Solution: Frequency of the Most Frequent Element

Statement

You are given an integer array, nums, and an integer k, representing the maximum number of operations you can perform. In each operation, you may select any index in nums and increment its value by 1.

Your task is to determine the maximum possible frequency of a single element in the final array after performing at most k operations. You can choose to increase any elements in a way that results in one particular element appearing as often as possible (within k operations). For example, if nums = [2, 2, 3] and k = 4, you can increment the first and the second element, 2, once to match the third element, 3, achieving a maximum frequency of 3.

Return the highest frequency that can be achieved for any element in nums after at most k operations.

The frequency of an element is the number of times it appears in an array.

Constraints:

• $1 \leq$ nums.length $\leq 10^3$

• $1 \leq$ nums[i] $\leq 10^3$

• $1 \leq$ k $\leq 10^3$

Solution

The best way to solve this problem is to convert smaller numbers into the largest possible number while using the least number of operations. To achieve this, we use the sliding window technique, which maintains a group of numbers that can be adjusted to match a target value (the largest number in the window). The window size is increased or decreased dynamically based on the sum of its elements, allowing us to quickly determine whether we can increase all numbers in the window to match the target value within the allowed k operations. The largest window where this condition holds gives us the maximum possible frequency.

In this solution, we first sort nums so that we always transform smaller numbers first, which requires fewer operations than modifying larger ones. This makes it easier to maintain a valid window where all elements can be equal to the rightmost (largest) element.

For any particular window, we decide to expand it further or shrink it based on the following condition:

$window ~ size \times target ~value$ $\leq$ $window ~sum$ + $k$

This means that for a window to be valid, the total sum (if all elements were target) should not exceed the actual sum of elements in the current window plus k allowed operations.

• If the condition is true and the window is valid, we add the next element of nums to the current window.

• Otherwise, if the condition is false, we shrink the window.

This process continues until we find the largest valid window within which all numbers can be equal. The illustration below helps us understand the concept better.