1846. Maximum Element After Decreasing and Rearranging

Problem

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 (Java)

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;
    }
}

Explain:

nope.

Complexity:

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