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103 lines
3.3 KiB
Markdown
103 lines
3.3 KiB
Markdown
# [1846. Maximum Element After Decreasing and Rearranging](https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/)
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## 题目
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You are given an array of positive integers `arr`. Perform some operations (possibly none) on `arr` so that it satisfies these conditions:
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- The value of the **first** element in `arr` must be `1`.
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- 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`.
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There are 2 types of operations that you can perform any number of times:
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- **Decrease** the value of any element of `arr` to a **smaller positive integer**.
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- **Rearrange** the elements of `arr` to be in any order.
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Return *the **maximum** possible value of an element in* `arr` *after performing the operations to satisfy the conditions*.
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**Example 1:**
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```
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Input: arr = [2,2,1,2,1]
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Output: 2
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Explanation:
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We can satisfy the conditions by rearrangingarr so it becomes[1,2,2,2,1].
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The largest element inarr is 2.
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```
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**Example 2:**
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```
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Input: arr = [100,1,1000]
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Output: 3
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Explanation:
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One possible way to satisfy the conditions is by doing the following:
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1. Rearrangearr so it becomes[1,100,1000].
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2. Decrease the value of the second element to 2.
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3. Decrease the value of the third element to 3.
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Nowarr = [1,2,3], whichsatisfies the conditions.
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The largest element inarr is 3.
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```
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**Example 3:**
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```
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Input: arr = [1,2,3,4,5]
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Output: 5
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Explanation: The array already satisfies the conditions, and the largest element is 5.
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```
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**Constraints:**
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- `1 <= arr.length <= 10^5`
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- `1 <= arr[i] <= 10^9`
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## 题目大意
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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：
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- arr 中 第一个 元素必须为 1 。
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- 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。
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你可以执行以下 2 种操作任意次：
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- 减小 arr 中任意元素的值，使其变为一个 更小的正整数 。
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- 重新排列 arr 中的元素，你可以以任意顺序重新排列。
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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 。
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## 解题思路
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- 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。
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## 代码
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```go
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package leetcode
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func maximumElementAfterDecrementingAndRearranging(arr []int) int {
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n := len(arr)
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count := make([]int, n+1)
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for _, v := range arr {
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count[min(v, n)]++
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}
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miss := 0
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for _, c := range count[1:] {
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if c == 0 {
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miss++
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} else {
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miss -= min(c-1, miss)
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}
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}
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return n - miss
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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``` |