What is the time complexity of Valid Arrangement of Pairs?

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

What data structures are used to solve Valid Arrangement of Pairs?

This problem uses Depth-First Search, Graph, Eulerian Circuit. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Valid Arrangement of Pairs?

Insight 1: The problem can be modeled as finding an Eulerian path in a directed graph. Each number in a pair represents a node, and each pair represents a directed edge from the first number to the second. Insight 2: To find an Eulerian path, we need to identify a starting node. This node will have an out-degree one greater than its in-degree (if the path isn't a cycle). If no such node exists, we can pick any node as the starting point (since it is an Eulerian cycle then).

Which companies ask LeetCode 2097. Valid Arrangement of Pairs in interviews?

This problem is frequently asked at Goldman Sachs, Snap.

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Valid Arrangement of Pairs - LeetCode 2097 Solution

Q: How to solve Valid Arrangement of Pairs?

The solution uses a Depth-First Search (DFS) approach to find an Eulerian path in the graph represented by the given pairs. First, we construct an adjacency list and a degree map to represent the graph and track the in-degrees and out-degrees of each node. Then, we identify the starting node for the Eulerian path. If a node with an out-degree one greater than its in-degree exists, we use it as the starting node. Otherwise, we arbitrarily choose the first node in the input pairs. Finally, we perform a DFS traversal from the starting node, adding nodes to a stack and removing edges as we go. When a node has no more outgoing edges, we pop it from the stack and add it to the path. After the DFS traversal, we reverse the path to obtain the correct Eulerian path and convert it to the desired arrangement of pairs.

Valid Arrangement of Pairs - Complete Solution Guide

Valid Arrangement of Pairs is LeetCode problem 2097, 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 2D integer array pairs where pairs[i] = [start i , end i ] . An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length , we have end i-1 == start i . Return any valid arrangement of pairs . Note: The inputs will be generated such that there exists a valid arrangement of pairs . Example 1: Input: pairs = [[5,1],[4,5],[11,9],[9,4]] Output: [[11,9],[9,4],[4,5],[5,1]] Explanation: This is a valid arrangement since end i-1 always equals start i . end

Detailed Explanation

The problem asks us to find a valid arrangement of pairs of integers, where a valid arrangement means that the second element of each pair equals the first element of the next pair. Given a list of pairs, `pairs`, we need to rearrange them such that `end_i-1 == start_i` for all `1 <= i < pairs.length`. The problem guarantees that a valid arrangement exists.

Solution Approach

The solution uses a Depth-First Search (DFS) approach to find an Eulerian path in the graph represented by the given pairs. First, we construct an adjacency list and a degree map to represent the graph and track the in-degrees and out-degrees of each node. Then, we identify the starting node for the Eulerian path. If a node with an out-degree one greater than its in-degree exists, we use it as the starting node. Otherwise, we arbitrarily choose the first node in the input pairs. Finally, we perform a DFS traversal from the starting node, adding nodes to a stack and removing edges as we go. When a node has no more outgoing edges, we pop it from the stack and add it to the path. After the DFS traversal, we reverse the path to obtain the correct Eulerian path and convert it to the desired arrangement of pairs.

Step-by-Step Algorithm

Step 1: Create an adjacency list (adj) and a degree map (degrees). The adjacency list stores the neighbors of each node, and the degree map stores the difference between the out-degree and in-degree of each node.
Step 2: Iterate through the input pairs and populate the adjacency list and degree map. For each pair (u, v), add v to the adjacency list of u, increment the out-degree of u, and increment the in-degree of v (equivalent to decrementing degrees[v]).
Step 3: Determine the starting node for the Eulerian path. Iterate through the degree map and find a node with a degree of 1 (out-degree > in-degree by 1). If no such node is found, choose the first node in the input pairs as the starting node (Eulerian cycle).
Step 4: Initialize an empty path list and a stack containing the starting node.
Step 5: Perform a DFS traversal using a while loop. While the stack is not empty:
Step 6: Get the current node 'u' from the top of the stack.
Step 7: If 'u' has outgoing edges (neighbors in the adjacency list), pop a neighbor 'v' from adj[u], push v onto the stack.
Step 8: If 'u' has no outgoing edges, pop 'u' from the stack and append it to the path list.
Step 9: Reverse the path list.
Step 10: Construct the result list of pairs from the reversed path. Iterate through the path list and create pairs (path[i], path[i+1]) for each consecutive pair of nodes in the path.
Step 11: Return the result list of pairs.

Key Insights

Insight 1: The problem can be modeled as finding an Eulerian path in a directed graph. Each number in a pair represents a node, and each pair represents a directed edge from the first number to the second.
Insight 2: To find an Eulerian path, we need to identify a starting node. This node will have an out-degree one greater than its in-degree (if the path isn't a cycle). If no such node exists, we can pick any node as the starting point (since it is an Eulerian cycle then).
Insight 3: We can use a stack-based Depth-First Search (DFS) approach to traverse the graph and construct the Eulerian path. As we visit each node, we remove outgoing edges. The order in which we pop nodes from the stack gives us the reverse of the Eulerian path, so we must reverse the resulting path.

Complexity Analysis

Time Complexity: O(E)

Space Complexity: O(E)

Topics

This problem involves: Depth-First Search, Graph, Eulerian Circuit.

Companies

Asked at: Goldman Sachs, Snap.

