[LeetCode] 1846. Maximum Element After Decreasing and Rearranging

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 <= 105
  • 1 <= arr[i] <= 109

减小和重新排列数组后的最大元素。

给你一个正整数数组 arr 。请你对 arr 执行一些操作(也可以不进行任何操作),使得数组满足以下条件:

arr 中 第一个 元素必须为 1 。
任意相邻两个元素的差的绝对值 小于等于 1 ,也就是说,对于任意的 1 <= i < arr.length (数组下标从 0 开始),都满足 abs(arr[i] - arr[i - 1]) <= 1 。abs(x) 为 x 的绝对值。
你可以执行以下 2 种操作任意次:

减小 arr 中任意元素的值,使其变为一个 更小的正整数 。
重新排列 arr 中的元素,你可以以任意顺序重新排列。
请你返回执行以上操作后,在满足前文所述的条件下,arr 中可能的 最大值 。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/maximum-element-after-decreasing-and-rearranging
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思路是贪心 + 排序。首先,排序了才能起码保证数组是非递减或者起码不是乱序的,同时只有排序了才能知道第一个元素是否是 1。从第二个元素开始,如果他与之前一个元素的差的绝对值大于 1 了,那么我们就试图改动那个较大的元素。最后返回数组的最后一个元素即可。

时间O(nlogn)

空间O(1)

Java实现

 1 class Solution {
 2     public int maximumElementAfterDecrementingAndRearranging(int[] arr) {
 3         Arrays.sort(arr);
 4         if (arr[0] != 1) {
 5             arr[0] = 1;
 6         }
 7         for (int i = 1; i < arr.length; i++) {
 8             if (Math.abs(arr[i] - arr[i - 1]) > 1) {
 9                 arr[i] = arr[i - 1] + 1;
10             }
11         }
12         return arr[arr.length - 1];
13     }
14 }

 

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