What is the time complexity of Find the City With the Smallest Number of Neighbors at a Threshold Distance?

The optimal solution has a time complexity of O(n^3).

What is the space complexity of Find the City With the Smallest Number of Neighbors at a Threshold Distance?

The space complexity is O(n^2).

What data structures are used to solve Find the City With the Smallest Number of Neighbors at a Threshold Distance?

This problem uses Dynamic Programming, Graph, Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find the City With the Smallest Number of Neighbors at a Threshold Distance?

Insight 1: The problem requires finding the shortest paths between all pairs of cities, which suggests using the Floyd-Warshall algorithm. Insight 2: After calculating the shortest path between all pairs of cities, we can iterate through each city and count how many other cities are reachable within the `distanceThreshold`.

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Find the City With the Smallest Number of Neighbors at a Threshold Distance - LeetCode 1334 Solution

Q: Which companies ask LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance in interviews?

This problem is frequently asked at Expedia.

Find the City With the Smallest Number of Neighbors at a Threshold Distance - Complete Solution Guide

Find the City With the Smallest Number of Neighbors at a Threshold Distance is LeetCode problem 1334, 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

There are n cities numbered from 0 to n-1 . Given the array edges where edges[i] = [from i , to i , weight i ] represents a bidirectional and weighted edge between cities from i and to i , and given the integer distanceThreshold . Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold , If there are multiple such cities, return the city with the greatest number. Notice that the distance of a path connecting cities i

Detailed Explanation

The problem asks us to find the city that has the fewest neighboring cities reachable within a given distance threshold. We are given a graph represented by edges (city pairs and their distances) and a maximum distance (`distanceThreshold`). We need to find the city for which the number of cities reachable within `distanceThreshold` is the smallest. If multiple cities satisfy this condition, we should return the city with the highest index.

Solution Approach

The provided solution uses the Floyd-Warshall algorithm to find the shortest path between all pairs of cities. Then, for each city, it counts the number of reachable cities within the given distance threshold. Finally, it returns the city with the smallest number of reachable cities (and the highest index in case of ties).

Step-by-Step Algorithm

Step 1: Initialize a distance matrix `dist` where `dist[i][j]` represents the shortest distance between city `i` and city `j`. Initially, set `dist[i][j]` to infinity (or a very large number) if there is no direct edge between `i` and `j`, and to the weight of the edge if there is a direct edge. Set `dist[i][i]` to 0.
Step 2: Populate the `dist` matrix with direct edge weights from the `edges` input.
Step 3: Apply the Floyd-Warshall algorithm: Iterate through all possible intermediate cities `k` from 0 to n-1. For each pair of cities `i` and `j`, update `dist[i][j]` to the minimum of its current value and `dist[i][k] + dist[k][j]`. This step finds the shortest path between all pairs of cities considering all possible intermediate nodes.
Step 4: Iterate through each city `i` from 0 to n-1 and count the number of reachable neighboring cities. A city `j` is reachable from city `i` if `i != j` and `dist[i][j] <= distanceThreshold`.
Step 5: Keep track of the city with the smallest number of reachable neighbors (`min_neighbors`) and its corresponding index (`result_city`). If a city has a smaller or equal number of reachable neighbors than the current minimum, update `min_neighbors` and `result_city`. Note that since we want the largest index in case of a tie, we only update if `reachable_neighbors <= min_neighbors` and the city index is greater than the current `result_city` or the `reachable_neighbors` is strictly less than `min_neighbors`
Step 6: Return the `result_city`.

Key Insights

Insight 1: The problem requires finding the shortest paths between all pairs of cities, which suggests using the Floyd-Warshall algorithm.
Insight 2: After calculating the shortest path between all pairs of cities, we can iterate through each city and count how many other cities are reachable within the `distanceThreshold`.
Insight 3: The city with the minimum number of reachable cities should be tracked. If a city has the same number of reachable cities as the current minimum, we should choose the city with the larger index.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

This problem involves: Dynamic Programming, Graph, Shortest Path.

Companies

Asked at: Expedia.

Find the City With the Smallest Number of Neighbors at a Threshold Distance | LeetCode 1334

Medium

Dynamic Programming Graph Shortest Path

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int findTheCity(int n, int[][] edges, int distanceThreshold) {
        int[][] dist = new int[n][n];

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dist[i][j] = Integer.MAX_VALUE;
            }
        }

        for (int i = 0; i < n; i++) {
            dist[i][i] = 0;
        }

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int w = edge[2];
            dist[u][v] = w;
            dist[v][u] = w;
        }

        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (dist[i][k] != Integer.MAX_VALUE && dist[k][j] != Integer.MAX_VALUE &&
                        dist[i][k] + dist[k][j] < dist[i][j]) {
                        dist[i][j] = dist[i][k] + dist[k][j];
                    }
                }
            }
        }

        int min_neighbors = Integer.MAX_VALUE;
        int result_city = -1;

        for (int i = 0; i < n; i++) {
            int reachable_neighbors = 0;
            for (int j = 0; j < n; j++) {
                if (i != j && dist[i][j] <= distanceThreshold) {
                    reachable_neighbors += 1;
                }
            }

            if (reachable_neighbors <= min_neighbors) {
                min_neighbors = reachable_neighbors;
                result_city = i;
            }
        }

        return result_city;
    }
}

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Problem Info

Description

Problem 1334: Find the City With the Smallest Number of Neighbors at a Threshold Distance - Detailed explanation and solution approaches.

