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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 {
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).