What is the time complexity of Maximum Length of Pair Chain?

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

What data structures are used to solve Maximum Length of Pair Chain?

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

What are the key insights for solving Maximum Length of Pair Chain?

Insight 1: Sorting the pairs based on their end times (right_i) is crucial. This allows us to greedily select pairs to form the longest chain. Insight 2: A greedy approach is applicable here. After sorting by end times, we pick the pair with the smallest end time that is still greater than the end time of the previous pair in the chain.

Which companies ask LeetCode 646. Maximum Length of Pair Chain in interviews?

This problem is frequently asked at Swiggy.

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Maximum Length of Pair Chain - LeetCode 646 Solution

Q: How to solve Maximum Length of Pair Chain?

The solution uses a greedy approach. First, the pairs are sorted in ascending order based on their end times. Then, we iterate through the sorted pairs. If the start time of the current pair is greater than the end time of the last pair added to the chain, we add the current pair to the chain and update the end time of the last pair added to the chain. This process continues until all pairs have been considered. The final chain length is the maximum length of the pair chain.

Maximum Length of Pair Chain - Complete Solution Guide

Maximum Length of Pair Chain is LeetCode problem 646, 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 array of n pairs pairs where pairs[i] = [left i , right i ] and left i < right i . A pair p2 = [c, d] follows a pair p1 = [a, b] if b < c . A chain of pairs can be formed in this fashion. Return the length longest chain which can be formed . You do not need to use up all the given intervals. You can select pairs in any order. Example 1: Input: pairs = [[1,2],[2,3],[3,4]] Output: 2 Explanation: The longest chain is [1,2] -> [3,4]. Example 2: Input: pairs = [[1,2],[7,8],[4,5]] Out

Detailed Explanation

The problem asks us to find the maximum length of a chain of pairs. A pair `p2 = [c, d]` follows a pair `p1 = [a, b]` if `b < c`. We are given an array of pairs `pairs` where each `pairs[i] = [left_i, right_i]` and `left_i < right_i`. The goal is to find the longest possible sequence of pairs where each pair follows the previous one.

Solution Approach

The solution uses a greedy approach. First, the pairs are sorted in ascending order based on their end times. Then, we iterate through the sorted pairs. If the start time of the current pair is greater than the end time of the last pair added to the chain, we add the current pair to the chain and update the end time of the last pair added to the chain. This process continues until all pairs have been considered. The final chain length is the maximum length of the pair chain.

Step-by-Step Algorithm

Step 1: Sort the input `pairs` array in ascending order based on the second element (end time) of each pair.
Step 2: Initialize `chain_length` to 0 and `current_end` to negative infinity (or the smallest possible integer value). This represents the end time of the last pair added to the chain.
Step 3: Iterate through the sorted `pairs` array. For each pair `[start, end]`, check if `start > current_end`.
Step 4: If `start > current_end`, it means the current pair can be added to the chain without violating the 'follows' condition. Increment `chain_length` by 1 and update `current_end` to `end`.
Step 5: After iterating through all the pairs, return the final `chain_length`.

Key Insights

Insight 1: Sorting the pairs based on their end times (right_i) is crucial. This allows us to greedily select pairs to form the longest chain.
Insight 2: A greedy approach is applicable here. After sorting by end times, we pick the pair with the smallest end time that is still greater than the end time of the previous pair in the chain.
Insight 3: The key property used in the greedy approach is that picking the pair with the smallest end time allows for the possibility of fitting more pairs into the chain later on.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Sorting.

Companies

Asked at: Swiggy.

Maximum Length of Pair Chain | LeetCode 646

Medium

Array Dynamic Programming Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, Comparator.comparingInt(a -> a[1]));
        
        int chain_length = 0;
        int current_end = Integer.MIN_VALUE;
        
        for (int[] pair : pairs) {
            int start = pair[0];
            int end = pair[1];
            
            if (start > current_end) {
                chain_length++;
                current_end = end;
            }
        }
        
        return chain_length;
    }
}

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Problem Info

Description

Problem 646: Maximum Length of Pair Chain - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Sorting

Companies

Swiggy

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:646

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: Sort the input `pairs` array in ascending order based on the second element (end time) of each pair.
2
Step 2: Initialize `chain_length` to 0 and `current_end` to negative infinity (or the smallest possible integer value). This represents the end time of the last pair added to the chain.
3
Step 3: Iterate through the sorted `pairs` array. For each pair `[start, end]`, check if `start > current_end`.
4
Step 4: If `start > current_end`, it means the current pair can be added to the chain without violating the 'follows' condition. Increment `chain_length` by 1 and update `current_end` to `end`.
5
Step 5: After iterating through all the pairs, return the final `chain_length`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting requires O(n log n) time. The iteration through the sorted list requires O(n) time. The space complexity is constant because no additional data structures dependent on the input size are used, apart from potentially modifying the existing array in place during the sorting step, or if the algorithm uses a sorting algorithm with space complexity O(1) at worst.

📖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

Another approach is to use dynamic programming. We can sort the pairs and create a `dp` array where `dp[i]` stores the maximum length of a chain ending at `pairs[i]`. We iterate through all the pairs and update the `dp` array accordingly.

Time: O(n^2)Space: O(n)

Trade-offs: The DP approach has a higher time complexity (O(n^2)) compared to the greedy approach (O(n log n)). However, it might be easier to understand for some beginners. DP is useful when the greedy approach doesn't guarantee optimal results, but in this case, the greedy approach does yield the optimal solution.

✨

Key Insights

1Insight 1: Sorting the pairs based on their end times (right_i) is crucial. This allows us to greedily select pairs to form the longest chain.
2Insight 2: A greedy approach is applicable here. After sorting by end times, we pick the pair with the smallest end time that is still greater than the end time of the previous pair in the chain.
3Insight 3: The key property used in the greedy approach is that picking the pair with the smallest end time allows for the possibility of fitting more pairs into the chain later on.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

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: Empty input array. The code handles this correctly because the loop won't execute, and the initial value of `chain_length` (0) will be returned.
Edge case 2: Pairs with overlapping intervals. The sorting and greedy selection ensure that only non-overlapping intervals are added to the chain to maximize its length. For instance, [[1, 2], [1, 3]] would result in only [1, 2] being picked, allowing potentially [3, 4] to come later.
Edge case 3: All pairs have the same ending time. In such cases, only the first pair would be selected because the next pairs would all have a start time less than or equal to current_end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the pairs based on their end times. Sorting by start times or not sorting at all will lead to incorrect results.

✗Common mistake 2: Not initializing `current_end` to negative infinity. This ensures that the first pair is always selected as the starting point of the chain if the input is not empty.

✗Common mistake 3: Not handling edge cases such as an empty input array or all pairs overlapping.

✅

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. We can consider each pair as a task with a start and end time. The goal is to schedule the maximum number of tasks that can be performed sequentially, where a task can only start after the previous task has finished.

🏦Real-world scenario 2: Meeting room scheduling. We have a set of meetings with start and end times. We want to find the maximum number of meetings that can be scheduled in the same room without any overlaps.

💡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: September 15, 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 13

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

@InterviewProJul 6

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Maximum Length of Pair Chain"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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