LeetCode #2366 — HARD

Minimum Replacements to Sort the Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Greedy

Example 1

[3,9,3]

Example 2

[1,2,3,4,5]

Core Insight

What unlocks the optimal approach

It is optimal to never make an operation to the last element of the array.
You can iterate from the second last element to the first. If the current value is greater than the previous bound, we want to break it into pieces so that the smaller one is as large as possible but not larger than the previous one.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2366: Minimum Replacements to Sort the Array
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;
    }
}

// Accepted solution for LeetCode #2366: Minimum Replacements to Sort the Array
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
}

# Accepted solution for LeetCode #2366: Minimum Replacements to Sort the Array
class Solution:
    def minimumReplacement(self, nums: List[int]) -> int:
        ans = 0
        n = len(nums)
        mx = nums[-1]
        for i in range(n - 2, -1, -1):
            if nums[i] <= mx:
                mx = nums[i]
                continue
            k = (nums[i] + mx - 1) // mx
            ans += k - 1
            mx = nums[i] // k
        return ans

// Accepted solution for LeetCode #2366: Minimum Replacements to Sort the Array
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
    }
}

// Accepted solution for LeetCode #2366: Minimum Replacements to Sort the Array
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.