2366.Minimum Replacements to Sort the Array
===

###### tags: `Hard` `Array` `Math` `Greedy`

[2366. Minimum Replacements to Sort the Array](https://leetcode.com/problems/minimum-replacements-to-sort-the-array/)

### 題目描述

You are given a **0-indexed** integer array `nums`. In one operation you can replace any element of the array with **any two** elements that **sum** to it.

* For example, consider `nums = [5,6,7]`. In one operation, we can replace `nums[1]` with `2` and `4` and convert `nums` to `[5,2,4,7]`.

Return *the minimum number of operations to make an array that is sorted in **non-decreasing** order.*

### 範例

**Example 1:**

Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.

**Example 2:**

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0.

**Constraints**:

* 1 <= `nums.length` <= 10^5^
* 1 <= `nums[i]` <= 10^9^

### 解答

#### C#
從最後一個數字往前找，如果比上一個數字大代表需要被分割，
- 最少分割次數就用 (目前數字 - 1)/上一個數字
- 分割後，排在前面的數字要盡量大，所以平均分配，用 目前數字/分割區塊
- 分割後，排在前面的數字變成下一輪的上一個數字

public class Solution {
    public long MinimumReplacement(int[] nums) {
        if (nums.Length == 1) return 0;
        
        int last = nums.Last();
        long splitCount = 0;

        for (int i = nums.Length - 2; i >= 0; i--) {
            if (last == 1) {
                splitCount += nums[i] - 1;
                continue;
            }

            if (nums[i] <= last) {
                last = nums[i];
                continue;
            }

            int splitTimes = (nums[i] - 1) / last;
            splitCount += splitTimes;      
            last = nums[i] / (splitTimes + 1);
        }

        return splitCount;
    }
}

>[name=Jim][time=Aug 30, 2023]
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