README_EN.md

2366. Minimum Replacements to Sort the Array

中文文档

Description

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.

 

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

Solutions

Python3

class Solution:
    def minimumReplacement(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        mi = nums[-1]
        for i in range(n - 2, -1, -1):
            v = nums[i]
            if v <= mi:
                mi = v
                continue
            k = (v + mi - 1) // mi
            ans += k - 1
            mi = v // k
        return ans

Java

class Solution {
    public long minimumReplacement(int[] nums) {
        long ans = 0;
        int n = nums.length;
        int mi = nums[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            int v = nums[i];
            if (v <= mi) {
                mi = v;
                continue;
            }
            int k = (v + mi - 1) / mi;
            ans += k - 1;
            mi = v / k;
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long minimumReplacement(vector<int>& nums) {
        long long ans = 0;
        int n = nums.size();
        int mi = nums[n - 1];
        for (int i = n - 2; ~i; --i) {
            int v = nums[i];
            if (v <= mi) {
                mi = v;
                continue;
            }
            int k = (v + mi - 1) / mi;
            ans += k - 1;
            mi = v / k;
        }
        return ans;
    }
};

Go

func minimumReplacement(nums []int) int64 {
	var ans int64
	n := len(nums)
	mi := nums[n-1]
	for i := n - 2; i >= 0; i-- {
		v := nums[i]
		if v <= mi {
			mi = v
			continue
		}
		k := (v + mi - 1) / mi
		ans += int64(k - 1)
		mi = v / k
	}
	return ans
}

TypeScript

...