LeetCode 2779: Maximum Beauty of an Array After Applying Operation Solution

Master LeetCode problem 2779 (Maximum Beauty of an Array After Applying Operation), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2779. Maximum Beauty of an Array After Applying Operation

Medium

Array Binary Search Sliding Window Sorting

Problem Explanation

Explanation:

To maximize the beauty of the array, we need to think about how we can increase the length of the longest subsequence consisting of equal elements. We can achieve this by making the elements equal to each other.

To do this, we can iterate through the array and for each element, we can adjust it to the number that will occur most frequently in the array after applying the operation. This means that we should adjust each element to be closer to the mode of the array.

We can achieve this by sorting the array and then iterating through it. For each element, we can calculate the frequency of that element in the array. We can then find the number that occurs most frequently in the array (the mode) and adjust the current element to be closer to this mode within the range [nums[i] - k, nums[i] + k].

By doing this for each element, we can maximize the beauty of the array.

Time complexity: O(nlogn) where n is the length of the array nums due to sorting.
Space complexity: O(n) for storing the frequency of elements.

Solution Code

import java.util.Arrays;

class Solution {
    public int maximumBeauty(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int[] freq = new int[100001];
        int res = 0;
        
        for (int i = 0; i < n; i++) {
            freq[nums[i]]++;
            if (freq[nums[i]] > 1) {
                res = Math.max(res, 2 * Math.min(freq[nums[i]], freq[nums[0]]));
            }
        }
        
        return res + k;
    }
}

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