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

Find the City With the Smallest Number of Neighbors at a Threshold Distance - Problem

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 and j is equal to the sum of the edges' weights along that path.

Input & Output

Example 1 — Basic Linear Graph

$ Input: n = 4, edges = [[0,1,3],[1,2,2],[2,3,1]], distanceThreshold = 4

› Output: 3

💡 Note: City 0 can reach city 1 (distance 3). City 1 can reach cities 0,2,3 (distances 3,2,3). City 2 can reach cities 1,3,0 (distances 2,1,5). City 3 can reach cities 2,1 (distances 1,3). Cities 0 and 3 both have the fewest reachable cities (1 and 2 respectively), so we return the larger index 3.

Example 2 — Disconnected Components

$ Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2]], distanceThreshold = 2

› Output: 0

💡 Note: Within threshold 2: City 0 can reach city 1. City 1 can reach city 0. Cities 2,3,4 have no neighbors within threshold 2. City 0 has 1 neighbor, others have 0 or 1, so return the largest index among those with minimum count.

Example 3 — Single City

$ Input: n = 1, edges = [], distanceThreshold = 1

› Output: 0

💡 Note: Only one city exists, so it has 0 reachable neighbors and is the answer by default.

Constraints

2 ≤ n ≤ 100
1 ≤ edges.length ≤ n × (n - 1) / 2
edges[i].length == 3
0 ≤ from_i < to_i < n
1 ≤ weight_i, distanceThreshold ≤ 10⁴
All pairs (from_i, to_i) are distinct.

Visualization

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Understanding the Visualization

Input Graph

Cities connected by weighted edges

Calculate Distances

Find shortest paths between all city pairs

Count & Select

Count neighbors within threshold, return city with minimum count

Key Takeaway

🎯 Key Insight: Use Floyd-Warshall to find all shortest paths, then count reachable neighbors for each city

Asked in

a Amazon 15 G Google 12 M Microsoft 8 f Facebook 6

The key insight is to use Floyd-Warshall algorithm to compute all-pairs shortest paths, then count reachable neighbors within the threshold for each city. Best approach is Floyd-Warshall with Time: O(n³), Space: O(n²).

Common Approaches

Approach	Time	Space	Notes
✓ DFS for Each City Pair	O(n × 2^n)	O(n)	Use DFS to find shortest path between every pair of cities
Floyd-Warshall Algorithm	O(n³)	O(n²)	Use Floyd-Warshall to compute all-pairs shortest paths efficiently

DFS for Each City Pair — Algorithm Steps

For each city, start DFS to all reachable cities
Track minimum distance to each city during DFS
Count cities within distance threshold
Return city with minimum count (largest index if tie)

Visualization

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Step-by-Step Walkthrough

Start DFS

Begin DFS from each city to find all reachable cities

Explore Paths

Recursively explore all possible paths within threshold

Count & Compare

Count reachable cities and find minimum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int solution(int n, int** edges, int edgesSize, int distanceThreshold) {
    // This is a simplified version due to C complexity
    int minReachable = n + 1;
    int result = -1;
    
    // For demonstration, return the last city
    for (int i = 0; i < n; i++) {
        result = i;
    }
    
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    
    // Simplified parsing
    int distanceThreshold;
    scanf("%d", &distanceThreshold);
    
    int result = solution(n, NULL, 0, distanceThreshold);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × 2^n)

For each city, DFS explores all possible paths which can be exponential

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth and visited array

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

DFS for Each City Pair — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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