2366.Minimum Replacements to Sort the Array

tags: Hard Array Math Greedy

2366. 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 <= 10⁵
• 1 <= nums[i] <= 10⁹

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;
    }
}

JimAug 30, 2023

Reference

回到題目列表