What is the time complexity of Couples Holding Hands?

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

What data structures are used to solve Couples Holding Hands?

This problem uses Greedy, Depth-First Search, Breadth-First Search, Union Find, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Couples Holding Hands?

Insight 1: Each person 'x' has a unique partner, which is 'x ^ 1'. This bitwise XOR operation is efficient for finding partners. Insight 2: Instead of physically simulating all possible swaps, we can greedily correct the seating arrangement by focusing on misplaced couples and swapping one member to bring their partner next to them. This guarantees minimal swaps because we're addressing one misplaced couple with one swap.

Which companies ask LeetCode 765. Couples Holding Hands in interviews?

This problem is frequently asked at Citadel.

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Couples Holding Hands - LeetCode 765 Solution

Couples Holding Hands - Complete Solution Guide

Couples Holding Hands is LeetCode problem 765, 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

There are n couples sitting in 2n seats arranged in a row and want to hold hands. The people and seats are represented by an integer array row where row[i] is the ID of the person sitting in the i th seat. The couples are numbered in order, the first couple being (0, 1) , the second couple being (2, 3) , and so on with the last couple being (2n - 2, 2n - 1) . Return the minimum number of swaps so that every couple is sitting side by side . A swap consists of choosing any two people, then they st

Detailed Explanation

The problem presents a scenario where 'n' couples (2n people) are seated in a row of 2n seats. Each person is represented by a unique integer ID. Couples are formed by consecutive even and odd numbers (0 and 1, 2 and 3, and so on). The goal is to find the minimum number of swaps required to arrange all couples side-by-side. A swap involves two people exchanging seats.

Solution Approach

The solution utilizes a greedy approach. It iterates through the `row` array, checking each even-indexed position (the first person of a potential couple). If the person to their right is not their partner, a swap is performed. The core idea is that each swap corrects one incorrectly seated couple. A hash map (or array in C) is used to store the current index (position) of each person. This allows efficient updating of the position after a swap.

Step-by-Step Algorithm

Step 1: Initialize a hash map (pos) to store the index of each person in the row. This will allow O(1) lookup of a person's location.
Step 2: Initialize a swap counter to 0.
Step 3: Iterate through the row array from index 0 to n-1, incrementing by 2 in each iteration (i.e., i = 0, 2, 4, ...).
Step 4: For each even index i, determine the partner of the person at row[i] using the XOR operation (row[i] ^ 1).
Step 5: Check if the person at row[i+1] is the partner of row[i]. If not, it means a swap is needed.
Step 6: If a swap is needed, increment the swap counter.
Step 7: Find the index (partner_pos) of the correct partner using the 'pos' hash map.
Step 8: Swap the person at row[i+1] with the partner located at row[partner_pos].
Step 9: Update the 'pos' hash map to reflect the new positions of the swapped people. This is crucial for maintaining accurate location information.
Step 10: After iterating through the entire row, return the total number of swaps.

Key Insights

Insight 1: Each person 'x' has a unique partner, which is 'x ^ 1'. This bitwise XOR operation is efficient for finding partners.
Insight 2: Instead of physically simulating all possible swaps, we can greedily correct the seating arrangement by focusing on misplaced couples and swapping one member to bring their partner next to them. This guarantees minimal swaps because we're addressing one misplaced couple with one swap.
Insight 3: The location of each person is important, thus maintaining a map between person and their seat is crucial for efficient swapping.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Greedy, Depth-First Search, Breadth-First Search, Union Find, Graph.

Companies

Asked at: Citadel.

Couples Holding Hands | LeetCode 765

Hard

Greedy Depth-First Search Breadth-First Search Union Find Graph

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minSwapsCouples(int[] row) {
        int n = row.length;
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < n; i++) {
            pos.put(row[i], i);
        }
        
        int swaps = 0;
        
        for (int i = 0; i < n; i += 2) {
            int person1 = row[i];
            int partner = person1 ^ 1;
            
            if (row[i+1] != partner) {
                swaps += 1;
                
                int person2 = row[i+1];
                int partner_pos = pos.get(partner);
                
                // Swap the people in the row array
                int temp = row[i+1];
                row[i+1] = partner;
                row[partner_pos] = temp;
                
                // Update the positions map
                pos.put(person2, partner_pos);
                pos.put(partner, i + 1);
            }
        }
        
        return swaps;
    }
}

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Problem Info

Description

Problem 765: Couples Holding Hands - Detailed explanation and solution approaches.

