Beats 97.64% of people with python..
\n\n### Approach\n\n1. Sorting the Array: The first step is to sort the array nums. This helps in using a sliding window approach to find the maximum frequency 
![Annotation 2024-05-15 070057.png](https://assets.leetcode.com/users/images/638391ad-54be-4c1c-997f-38a6c4188afa_1715867086.411212.png)\n\n### Approach\n\n1. **Sorting the Array**: The first step is to sort the array `nums`. This helps in using a sliding window approach to find the maximum frequency of any element that can be achieved by performing up to `k` operations.\n\n2. **Sliding Window**: We maintain a sliding window `[i, j]` where `i` is the start index and `j` is the end index of the window.\n   - For each element `x` at index `j`, we calculate the number of operations needed to make all elements in the current window equal to the median element `nums[(i + j) // 2]`.\n   - `k -= x - nums[(i + j) // 2]` reduces the remaining operations by the cost of making `x` equal to the median.\n   - If the operations exceed `k` (`k < 0`), we shrink the window from the left by incrementing `i` and adjust `k` to reflect the reduced window size.\n\n3. **Return the Maximum Frequency**: The difference `j - i + 1` represents the size of the largest window where the elements can be made equal using at most `k` operations.\n\n### Explanation with Example\n\nConsider `nums = [1, 2, 4]` and `k = 5`.\n\n1. Sort the array: `nums = [1, 2, 4]`.\n2. Initialize `i = 0`, `k = 5`.\n3. Iterate through `nums`:\n   - For `j = 0`, `x = 1`: `k = 5 - (1 - 1) = 5`.\n   - For `j = 1`, `x = 2`: `k = 5 - (2 - 1) = 4`.\n   - For `j = 2`, `x = 4`: `k = 4 - (4 - 2) = 2`.\n4. The window `[1, 2, 4]` can be transformed into `[2, 2, 2]` using 3 operations. The maximum frequency is 3 (i.e., `3 - 0 + 1 = 3`).\n\n### Time Complexity\n\n- **Sorting**: `O(n log n)`, where `n` is the number of elements in `nums`.\n- **Sliding Window**: `O(n)`, since each element is processed at most twice (once when `j` is incremented, and potentially once when `i` is incremented).\n\nOverall, the time complexity is dominated by the sorting step, making it **`O(n log n)`**.\n\n### Space Complexity\n\n- The space complexity is **`O(1)`** excluding the input space since we are using a fixed amount of extra space for variables (`i`, `j`, and `k`). Sorting `nums` in place does not require additional space beyond the input array.\n\n# Code\n```\nclass Solution:\n    def maxFrequencyScore(self, nums: List[int], k: int) -> int:\n        nums.sort()\n        i = 0\n        for j, x in enumerate(nums):\n            k -= x - nums[(i + j) // 2]\n            if k < 0:\n                i += 1\n                k += nums[(i + j) // 2] - nums[i - 1]\n        return j - i + 1\n```\n![7be995b4-0cf4-4fa3-b48b-8086738ea4ba_1699897744.9062278.jpeg](https://assets.leetcode.com/users/images/860cf95f-99bf-41f7-803f-cd96462c1f8a_1715867013.7423081.jpeg)\n
['Python3']