Frequency of the Most Frequent Element

Frequency of the Most Frequent Element

What is this problem about?

The Frequency of the Most Frequent Element coding problem gives you an integer array nums and an integer k. In one operation, you can pick an element and increment it by 1. You are allowed at most k operations. Your goal is to find the maximum possible frequency of any single element in the array after performing these operations.

Why is this asked in interviews?

Companies like Amazon, Google, and Meta use this Sliding Window interview pattern problem to test your ability to optimize range queries. Because you can only increment numbers (not decrement), you want to group numbers that are already close to each other. It evaluates whether you can use sorting and a sliding window to check if a group of numbers can be equalized within the k budget.

Algorithmic pattern used

This problem is solved using Sorting and a Sliding Window.

1. Sort: Sort the array so that elements close in value are adjacent.
2. Slide: Use two pointers, left and right. The element at nums[right] is the target value we want to increment the others to match.
3. Cost Check: The total operations needed to make all elements in the window [left, right] equal to nums[right] is: (length_of_window * nums[right]) - sum_of_window.
4. Shrink: If this cost is greater than k, the window is too large. Shrink it by moving the left pointer forward and updating the sum.
5. Track the maximum window size.

Example explanation

nums = [1, 2, 4], $k = 5$

1. Sort: [1, 2, 4].
2. right = 0: target 1, window [1]. Cost: $1 imes 1 - 1 = 0 \le 5$. Max freq = 1.
3. right = 1: target 2, window [1, 2]. Cost: $2 imes 2 - (1+2) = 4 - 3 = 1 \le 5$. Max freq = 2.
4. right = 2: target 4, window [1, 2, 4]. Cost: $3 imes 4 - (1+2+4) = 12 - 7 = 5 \le 5$. Max freq = 3. Result: 3. All numbers can become 4.

Common mistakes candidates make

• Not Sorting: Trying to find subsets without sorting, making it impossible to use a sliding window.
• Iterating to find cost: Using a loop inside the sliding window to calculate the cost, leading to $O(N^2)$ time. You must use a running sum (prefix sum) to calculate the cost in $O(1)$.
• Integer Overflow: The sum_of_window and length * target calculations can exceed 32-bit integers. Use 64-bit integers (long) for these variables.

Interview preparation tip

The formula cost = (count * target) - sum_of_elements is a crucial pattern for any problem involving "making elements equal." Memorize it, as it instantly turns an $O(N)$ check into an $O(1)$ calculation inside your sliding window.