# Maximize Element Frequency with Limited Increments Implement `max_frequency_after_increments(nums: list[int], k: int) -> int`. In one operation, choose an element of `nums` and increase it by `1`. You may perform at most `k` operations in total. Return the largest possible frequency of any value after the operations. ### Input Domain - `1 <= len(nums) <= 100,000`. - `1 <= nums[i] <= 100,000`. - `0 <= k <= 10^14`. - Values and intermediate operation counts fit in signed 64-bit integers. ### Output Rules - Return only the maximum attainable frequency. - Elements may be increased but never decreased. - If several target values attain the same maximum, the returned scalar is unchanged, so no tie-break is needed. ### Constraints - Target time complexity is `O(n log n)` or better. - Additional space may be `O(n)`. ### Examples #### Example 1 Input: `nums = [1,3,5,7,8,9,10,15], k = 6` Output: `4` Raising `7`, `8`, and `9` to `10` costs `3 + 2 + 1 = 6`, producing four copies of `10`. #### Example 2 Input: `nums = [1,3,5,10,10,10,10,15], k = 0` Output: `4` No increments are available, and `10` already appears four times. ```hint Equalize toward an existing value After ordering the values, consider a contiguous group raised to match its largest member and maintain the cost of that choice. ```