leetcode - 2366. Minimum Replacements to Sort the Array

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 <= 10^5
1 <= nums[i] <= 10^9

Solution

Greedy, every time, split the current number to be less or equal to the number at the right. In order to calculate the number of the operation, use nums[i] to divide nums[i + 1], then the number of operation is the result of the division - 1.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def minimumReplacement(self, nums: List[int]) -> int:
        res = 0
        for i in range(len(nums) - 2, -1, -1):
            if nums[i] > nums[i + 1]:
                num_of_diviors = (nums[i] // nums[i + 1]) + (1 if nums[i] % nums[i + 1] else 0)
                nums[i] //= num_of_diviors
                res += num_of_diviors - 1
        return res