\n                // Check if the cost is within the allowed limit\n                if (cost <= max_cost) {\n                    sat = true;\n                    break;\n                }\n            }\n\n            // Adjust the search range based on whether the condition is satisfied\n            if (sat) {\n                lower = window_len + 1;\n            } else {\n                upper = window_len;\n            }\n        }\n\n        // Return the maximum window length that satisfies the condition\n        return upper - 1;\n    }\n};\n\n```
['C++']