What is the time complexity of Destination City?

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

What data structures are used to solve Destination City?

This problem uses Array, Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Destination City?

Insight 1: Using a set to efficiently track starting cities avoids redundant searches. Sets offer O(1) average time complexity for insertion and lookup. Insight 2: The destination city is the only city present in the list of end cities (cities that are destinations in the paths) but not in the list of starting cities.

Which companies ask LeetCode 1436. Destination City in interviews?

This problem is frequently asked at Yandex, Yelp.

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Destination City - LeetCode 1436 Solution

Destination City - Complete Solution Guide

Destination City is LeetCode problem 1436, a Easy 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 the array paths , where paths[i] = [cityA i , cityB i ] means there exists a direct path going from cityA i to cityB i . Return the destination city, that is, the city without any path outgoing to another city. It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city. Example 1: Input: paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]] Output: "Sao Paulo" Explanation: Starting at "London" cit

Detailed Explanation

The problem asks you to find the destination city in a list of paths. Each path is represented as a list of two strings: the starting city and the destination city. The destination city is the one that is never a starting city in any path. The input is a list of these paths, and the output is a single string representing the destination city. It's guaranteed that there will be exactly one destination city and that the paths form a directed acyclic graph (DAG) – meaning there are no cycles.

Solution Approach

The provided solutions utilize a set to store the starting cities. They iterate through the paths, adding each starting city to the set. Then, they iterate again, checking if each ending city is present in the set of starting cities. If an ending city is not found in the set, it is the destination city and is returned. This approach leverages the efficiency of set lookups.

Step-by-Step Algorithm

Step 1: Initialize a set (HashSet in Java, unordered_set in C++, set in Python) to store the starting cities.
Step 2: Iterate through the input list of paths. For each path, add the starting city (the first element in the path) to the set of starting cities.
Step 3: Iterate through the input list of paths again. For each path, check if the ending city (the second element in the path) is present in the set of starting cities.
Step 4: If the ending city is NOT in the set of starting cities, it means that there is no path leaving from this city and it's the destination city. Return this city.
Step 5: If the loop completes without finding such a city (meaning the set was empty or there's an issue with the input), return an appropriate value (empty string or null).

Key Insights

Insight 1: Using a set to efficiently track starting cities avoids redundant searches. Sets offer O(1) average time complexity for insertion and lookup.
Insight 2: The destination city is the only city present in the list of end cities (cities that are destinations in the paths) but not in the list of starting cities.
Insight 3: Handling the edge case of an empty input list of paths needs to be considered. The code should return an appropriate value (an empty string or null) in such situations.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Yandex, Yelp.

Destination City | LeetCode 1436

Easy

Array Hash Table String

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

Solution Code

Available in:

import java.util.List;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public String destCity(List<List<String>> paths) {
        Set<String> startingCities = new HashSet<>();
        for (List<String> path : paths) {
            startingCities.add(path.get(0));
        }

        for (List<String> path : paths) {
            String destination = path.get(1);
            if (!startingCities.contains(destination)) {
                return destination;
            }
        }

        return "";
    }
}

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Problem Info

Description

Problem 1436: Destination City - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

YandexYelp

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1436

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 a set (HashSet in Java, unordered_set in C++, set in Python) to store the starting cities.
2
Step 2: Iterate through the input list of paths. For each path, add the starting city (the first element in the path) to the set of starting cities.
3
Step 3: Iterate through the input list of paths again. For each path, check if the ending city (the second element in the path) is present in the set of starting cities.
4
Step 4: If the ending city is NOT in the set of starting cities, it means that there is no path leaving from this city and it's the destination city. Return this city.
5
Step 5: If the loop completes without finding such a city (meaning the set was empty or there's an issue with the input), return an appropriate value (empty string or null).

💡

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 algorithm iterates through the list of paths twice. Set operations (insertion and lookup) take O(1) average time. Therefore, the overall time complexity is O(n). The space used is proportional to the number of unique starting cities, which is at most the number of paths in the worst case, leading to O(n) space complexity.

📖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

Using a HashMap/Dictionary

Instead of two sets, use a single HashMap/Dictionary where keys are cities and values are booleans indicating if the city is a start or end city. Iterate and update the map. Find the key that is only an end city.

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

Trade-offs: This approach might be slightly less efficient than using two sets because of the overhead of HashMap operations. However, it might be easier to read and understand for some.

✨

Key Insights

1Insight 1: Using a set to efficiently track starting cities avoids redundant searches. Sets offer O(1) average time complexity for insertion and lookup.
2Insight 2: The destination city is the only city present in the list of end cities (cities that are destinations in the paths) but not in the list of starting cities.
3Insight 3: Handling the edge case of an empty input list of paths needs to be considered. The code should return an appropriate value (an empty string or null) in such situations.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty input list: The solutions handle this by returning an empty string or NULL. This is important because an empty input list should result in a specific output indicating no destination is found.
Edge case 2: Only one path: The algorithms correctly identify the second city in the single path as the destination city.
Edge case 3: Multiple paths with the same destination: The algorithm correctly identifies the destination city even if the same city appears as a destination in several paths.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the empty input case correctly. Forgetting to return an appropriate value (e.g., null or empty string) if the input list is empty leads to undefined behavior or errors.

✗Common mistake 2: Inefficiently searching for the destination city. Using linear search instead of sets for checking if a city is a starting city results in a slower algorithm with O(n^2) time complexity.

✅

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: Network routing: Finding the final destination in a network or transport system.

🏦Real-world scenario 2: Workflow tracking: Determining the end point of a process involving multiple steps.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: November 24, 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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Destination City"?

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

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