What is the time complexity of Minimum Replacements to Sort the Array?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Replacements to Sort the Array?

This problem uses Array, Math, Greedy. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Replacements to Sort the Array?

Insight 1: The key is to iterate from the end of the array to the beginning. This allows us to determine a maximum allowed value for the current element based on the elements to its right which have already been processed. Insight 2: We can use a greedy approach. For each element, we determine the number of parts it needs to be split into such that each part is less than or equal to the maximum allowed value from the right.

Which companies ask LeetCode 2366. Minimum Replacements to Sort the Array in interviews?

This problem is frequently asked at Expedia, PayPal.

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Minimum Replacements to Sort the Array - LeetCode 2366 Solution

Minimum Replacements to Sort the Array - Complete Solution Guide

Minimum Replacements to Sort the Array is LeetCode problem 2366, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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],

Detailed Explanation

The problem asks us to find the minimum number of operations required to sort a given array `nums` in non-decreasing order. An operation involves replacing an element of the array with two elements that sum to the original element. The goal is to transform the input array into a non-decreasing sequence using these replacement operations while minimizing the number of operations performed.

Solution Approach

The solution uses a greedy approach to minimize the number of replacement operations. We iterate through the array from right to left. At each element `nums[i]`, we check if it is greater than the maximum allowed value `next_allowed_val` derived from the previously processed (to the right) elements. If it is, we calculate the number of parts to split `nums[i]` into such that each part is less than or equal to `next_allowed_val`. The number of operations is the number of parts minus 1. Then, we update `next_allowed_val` to the new maximum allowed value for elements further to the left based on this split.

Step-by-Step Algorithm

Step 1: Initialize `operations` to 0. This variable will store the total number of operations.
Step 2: Initialize `next_allowed_val` to the last element of the array `nums[n-1]`. This is because the last element is already sorted relative to what's to the right of it (nothing).
Step 3: Iterate through the array from `n-2` down to 0.
Step 4: For each element `nums[i]`, check if it's greater than `next_allowed_val`. If not, it's already in the correct order relative to elements to the right, so update `next_allowed_val` to `nums[i]` and continue to the next element.
Step 5: If `nums[i]` is greater than `next_allowed_val`, calculate the minimum number of parts `num_parts` that `nums[i]` needs to be split into. This is achieved using ceiling division: `num_parts = (nums[i] + next_allowed_val - 1) // next_allowed_val`.
Step 6: Add `num_parts - 1` to the `operations` count, since we're making `num_parts - 1` replacements.
Step 7: Update `next_allowed_val` to `nums[i] // num_parts`. This gives the maximum allowed value of the split parts. This is the largest possible value any element to the left can have to ensure the array remains non-decreasing after splitting `nums[i]`.
Step 8: After the loop finishes, return the total `operations`.

Key Insights

Insight 1: The key is to iterate from the end of the array to the beginning. This allows us to determine a maximum allowed value for the current element based on the elements to its right which have already been processed.
Insight 2: We can use a greedy approach. For each element, we determine the number of parts it needs to be split into such that each part is less than or equal to the maximum allowed value from the right.
Insight 3: Integer division plays a crucial role in finding the number of parts and the value of each part. Using ceiling division is essential for getting the minimum number of parts required.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Greedy.

Companies

Asked at: Expedia, PayPal.

Minimum Replacements to Sort the Array | LeetCode 2366

Hard

Array Math Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public long minimumReplacement(int[] nums) {
        int n = nums.length;
        
        long operations = 0;
        long next_allowed_val = nums[n - 1];
        
        for (int i = n - 2; i >= 0; i--) {
            long current_val = nums[i];
            
            if (current_val <= next_allowed_val) {
                next_allowed_val = current_val;
                continue;
            }
            
            long num_parts = (current_val + next_allowed_val - 1) / next_allowed_val;
            
            operations += num_parts - 1;
            
            next_allowed_val = current_val / num_parts;
            
        }
        
        return operations;
    }
}

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Problem Info

Description

Problem 2366: Minimum Replacements to Sort the Array - Detailed explanation and solution approaches.