Valid Arrangement of Pairs | LeetCode 2097

Hard

Depth-First Search Graph Eulerian Circuit

Time: O(E)Space: O(E)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[][] validArrangement(int[][] pairs) {
        Map<Integer, List<Integer>> adj = new HashMap<>();
        Map<Integer, Integer> degrees = new HashMap<>();

        for (int[] pair : pairs) {
            int u = pair[0];
            int v = pair[1];
            adj.computeIfAbsent(u, k -> new ArrayList<>()).add(v);
            degrees.put(u, degrees.getOrDefault(u, 0) + 1);
            degrees.put(v, degrees.getOrDefault(v, 0) - 1);
        }

        int startNode = -1;
        for (int node : degrees.keySet()) {
            if (degrees.get(node) == 1) {
                startNode = node;
                break;
            }
        }

        if (startNode == -1) {
            startNode = pairs[0][0];
        }

        List<Integer> path = new ArrayList<>();
        Stack<Integer> stack = new Stack<>();
        stack.push(startNode);

        while (!stack.isEmpty()) {
            int u = stack.peek();
            if (adj.containsKey(u) && !adj.get(u).isEmpty()) {
                int v = adj.get(u).remove(0);
                if(adj.get(u).isEmpty()){
                    adj.remove(u);
                }
                stack.push(v);
            } else {
                stack.pop();
                path.add(u);
            }
        }

        Collections.reverse(path);

        int[][] result = new int[path.size() - 1][2];
        for (int i = 0; i < path.size() - 1; i++) {
            result[i][0] = path.get(i);
            result[i][1] = path.get(i + 1);
        }

        return result;
    }
}

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Problem Info

Description

Problem 2097: Valid Arrangement of Pairs - Detailed explanation and solution approaches.

Topics

Depth-First Search Graph Eulerian Circuit

Companies

Goldman SachsSnap

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2097

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

11 steps

1
Step 1: Create an adjacency list (adj) and a degree map (degrees). The adjacency list stores the neighbors of each node, and the degree map stores the difference between the out-degree and in-degree of each node.
2
Step 2: Iterate through the input pairs and populate the adjacency list and degree map. For each pair (u, v), add v to the adjacency list of u, increment the out-degree of u, and increment the in-degree of v (equivalent to decrementing degrees[v]).
3
Step 3: Determine the starting node for the Eulerian path. Iterate through the degree map and find a node with a degree of 1 (out-degree > in-degree by 1). If no such node is found, choose the first node in the input pairs as the starting node (Eulerian cycle).
4
Step 4: Initialize an empty path list and a stack containing the starting node.
5
Step 5: Perform a DFS traversal using a while loop. While the stack is not empty:
6
Step 6: Get the current node 'u' from the top of the stack.
7
Step 7: If 'u' has outgoing edges (neighbors in the adjacency list), pop a neighbor 'v' from adj[u], push v onto the stack.
8
Step 8: If 'u' has no outgoing edges, pop 'u' from the stack and append it to the path list.
9
Step 9: Reverse the path list.
10
Step 10: Construct the result list of pairs from the reversed path. Iterate through the path list and create pairs (path[i], path[i+1]) for each consecutive pair of nodes in the path.
11
Step 11: Return the result list of pairs.

💡

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(E)

💾

Space Complexity

O(E)

📝

Why?

The time complexity is dominated by the DFS traversal of the graph. Since we visit each edge once, the time complexity is O(E). The space complexity is determined by the size of the adjacency list, stack, and path, all of which are proportional to the number of edges E.

📖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

Hierholzer's Algorithm

Hierholzer's algorithm is a classic and efficient algorithm for finding Eulerian paths and cycles in a graph. It avoids recursion by iteratively exploring cycles within the graph and merging them until a single Eulerian path/cycle is found.

Time: O(E)Space: O(E)

Trade-offs: Hierholzer's Algorithm can be slightly more complex to implement but might offer better performance in certain scenarios because it avoids explicit recursion and stack management like the solution using DFS.

✨

Key Insights

1Insight 1: The problem can be modeled as finding an Eulerian path in a directed graph. Each number in a pair represents a node, and each pair represents a directed edge from the first number to the second.
2Insight 2: To find an Eulerian path, we need to identify a starting node. This node will have an out-degree one greater than its in-degree (if the path isn't a cycle). If no such node exists, we can pick any node as the starting point (since it is an Eulerian cycle then).
3Insight 3: We can use a stack-based Depth-First Search (DFS) approach to traverse the graph and construct the Eulerian path. As we visit each node, we remove outgoing edges. The order in which we pop nodes from the stack gives us the reverse of the Eulerian path, so we must reverse the resulting path.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: The input graph has an Eulerian cycle (no node with out-degree greater than in-degree). The solution correctly handles this by choosing an arbitrary starting node, as any node can be the starting point in an Eulerian cycle.
Edge case 2: The input contains pairs with large integer values. The solution uses `defaultdict` in Python and `HashMap` in Java and `unordered_map` in C++ to handle large integer values as nodes without issues.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to reverse the path after the DFS traversal. The stack pops nodes in reverse order of the Eulerian path, so reversing the path is essential for obtaining the correct arrangement.

✗Common mistake 2: Incorrectly identifying the starting node. The starting node should have an out-degree greater than its in-degree (by one) if it exists, otherwise any node can be chosen. Missing this leads to incorrect traversal.

✗Common mistake 3: Not removing the edge after visiting it, which leads to infinite loop.

✅

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: DNA sequencing. Determining the order of DNA fragments can be modeled as finding an Eulerian path in a graph where fragments overlap.

🏦Real-world scenario 2: Route planning for delivery services. Finding a route that traverses each street in a city once can be modeled as finding an Eulerian path in a graph where streets are edges and intersections are nodes.

💡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

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.

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 26, 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use BFS vs DFS for graph problems?

Use BFS for shortest path in unweighted graphs, level-order processing, or when exploring neighbors systematically. Use DFS for exhaustive search, detecting cycles, topological sort, or when you need to explore complete paths. DFS uses less memory for very wide graphs.

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 "Valid Arrangement of Pairs"?

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

🤖 Need help with this problem?