LeetCode #2779 — MEDIUM

Maximum Beauty of an Array After Applying Operation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

In one operation, you can do the following:

Choose an index i that hasn't been chosen before from the range [0, nums.length - 1].
Replace nums[i] with any integer from the range [nums[i] - k, nums[i] + k].

The beauty of the array is the length of the longest subsequence consisting of equal elements.

Return the maximum possible beauty of the array nums after applying the operation any number of times.

Note that you can apply the operation to each index only once.

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the order of the remaining elements.

Example 1:

Input: nums = [4,6,1,2], k = 2
Output: 3
Explanation: In this example, we apply the following operations:
- Choose index 1, replace it with 4 (from range [4,8]), nums = [4,4,1,2].
- Choose index 3, replace it with 4 (from range [0,4]), nums = [4,4,1,4].
After the applied operations, the beauty of the array nums is 3 (subsequence consisting of indices 0, 1, and 3).
It can be proven that 3 is the maximum possible length we can achieve.

Example 2:

Input: nums = [1,1,1,1], k = 10
Output: 4
Explanation: In this example we don't have to apply any operations.
The beauty of the array nums is 4 (whole array).

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums and a non-negative integer k. In one operation, you can do the following: Choose an index i that hasn't been chosen before from the range [0, nums.length - 1]. Replace nums[i] with any integer from the range [nums[i] - k, nums[i] + k]. The beauty of the array is the length of the longest subsequence consisting of equal elements. Return the maximum possible beauty of the array nums after applying the operation any number of times. Note that you can apply the operation to each index only once. A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the order of the remaining elements.

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

[4,6,1,2]
2

Example 2

[1,1,1,1]
10

Core Insight

What unlocks the optimal approach

Sort the array.
The problem becomes the following: find maximum subarray A[i … j] such that A[j] - A[i] ≤ 2 * k.

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

Upper-end input sizes

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 #2779: Maximum Beauty of an Array After Applying Operation
class Solution {
    public int maximumBeauty(int[] nums, int k) {
        int m = Arrays.stream(nums).max().getAsInt() + k * 2 + 2;
        int[] d = new int[m];
        for (int x : nums) {
            d[x]++;
            d[x + k * 2 + 1]--;
        }
        int ans = 0, s = 0;
        for (int x : d) {
            s += x;
            ans = Math.max(ans, s);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2779: Maximum Beauty of an Array After Applying Operation
func maximumBeauty(nums []int, k int) (ans int) {
	m := slices.Max(nums)
	m += k*2 + 2
	d := make([]int, m)
	for _, x := range nums {
		d[x]++
		d[x+k*2+1]--
	}
	s := 0
	for _, x := range d {
		s += x
		if ans < s {
			ans = s
		}
	}
	return
}

# Accepted solution for LeetCode #2779: Maximum Beauty of an Array After Applying Operation
class Solution:
    def maximumBeauty(self, nums: List[int], k: int) -> int:
        m = max(nums) + k * 2 + 2
        d = [0] * m
        for x in nums:
            d[x] += 1
            d[x + k * 2 + 1] -= 1
        return max(accumulate(d))

// Accepted solution for LeetCode #2779: Maximum Beauty of an Array After Applying Operation
// 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 #2779: Maximum Beauty of an Array After Applying Operation
// class Solution {
//     public int maximumBeauty(int[] nums, int k) {
//         int m = Arrays.stream(nums).max().getAsInt() + k * 2 + 2;
//         int[] d = new int[m];
//         for (int x : nums) {
//             d[x]++;
//             d[x + k * 2 + 1]--;
//         }
//         int ans = 0, s = 0;
//         for (int x : d) {
//             s += x;
//             ans = Math.max(ans, s);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2779: Maximum Beauty of an Array After Applying Operation
function maximumBeauty(nums: number[], k: number): number {
    const m = Math.max(...nums) + k * 2 + 2;
    const d: number[] = Array(m).fill(0);
    for (const x of nums) {
        d[x]++;
        d[x + k * 2 + 1]--;
    }
    let ans = 0;
    let s = 0;
    for (const x of d) {
        s += x;
        ans = Math.max(ans, s);
    }
    return ans;
}

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(M + 2 × k + n)

Space

O(M + 2 × k)

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.