Topics

Array Math Greedy

Companies

ExpediaPayPal

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2366

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `operations` to 0. This variable will store the total number of operations.
2
Step 2: Initialize `next_allowed_val` to the last element of the array `nums[n-1]`. This is because the last element is already sorted relative to what's to the right of it (nothing).
3
Step 3: Iterate through the array from `n-2` down to 0.
4
Step 4: For each element `nums[i]`, check if it's greater than `next_allowed_val`. If not, it's already in the correct order relative to elements to the right, so update `next_allowed_val` to `nums[i]` and continue to the next element.
5
Step 5: If `nums[i]` is greater than `next_allowed_val`, calculate the minimum number of parts `num_parts` that `nums[i]` needs to be split into. This is achieved using ceiling division: `num_parts = (nums[i] + next_allowed_val - 1) // next_allowed_val`.
6
Step 6: Add `num_parts - 1` to the `operations` count, since we're making `num_parts - 1` replacements.
7
Step 7: Update `next_allowed_val` to `nums[i] // num_parts`. This gives the maximum allowed value of the split parts. This is the largest possible value any element to the left can have to ensure the array remains non-decreasing after splitting `nums[i]`.
8
Step 8: After the loop finishes, return the total `operations`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the for loop iterates through each element of the array once. The operations inside the loop are constant time operations. The space complexity is O(1) because we are only using a few variables to store intermediate results and the number of variables does not depend on the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A dynamic programming approach could be considered, but it would likely be less efficient. You could define dp[i] as the minimum operations to sort nums[0...i]. However, determining the optimal split for each nums[i] would require iterating through possible split sizes, which can be expensive.

Time: O(n^2) or higher, depending on the DP formulation.Space: O(n) for the DP table.

Trade-offs: The DP approach is significantly slower and requires more space than the greedy approach. The greedy approach is optimal for this problem.

✨

Key Insights

1Insight 1: The key is to iterate from the end of the array to the beginning. This allows us to determine a maximum allowed value for the current element based on the elements to its right which have already been processed.
2Insight 2: We can use a greedy approach. For each element, we determine the number of parts it needs to be split into such that each part is less than or equal to the maximum allowed value from the right.
3Insight 3: Integer division plays a crucial role in finding the number of parts and the value of each part. Using ceiling division is essential for getting the minimum number of parts required.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Array is already sorted (non-decreasing). The algorithm will correctly return 0 because `nums[i] <= next_allowed_val` will always be true and `operations` will not be incremented.
Edge case 2: Array has only one element. The loop will not execute, and `operations` remains 0, which is the correct answer.
Edge case 3: Array has descending order. This will result in the maximum number of splits and operations. The algorithm will handle this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Iterating from left to right. This makes it difficult to determine the maximum allowed value for each element. It is better to process from right to left because the value to the right will impose the constrain to the elements to the left.

✗Common mistake 2: Using incorrect integer division, especially not using ceiling division when calculating `num_parts`. This can lead to incorrect results because the calculated parts will not satisfy the non-decreasing constraint. Using `(current_val + next_allowed_val - 1) // next_allowed_val` is important.

✗Common mistake 3: Incorrectly updating `next_allowed_val` after each replacement. The value of `next_allowed_val` must represent the largest possible element that could appear after replacement to keep the sequence sorted. So `current_val // num_parts` is the correct calculation.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling. You might have tasks of varying sizes that need to be scheduled in a certain order. You can split large tasks into smaller subtasks to fit them within specific time slots while minimizing the number of splits.

🏦Real-world scenario 2: Resource allocation. You have a limited number of resources and need to allocate them to different projects. You can split a large resource request into smaller requests to better utilize the available resources and minimize the number of divisions.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Replacements to Sort the Array"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?