Topics

Dynamic Programming Graph Shortest Path

Companies

Expedia

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1334

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

6 steps

1
Step 1: Initialize a distance matrix `dist` where `dist[i][j]` represents the shortest distance between city `i` and city `j`. Initially, set `dist[i][j]` to infinity (or a very large number) if there is no direct edge between `i` and `j`, and to the weight of the edge if there is a direct edge. Set `dist[i][i]` to 0.
2
Step 2: Populate the `dist` matrix with direct edge weights from the `edges` input.
3
Step 3: Apply the Floyd-Warshall algorithm: Iterate through all possible intermediate cities `k` from 0 to n-1. For each pair of cities `i` and `j`, update `dist[i][j]` to the minimum of its current value and `dist[i][k] + dist[k][j]`. This step finds the shortest path between all pairs of cities considering all possible intermediate nodes.
4
Step 4: Iterate through each city `i` from 0 to n-1 and count the number of reachable neighboring cities. A city `j` is reachable from city `i` if `i != j` and `dist[i][j] <= distanceThreshold`.
5
Step 5: Keep track of the city with the smallest number of reachable neighbors (`min_neighbors`) and its corresponding index (`result_city`). If a city has a smaller or equal number of reachable neighbors than the current minimum, update `min_neighbors` and `result_city`. Note that since we want the largest index in case of a tie, we only update if `reachable_neighbors <= min_neighbors` and the city index is greater than the current `result_city` or the `reachable_neighbors` is strictly less than `min_neighbors`
6
Step 6: Return the `result_city`.

💡

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^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The Floyd-Warshall algorithm has three nested loops, each iterating `n` times, resulting in a time complexity of O(n^3). The distance matrix `dist` requires O(n^2) space to store the shortest distances between all pairs of cities.

📖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

Dijkstra's Algorithm

Instead of Floyd-Warshall, Dijkstra's algorithm could be used to find the shortest paths from each city to all other cities. Dijkstra's would be run `n` times, once for each starting city.

Time: O(n * (E + V)logV) or O(n^2 * log n) if using priority queue, O(n^3) if using simple array implementation, where E is number of edges and V is number of vertices (cities)Space: O(E + V)

Trade-offs: If the graph is sparse (E << V^2), Dijkstra's algorithm run `n` times can potentially be faster than Floyd-Warshall. However, in the worst case and for dense graphs, Floyd-Warshall is simpler to implement and its complexity is comparable.

Bellman-Ford Algorithm

Bellman-Ford algorithm could be used to find the shortest paths from each city to all other cities. Bellman-Ford would be run `n` times, once for each starting city.

Time: O(n * n * E) where E is number of edges and n is number of vertices (cities)Space: O(V)

Trade-offs: Bellman-Ford Algorithm can detect negative weight cycle but its time complexity is not as good as Floyd Warshall.

✨

Key Insights

1Insight 1: The problem requires finding the shortest paths between all pairs of cities, which suggests using the Floyd-Warshall algorithm.
2Insight 2: After calculating the shortest path between all pairs of cities, we can iterate through each city and count how many other cities are reachable within the `distanceThreshold`.
3Insight 3: The city with the minimum number of reachable cities should be tracked. If a city has the same number of reachable cities as the current minimum, we should choose the city with the larger index.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

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

Edge Cases

Edge case 1: No edges are provided. The algorithm correctly handles this by initializing the distance matrix with infinity and then setting the distance from a city to itself to 0. In this scenario, no cities will be reachable within the threshold (unless the threshold is very large making all distances 0) and the algorithm will likely return the last city `n-1` or city with the smallest number of reachable cities given the implementation.
Edge case 2: `distanceThreshold` is very small (e.g., 0 or 1) compared to edge weights. In this case, most cities may not be reachable from each other. The algorithm correctly calculates and compares only the cities reachable within this small threshold.
Edge case 3: Negative edge weights are not allowed as per problem constraint. If present, Floyd-Warshall might result in incorrect shortest paths or negative cycles.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly initializing the distance matrix with infinity (or a large number) for unreachable cities. This can lead to incorrect shortest path calculations.

✗Common mistake 2: Not considering the condition `i != j` when counting reachable neighbors, which will incorrectly include the city itself in the count.

✗Common mistake 3: Not handling the case where multiple cities have the same minimum number of reachable neighbors. The problem requires returning the city with the largest index in this case.

✅

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: Finding the optimal location for a distribution center in a city to minimize the average distance to customers. The cities represent locations, edges represent roads, and weights represent distances.

🏦Real-world scenario 2: Network routing, where the algorithm helps determine the closest server to a client based on network latency or bandwidth.

🎮Real-world scenario 3: Social network analysis, determining the most central person (vertex) in a social network based on connectivity.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: August 14, 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

❓

Frequently Asked Questions

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."

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

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 "Find the City With the Smallest Number of Neighbors at a Threshold Distance"?

For this medium-level Dynamic Programming 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.

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