Maximum Frequency of an Element After Performing Operations II

Problem Description

You are given an integer array nums and two integers k and numOperations. You must perform an operation numOperations times on nums, where in each operation you select an index i that was not selected in any previous operations and add an integer in the range [-k, k] to nums[i]. Return the maximum possible frequency of any element in nums after performing the operations.

Key Insights

• The goal is to maximize the frequency of the most common element in the array after a limited number of operations.
• Each operation allows modifying a unique element by a value within the range of [-k, k].
• After sorting the array, we can use a sliding window approach to efficiently calculate the maximum frequency of any element.
• The challenge lies in balancing the operations effectively among the elements.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the array.
Space Complexity: O(1) - we can achieve the solution without using extra space proportional to the input size.

Solution

To solve this problem, we can leverage a two-pointer (sliding window) technique after sorting the input array. By keeping track of the maximum frequency of a particular element, we can determine how many operations we need to perform to make other elements equal to it. The key steps are:

1. Sort the array to bring similar elements close together.
2. Use two pointers to create a window that represents elements we can potentially equalize to the current element being considered.
3. Calculate the required operations to make all elements in the window equal to the right pointer's element, and ensure it does not exceed numOperations.
4. Update the maximum frequency found during the iteration.

Code Solutions