LeetCode #2968 — HARD

Apply Operations to Maximize Frequency Score

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed integer array nums and an integer k.

You can perform the following operation on the array at most k times:

Choose any index i from the array and increase or decrease nums[i] by 1.

The score of the final array is the frequency of the most frequent element in the array.

Return the maximum score you can achieve.

The frequency of an element is the number of occurences of that element in the array.

Example 1:

Input: nums = [1,2,6,4], k = 3
Output: 3
Explanation: We can do the following operations on the array:
- Choose i = 0, and increase the value of nums[0] by 1. The resulting array is [2,2,6,4].
- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,3].
- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,2].
The element 2 is the most frequent in the final array so our score is 3.
It can be shown that we cannot achieve a better score.

Example 2:

Input: nums = [1,4,4,2,4], k = 0
Output: 3
Explanation: We cannot apply any operations so our score will be the frequency of the most frequent element in the original array, which is 3.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
0 <= k <= 10¹⁴

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums and an integer k. You can perform the following operation on the array at most k times: Choose any index i from the array and increase or decrease nums[i] by 1. The score of the final array is the frequency of the most frequent element in the array. Return the maximum score you can achieve. The frequency of an element is the number of occurences of that element in the array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[1,2,6,4]
3

Example 2

[1,4,4,2,4]
0

Core Insight

What unlocks the optimal approach

If you sort the original array, it is optimal to apply the operations on one subarray such that all the elements of that subarray become equal.
You can use binary search to find the longest subarray where we can make the elements equal in at most <code>k</code> operations.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
class Solution {
    public int maxFrequencyScore(int[] nums, long k) {
        Arrays.sort(nums);
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 1; i <= n; i++) {
            s[i] = s[i - 1] + nums[i - 1];
        }
        int l = 0, r = n;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            boolean ok = false;

            for (int i = 0; i <= n - mid; i++) {
                int j = i + mid;
                int x = nums[(i + j) / 2];
                long left = ((i + j) / 2 - i) * (long) x - (s[(i + j) / 2] - s[i]);
                long right = (s[j] - s[(i + j) / 2]) - ((j - (i + j) / 2) * (long) x);
                if (left + right <= k) {
                    ok = true;
                    break;
                }
            }

            if (ok) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        return l;
    }
}

// Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
func maxFrequencyScore(nums []int, k int64) int {
	sort.Ints(nums)
	n := len(nums)
	s := make([]int64, n+1)
	for i := 1; i <= n; i++ {
		s[i] = s[i-1] + int64(nums[i-1])
	}

	l, r := 0, n
	for l < r {
		mid := (l + r + 1) >> 1
		ok := false

		for i := 0; i <= n-mid; i++ {
			j := i + mid
			x := int64(nums[(i+j)/2])
			left := (int64((i+j)/2-i) * x) - (s[(i+j)/2] - s[i])
			right := (s[j] - s[(i+j)/2]) - (int64(j-(i+j)/2) * x)

			if left+right <= k {
				ok = true
				break
			}
		}

		if ok {
			l = mid
		} else {
			r = mid - 1
		}
	}

	return l
}

# Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
class Solution:
    def maxFrequencyScore(self, nums: List[int], k: int) -> int:
        nums.sort()
        s = list(accumulate(nums, initial=0))
        n = len(nums)
        l, r = 0, n
        while l < r:
            mid = (l + r + 1) >> 1
            ok = False
            for i in range(n - mid + 1):
                j = i + mid
                x = nums[(i + j) // 2]
                left = ((i + j) // 2 - i) * x - (s[(i + j) // 2] - s[i])
                right = (s[j] - s[(i + j) // 2]) - ((j - (i + j) // 2) * x)
                if left + right <= k:
                    ok = True
                    break
            if ok:
                l = mid
            else:
                r = mid - 1
        return l

// Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
// class Solution {
//     public int maxFrequencyScore(int[] nums, long k) {
//         Arrays.sort(nums);
//         int n = nums.length;
//         long[] s = new long[n + 1];
//         for (int i = 1; i <= n; i++) {
//             s[i] = s[i - 1] + nums[i - 1];
//         }
//         int l = 0, r = n;
//         while (l < r) {
//             int mid = (l + r + 1) >> 1;
//             boolean ok = false;
// 
//             for (int i = 0; i <= n - mid; i++) {
//                 int j = i + mid;
//                 int x = nums[(i + j) / 2];
//                 long left = ((i + j) / 2 - i) * (long) x - (s[(i + j) / 2] - s[i]);
//                 long right = (s[j] - s[(i + j) / 2]) - ((j - (i + j) / 2) * (long) x);
//                 if (left + right <= k) {
//                     ok = true;
//                     break;
//                 }
//             }
// 
//             if (ok) {
//                 l = mid;
//             } else {
//                 r = mid - 1;
//             }
//         }
// 
//         return l;
//     }
// }

// Accepted solution for LeetCode #2968: Apply Operations to Maximize Frequency Score
function maxFrequencyScore(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; i++) {
        s[i] = s[i - 1] + nums[i - 1];
    }

    let l: number = 0;
    let r: number = n;
    while (l < r) {
        const mid: number = (l + r + 1) >> 1;
        let ok: boolean = false;
        for (let i = 0; i <= n - mid; i++) {
            const j = i + mid;
            const x = nums[Math.floor((i + j) / 2)];
            const left = (Math.floor((i + j) / 2) - i) * x - (s[Math.floor((i + j) / 2)] - s[i]);
            const right = s[j] - s[Math.floor((i + j) / 2)] - (j - Math.floor((i + j) / 2)) * x;
            if (left + right <= k) {
                ok = true;
                break;
            }
        }
        if (ok) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.