#2366

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Replacements to Sort the Array

Minimize the number of operations to make the array sorted in non-decreasing order by replacing elements with sums of two values.

array math greedy

Problem Statement

You are given a 0-indexed integer array nums. In one operation, you can replace any element of the array with two elements that sum to the original value. The goal is to perform the minimum number of operations to make the array sorted in non-decreasing order.

Return the minimum number of operations required to sort the array. A non-decreasing order means that each element is less than or equal to the next element.

Examples

Example 1

Input: nums = [3,9,3]

Output: 2

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

The array is already in non-decreasing order. Therefore, we return 0.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109

Solution Approach

Greedy Strategy

Start from the end of the array and work backward. At each step, choose the largest possible split of a number that allows it to fit in the desired order while minimizing the number of operations. This greedy approach ensures that no unnecessary splits are performed.

Invariant Validation

Maintain the invariant that, at each step, the elements from right to left are progressively less than or equal to the element before them. This helps in determining where replacements can be made efficiently.

Avoid Operations on Last Element

A key observation is that no operation should be performed on the last element of the array unless absolutely necessary, as it already marks the final sorted position.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the specific approach, but typically, this algorithm runs in linear time, O(n), where n is the length of the array. The space complexity is O(1) for the in-place solution, though extra space may be required for storing the results of splits if necessary.

What Interviewers Usually Probe

Look for the candidate's understanding of greedy algorithms and how they manage state during iteration.
Pay attention to how the candidate avoids unnecessary operations, especially regarding the last element.
Check whether the candidate correctly validates the invariant while implementing their solution.

Common Pitfalls or Variants

Common pitfalls

Not considering edge cases, like already sorted arrays or arrays where no replacements are needed.
Performing operations on the last element when it is unnecessary, leading to extra work.
Failing to manage the array state correctly during the greedy approach, causing invalid splits or incorrect results.

Follow-up variants

Allowing more than two elements in a replacement, which changes the greedy strategy.
Limiting the number of replacements allowed, requiring an optimization to minimize operations further.
Introducing additional constraints, like maximum number of total elements, to affect the problem's scalability.

FAQ

What is the main strategy for solving Minimum Replacements to Sort the Array?

The main strategy is to use a greedy approach, starting from the end of the array and performing the minimal number of operations to maintain the sorted order.

How do I avoid unnecessary operations in this problem?

By avoiding operations on the last element and ensuring each replacement maintains the non-decreasing order, you minimize unnecessary operations.

How does GhostInterview help with this problem?

GhostInterview helps by providing a clear explanation of the greedy approach, pointing out key observations, and breaking down the problem into manageable steps.

What is the time complexity of the solution for this problem?

The time complexity is typically O(n), where n is the length of the array, since we iterate through the array once in a greedy manner.

What should I keep in mind when solving Minimum Replacements to Sort the Array?

Focus on minimizing the number of operations by using a greedy strategy and validating that the array remains sorted after each replacement.

terminal

Solution

Solution 1: Greedy Approach

We observe that to make the array $nums$ non-decreasing or monotonically increasing, the elements towards the end of the array should be as large as possible. Therefore, it is unnecessary to replace the last element $nums[n-1]$ of the array $nums$ with multiple smaller numbers.

1

2

3

4

5

6

7

8

9

10

11

12

13

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

Continue Practicing

#2358 maximum number of groups entering a competition #2195 append k integers with minimal sum #2541 minimum operations to make array equal ii