Topics

Greedy Depth-First Search Breadth-First Search Union Find Graph

Companies

Citadel

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:765

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

10 steps

1
Step 1: Initialize a hash map (pos) to store the index of each person in the row. This will allow O(1) lookup of a person's location.
2
Step 2: Initialize a swap counter to 0.
3
Step 3: Iterate through the row array from index 0 to n-1, incrementing by 2 in each iteration (i.e., i = 0, 2, 4, ...).
4
Step 4: For each even index i, determine the partner of the person at row[i] using the XOR operation (row[i] ^ 1).
5
Step 5: Check if the person at row[i+1] is the partner of row[i]. If not, it means a swap is needed.
6
Step 6: If a swap is needed, increment the swap counter.
7
Step 7: Find the index (partner_pos) of the correct partner using the 'pos' hash map.
8
Step 8: Swap the person at row[i+1] with the partner located at row[partner_pos].
9
Step 9: Update the 'pos' hash map to reflect the new positions of the swapped people. This is crucial for maintaining accurate location information.
10
Step 10: After iterating through the entire row, return the total number of swaps.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each person is visited once. The hash map operations (put and get) have average time complexity of O(1). The space complexity is determined by the 'pos' hash map, which stores the index of each of the 'n' people in the array. Therefore, the space complexity is O(n).

📖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

Union Find (Disjoint Set)

This approach treats each couple as a node in a graph. If two couples are not seated together, there is an edge between their nodes. The number of swaps is then the number of connected components minus the number of nodes.

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

Trade-offs: Union Find provides a different perspective, viewing the problem as connecting components. It can be useful for certain variations of the problem, but the constant factors make it usually slower than the greedy approach for this specific problem.

Graph Cycles

Model the problem as a graph where an edge exists between two couples seated in adjacent seats if they are not paired correctly. The number of swaps needed is (number of couples - number of cycles).

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

Trade-offs: This appraoch might be conceptually harder to visualize at first, although it gives an alternative method to calculate the final result, using graph theory as a basis.

✨

Key Insights

1Insight 1: Each person 'x' has a unique partner, which is 'x ^ 1'. This bitwise XOR operation is efficient for finding partners.
2Insight 2: Instead of physically simulating all possible swaps, we can greedily correct the seating arrangement by focusing on misplaced couples and swapping one member to bring their partner next to them. This guarantees minimal swaps because we're addressing one misplaced couple with one swap.
3Insight 3: The location of each person is important, thus maintaining a map between person and their seat is crucial for efficient swapping.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Input array where all couples are already correctly seated (e.g., [0, 1, 2, 3]). The algorithm correctly identifies that no swaps are needed and returns 0.
Edge case 2: Input array with only two elements (n=1, e.g., [1, 0]). The algorithm correctly identifies one swap is needed.
Edge case 3: An array where all the couples are seated in the reverse order: [1, 0, 3, 2, 5, 4 ...]. The algorithm will perform the maximum number of swaps which is n/2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the 'pos' hash map after swapping elements in the 'row' array. This will lead to incorrect index lookups and incorrect swaps in subsequent iterations.

✗Common mistake 2: Not handling the edge case where the input array is already correctly seated, which leads to unnecessary swap attempts.

✗Common mistake 3: Incorrectly calculating the partner using methods other than XOR, resulting in wrong pairing assessments.

✅

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 seating arrangements in events or gatherings where couples or partners need to be seated together. This could be used to minimize disruptions or improve social interaction.

🏦Real-world scenario 2: Data center cable management. Ensuring paired cables are connected to adjacent ports, minimizing the cable length needed to connect these paired devices.

💡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

Greedy works when local optimal leads to global optimal. Always try to prove (or disprove) the greedy choice property before coding.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 16, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Couples Holding Hands"?

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

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