Formatted question description: https://leetcode.ca/all/1846.html

# 1846. Maximum Element After Decreasing and Rearranging

Medium

## Description

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

• The value of the first element in arr must be 1.
• The absolute difference between any 2 adjacent elements must be less than or equal to 1. In other words, abs(arr[i] - arr[i - 1]) <= 1 for each i where 1 <= i < arr.length (0-indexed). abs(x) is the absolute value of x.

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

• Decrease the value of any element of arr to a smaller positive integer.
• Rearrange the elements of arr to be in any order.

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:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 10^9

## Solution

Sort arr and let arr = 1. For 1 <= i < arr.length, set arr[i] to the minimum of arr[i] and arr[i - 1] + 1. Finally, return arr[arr.length - 1].

class Solution {
public int maximumElementAfterDecrementingAndRearranging(int[] arr) {
Arrays.sort(arr);
int length = arr.length;
arr = 1;
for (int i = 1; i < length; i++)
arr[i] = Math.min(arr[i], arr[i - 1] + 1);
return arr[length - 1];
}
}