Formatted question description: https://leetcode.ca/all/1846.html
1846. Maximum Element After Decreasing and Rearranging
Level
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 be1
. - 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 eachi
where1 <= i < arr.length
(0-indexed).abs(x)
is the absolute value ofx
.
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:
- Rearrange arr so it becomes [1,100,1000].
- Decrease the value of the second element to 2.
- 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[0] = 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[0] = 1;
for (int i = 1; i < length; i++)
arr[i] = Math.min(arr[i], arr[i - 1] + 1);
return arr[length - 1];
}
}