What is the time complexity of Max Chunks To Make Sorted?

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

What data structures are used to solve Max Chunks To Make Sorted?

This problem uses Array, Stack, Greedy, Sorting, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Max Chunks To Make Sorted?

Insight 1: The key insight is that if the maximum value encountered up to a certain index `i` is equal to `i`, then all the elements from 0 to `i` must be present in the subarray from index 0 to `i`. Therefore, we can form a chunk at index `i`. Insight 2: We don't need to actually sort any subarrays. We only need to keep track of the running maximum and compare it to the current index.

Which companies ask LeetCode 769. Max Chunks To Make Sorted in interviews?

This problem is frequently asked at Poshmark.

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Max Chunks To Make Sorted - LeetCode 769 Solution

Max Chunks To Make Sorted - Complete Solution Guide

Max Chunks To Make Sorted is LeetCode problem 769, a Medium 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 an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1] . We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array. Return the largest number of chunks we can make to sort the array . Example 1: Input: arr = [4,3,2,1,0] Output: 1 Explanation: Splitting into two or more chunks will not return the required result. For example, splittin

Detailed Explanation

The problem asks us to find the maximum number of chunks we can split an array `arr` into such that when each chunk is sorted individually and then concatenated, the resulting array is sorted. The array `arr` is guaranteed to be a permutation of the integers from 0 to n-1, where n is the length of the array. This means each number from 0 to n-1 appears exactly once in `arr`.

Solution Approach

The solution iterates through the array, keeping track of the maximum value seen so far (`current_max`). In each iteration, we update `current_max` to be the maximum of the current element and the previous `current_max`. If `current_max` is equal to the index `i`, it means that all numbers from 0 to `i` are present in the subarray from index 0 to `i`. This guarantees that we can form a chunk at index `i`. Therefore, we increment the chunk count. This approach efficiently determines the maximum number of chunks possible.

Step-by-Step Algorithm

Step 1: Initialize `chunks` to 0 and `current_max` to -1.
Step 2: Iterate through the array `arr` from index `i = 0` to `n - 1`.
Step 3: In each iteration, update `current_max` as `current_max = max(current_max, arr[i])`.
Step 4: If `current_max` is equal to `i`, increment `chunks` by 1.
Step 5: After iterating through the entire array, return `chunks`.

Key Insights

Insight 1: The key insight is that if the maximum value encountered up to a certain index `i` is equal to `i`, then all the elements from 0 to `i` must be present in the subarray from index 0 to `i`. Therefore, we can form a chunk at index `i`.
Insight 2: We don't need to actually sort any subarrays. We only need to keep track of the running maximum and compare it to the current index.
Insight 3: Since the array is a permutation of [0, n-1], a chunk ending at index `i` can be formed if and only if the maximum element in `arr[0...i]` is equal to `i`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Stack, Greedy, Sorting, Monotonic Stack.

Companies

Asked at: Poshmark.

Max Chunks To Make Sorted | LeetCode 769

Medium

Array Stack Greedy Sorting Monotonic Stack

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

Solution Code

Available in:

class Solution {
    public int maxChunksToSorted(int[] arr) {
        int chunks = 0;
        int current_max = -1;
        for (int i = 0; i < arr.length; i++) {
            current_max = Math.max(current_max, arr[i]);
            if (current_max == i) {
                chunks += 1;
            }
        }
        return chunks;
    }
}

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Problem Info

Description

Problem 769: Max Chunks To Make Sorted - Detailed explanation and solution approaches.

Topics

Array Stack Greedy Sorting Monotonic Stack

Companies

Poshmark

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:769

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

5 steps

1
Step 1: Initialize `chunks` to 0 and `current_max` to -1.
2
Step 2: Iterate through the array `arr` from index `i = 0` to `n - 1`.
3
Step 3: In each iteration, update `current_max` as `current_max = max(current_max, arr[i])`.
4
Step 4: If `current_max` is equal to `i`, increment `chunks` by 1.
5
Step 5: After iterating through the entire array, return `chunks`.

💡

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 directly proportional to the size of the input array due to the single for loop. The space complexity is constant as we only utilize a fixed number of variables regardless of the input array's 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

Sorting and Prefix Max

Another approach is to create a sorted copy of the array and calculate prefix maximums for both the original and the sorted arrays. Then, compare the prefix maximums at each index to identify chunk boundaries.

Time: O(n log n) due to the sorting step.Space: O(n) for storing the sorted array and prefix maximums.

Trade-offs: This approach is less efficient than the O(n) approach. It's only useful if you need the sorted array for other purposes.

✨

Key Insights

1Insight 1: The key insight is that if the maximum value encountered up to a certain index `i` is equal to `i`, then all the elements from 0 to `i` must be present in the subarray from index 0 to `i`. Therefore, we can form a chunk at index `i`.
2Insight 2: We don't need to actually sort any subarrays. We only need to keep track of the running maximum and compare it to the current index.
3Insight 3: Since the array is a permutation of [0, n-1], a chunk ending at index `i` can be formed if and only if the maximum element in `arr[0...i]` is equal to `i`.

📚

Key Terms Explained

For Students

Array

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

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `arr = [0]`. The solution correctly returns 1 because `current_max` will be 0 at index 0, which is equal to the index. `chunks` will be incremented to 1.
Edge case 2: `arr = [1, 0, 2, 3, 4]`. The solution correctly identifies the chunks [1, 0], [2], [3], [4] and returns 4. `current_max` becomes 1 at index 0. At index 1, `current_max` becomes 1, which equals the index, so `chunks` becomes 1. Then `current_max` becomes 2 at index 2, so `chunks` becomes 2. This pattern continues, correctly identifying 4 chunks.
Edge case 3: `arr = [4, 3, 2, 1, 0]`. The solution correctly identifies only one chunk. The `current_max` will be updated in each iteration. Only when it reaches index 4 will the value of `current_max` equal index, thus chunk count is 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to sort the chunks and then comparing to the overall sorted array. This would be much more complex and inefficient. The key is to recognize that we don't actually need to sort anything, only track the maximum seen so far.

✗Common mistake 2: Forgetting to initialize `current_max` to -1. This could lead to incorrect results when the first element of the array is 0.

✅

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: Optimizing resource allocation. Imagine splitting a large task into smaller, independent subtasks that can be processed in parallel. This algorithm can help determine how to divide the task to maximize parallelism while ensuring the overall task is completed correctly.

🏦Real-world scenario 2: Data partitioning in distributed systems. When distributing data across multiple nodes, it's often necessary to partition the data in a way that preserves certain orderings or relationships. This algorithm can be adapted to help find optimal partition boundaries.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 20, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Max Chunks To Make Sorted"?

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

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