Frequency of the Most Frequent Element

Problem Description

You are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1. Return the maximum possible frequency of an element after performing at most k operations.

Key Insights

• The goal is to maximize the frequency of any element in the array.
• Incrementing smaller numbers to match larger ones is key to increasing the frequency.
• A sliding window approach can be used to efficiently calculate the number of operations required to increase elements to a certain target value within the window.
• Sorting the array helps in easily determining the target values for increments.

Space and Time Complexity

Time Complexity: O(n log n) due to sorting, where n is the length of the nums array.
Space Complexity: O(1) for the sliding window approach, excluding the input storage.

Solution

To solve the problem, we will use the following approach:

1. Sort the array: This allows us to consider increasing the smaller numbers to match larger numbers efficiently.
2. Sliding Window: We will maintain a window of numbers and compute the total operations needed to make all numbers in the window equal to the largest number in that window.
3. Calculate Operations: For each position in the sorted array, we calculate how many increments would be needed to make all previous numbers equal to the current number.
4. Check Feasibility: If the number of required increments exceeds k, we move the left boundary of the window to reduce the size until we meet the conditions.

This ensures that we maximize the frequency of the most frequent element after at most k operations.

Code Solutions