2779. Maximum Beauty of an Array After Applying Operation

Problem

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

Solution (Java)

class Solution {
    public int maximumBeauty(int[] A, int k) {
        Arrays.sort(A);
        int i = 0, j, n = A.length;
        for (j = 0; j < n; ++j)
            if (A[j] - A[i] > k * 2)
                i++;
        return j - i;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).