LeetCode-in-Java

1846. Maximum Element After Decreasing and Rearranging

Medium

You are given an array of positive integers arr. Perform some operations (possibly none) on arr so that it satisfies these conditions:

There are 2 types of operations that you can perform any number of times:

Return the maximum possible value of an element in arr after performing the operations to satisfy the conditions.

Example 1:

Input: arr = [2,2,1,2,1]

Output: 2

Explanation:

We can satisfy the conditions by rearranging arr so it becomes [1,2,2,2,1].

The largest element in arr is 2.

Example 2:

Input: arr = [100,1,1000]

Output: 3

Explanation:

One possible way to satisfy the conditions is by doing the following:

  1. Rearrange arr so it becomes [1,100,1000].

  2. Decrease the value of the second element to 2.

  3. Decrease the value of the third element to 3.

Now arr = [1,2,3], which satisfies the conditions.

The largest element in arr is 3.

Example 3:

Input: arr = [1,2,3,4,5]

Output: 5

Explanation: The array already satisfies the conditions, and the largest element is 5.

Constraints:

Solution

public class Solution {
    public int maximumElementAfterDecrementingAndRearranging(int[] arr) {
        int[] count = new int[arr.length + 1];
        for (int j : arr) {
            count[Math.min(j, arr.length)]++;
        }
        int ans = 1;
        for (int i = 1; i < count.length; i++) {
            ans = Math.min(i, ans + count[i]);
        }
        return ans;
    }
}