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Formatted question description: https://leetcode.ca/all/2366.html

2366. Minimum Replacements to Sort the Array

  • Difficulty: Hard.
  • Related Topics: Array, Math, Greedy.
  • Similar Questions: Minimum Operations to Make the Array Increasing.

Problem

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

Solution

  • class Solution {
        public long minimumReplacement(int[] nums) {
            int limit = nums[nums.length - 1];
            long ans = 0;
            for (int i = nums.length - 2; i >= 0; i--) {
                int replacements = nums[i] / limit - 1;
                if (nums[i] % limit != 0) {
                    replacements++;
                }
                ans += replacements;
                limit = nums[i] / (replacements + 1);
            }
            return ans;
        }
    }
    
    ############
    
    class Solution {
        public long minimumReplacement(int[] nums) {
            long ans = 0;
            int n = nums.length;
            int mx = nums[n - 1];
            for (int i = n - 2; i >= 0; --i) {
                if (nums[i] <= mx) {
                    mx = nums[i];
                    continue;
                }
                int k = (nums[i] + mx - 1) / mx;
                ans += k - 1;
                mx = nums[i] / k;
            }
            return ans;
        }
    }
    
  • 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
    
    ############
    
    # 2366. Minimum Replacements to Sort the Array
    # https://leetcode.com/problems/minimum-replacements-to-sort-the-array
    
    class Solution:
        def minimumReplacement(self, nums: List[int]) -> int:
            n = len(nums)
            res = 0
            prev = nums[-1]
            
            for i in range(n - 2, -1, -1):
                if nums[i] > prev:
                    sections = ceil(nums[i] / prev)
                    res += sections - 1
                    prev = nums[i] // sections
                
                else:
                    prev = nums[i]
                
            return res
    
    
  • class Solution {
    public:
        long long minimumReplacement(vector<int>& nums) {
            long long ans = 0;
            int n = nums.size();
            int mx = nums[n - 1];
            for (int i = n - 2; i >= 0; --i) {
                if (nums[i] <= mx) {
                    mx = nums[i];
                    continue;
                }
                int k = (nums[i] + mx - 1) / mx;
                ans += k - 1;
                mx = nums[i] / k;
            }
            return ans;
        }
    };
    
  • func minimumReplacement(nums []int) (ans int64) {
    	n := len(nums)
    	mx := nums[n-1]
    	for i := n - 2; i >= 0; i-- {
    		if nums[i] <= mx {
    			mx = nums[i]
    			continue
    		}
    		k := (nums[i] + mx - 1) / mx
    		ans += int64(k - 1)
    		mx = nums[i] / k
    	}
    	return
    }
    
  • function minimumReplacement(nums: number[]): number {
        const n = nums.length;
        let mx = nums[n - 1];
        let ans = 0;
        for (let i = n - 2; i >= 0; --i) {
            if (nums[i] <= mx) {
                mx = nums[i];
                continue;
            }
            const k = Math.ceil(nums[i] / mx);
            ans += k - 1;
            mx = Math.floor(nums[i] / k);
        }
        return ans;
    }
    
    
  • impl Solution {
        #[allow(dead_code)]
        pub fn minimum_replacement(nums: Vec<i32>) -> i64 {
            if nums.len() == 1 {
                return 0;
            }
    
            let n = nums.len();
            let mut max = *nums.last().unwrap();
            let mut ret = 0;
    
            for i in (0..=n - 2).rev() {
                if nums[i] <= max {
                    max = nums[i];
                    continue;
                }
                // Otherwise make the substitution
                let k = (nums[i] + max - 1) / max;
                ret += (k - 1) as i64;
                // Update the max value, which should be the minimum among the substitution
                max = nums[i] / k;
            }
    
            ret
        }
    }
    

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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