1846. Maximum Element After Decreasing and Rearranging

1846. Maximum Element After Decreasing and Rearranging #

题目 #

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 rearrangingarr so it becomes[1,2,2,2,1].
The largest element inarr 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. Rearrangearr 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.
Nowarr = [1,2,3], whichsatisfies the conditions.
The largest element inarr 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

题目大意 #

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 。

解题思路 #

  • 正整数数组 arr 第一个元素必须为 1,且两两元素绝对值小于等于 1,那么 arr 最大值肯定不大于 n。采用贪心的策略,先统计所有元素出现的次数,大于 n 的元素出现次数都累加到 n 上。然后从 1 扫描到 n,遇到“空隙”(出现次数为 0 的元素),便将最近一个出现次数大于 1 的元素“挪”过来填补“空隙”。题目所求最大值出现在,“填补空隙”之后,数组从左往右连续的最右端。

代码 #

package leetcode

func maximumElementAfterDecrementingAndRearranging(arr []int) int {
	n := len(arr)
	count := make([]int, n+1)
	for _, v := range arr {
		count[min(v, n)]++
	}
	miss := 0
	for _, c := range count[1:] {
		if c == 0 {
			miss++
		} else {
			miss -= min(c-1, miss)
		}
	}
	return n - miss
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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