The Maximum Beauty of an Array After Applying Operation interview question focuses on finding the maximum number of identical elements you can achieve in an array after modifying each element once. You are given an array of integers and a non-negative integer k. For each element in the array, you can choose any value within the range [element - k, element + k]. The "beauty" of the array is defined as the maximum frequency of any value in the array after these modifications.
Essentially, this is a range overlap problem. Each number x in the original array represents an interval [x - k, x + k]. The goal is to find a single point on the number line that is covered by the maximum number of these intervals.
This coding problem is a favorite among companies like Amazon and Google because it tests a candidate's ability to transform a seemingly complex modification problem into a classic interval or windowing problem. It evaluates:
x by k is equivalent to saying x can represent any value in a specific range?The most effective interview pattern for this problem involves Sorting combined with a Sliding Window or Two Pointers approach.
By sorting the array, we can represent the problem as finding the longest contiguous subarray where the difference between the maximum and minimum elements is at most 2k. Why 2k? Because if two numbers a and b (where a < b) can both be changed to the same value v, then a + k >= v and b - k <= v. This implies b - a <= 2k.
Imagine an array [4, 6, 1, 2] and k = 2.
[1, 2, 4, 6].2 * k = 4.[1]. Diff = 0 (<= 4). Length = 1.[1, 2]. Diff = 1 (<= 4). Length = 2.[1, 2, 4]. Diff = 3 (<= 4). Length = 3.[1, 2, 4, 6]. Diff = 5 (> 4). Too wide![2, 4, 6]. Diff = 4 (<= 4). Length = 3.
The maximum beauty is 3.[min(arr)-k, max(arr)+k] is highly inefficient if the range of numbers is large.k instead of 2k as the threshold for the sliding window.When you see a problem asking for the "maximum number of elements" that can satisfy a condition involving a "range" or "buffer" (k), always think about how those ranges overlap. Visualizing these as intervals on a 1D line often reveals that sorting and two-pointers will provide the most efficient solution.