Count nodes with maximum reachable neighbours at a d distance

Given a graph with n nodes and m edges, each edges[i] = [u, v, weight] and d as the maximum distance to reach the neighbor nodes, the task is to find the total number of nodes with maximum reachable neighbors.

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

Example 1

Output: 2
Explanation: node 0 -> [node 1, node 2] 
node 1 -> [node 0, node 2, node 3] 
node 2 -> [node 0, node 1, node 3] 
node 3 -> [node 1, node 2] 

These are the nodes that are reachable with a maximum distance of d. Thereby out of these node1 and node2 can reach maximum neighbors (3 neighbors). So the answer is node1 and node2, i.e. 2. As we need to return the count of the total number of such nodes.

Approach: To solve the problem follow the below observations:

Here in the above graph, We can observe that Node 1 and 2 can reach to maximum no. of neighbors (here it is all the 3 other nodes). So total there 2 number of nodes that can reach the maximum no. of neighbors (here it is 3) with a limited distance of d (here it is 4).

Follow the below steps to solve the above approach:

• Create an adjacency list for the given graph.
• Run the Floyd warshall algorithm to find all sources shortest distance.
• Then iterate for every node against every adjacent node.
• Count the number of adjacent nodes that are reachable with less than or equal to distance d.
• Then, have a separate count of answers and maxCount (Maximum neighbors), comparing to it increment the answer counter.

Below is the Implementation of the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

int findTheCity(int n, vector<vector<int> >& edges, int d)
{

    // 2D matrix n*n
    vector<vector<int> > dist(n, vector<int>(n, 1e9));
    for (auto e : edges) {

        // nodes
        int u = e[0];
        int v = e[1];
        int d = e[2];

        // edge wt
        // bidirectional edges u-v and v-u
        dist[u][v] = d;
        dist[v][u] = d;
    }
    for (int i = 0; i < n; i++) {

        // Self loops
        dist[i][i] = 0;
    }

    // Floyd Warshall Algo
    for (int k = 0; k < n; k++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dist[i][j] = min(dist[i][j],
                                 dist[i][k] + dist[k][j]);
            }
        }
    }

    // To store the maximum neighbors
    // possible
    int maxCtr = INT_MIN;

    // Stores the count of total such nodes
    // having max neighbours with
    // atmax d distance
    int ansctr = 0;

    for (int node = 0; node < n; node++) {

        // How many neighbours within limit
        // for that node
        int ctr = 0;
        for (int adjNode = 0; adjNode < n; adjNode++) {
            if (dist[node][adjNode] <= d) {
                ctr++;
            }
        }
        if (ctr > maxCtr) {

            // Update as we got a node that
            // has higher neighbours
            maxCtr = ctr;
            ansctr = 1;
        }

        // Equal to max, so increment counter
        // as of now this is the max
        else if (ctr == maxCtr)
            ansctr++;
    }
    return ansctr;
}

// Driver Code
int main()
{

    // edges [u, v, wt]
    vector<vector<int> > edges = {
        { 0, 1, 3 }, { 1, 2, 1 }, { 1, 3, 4 }, { 2, 3, 1 }
    };

    // Maximum distance to reach the neighbour
    int d = 4;

    // Number of nodes in graph
    int n = 4;

    int ans = findTheCity(n, edges, d);
    cout << ans << "\n";
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;

public class Main {
    static int findTheCity(int n, int[][] edges, int d)
    {
        // 2D matrix n*n
        int[][] dist = new int[n][n];
        for (int i = 0; i < n; i++) {
            Arrays.fill(dist[i], Integer.MAX_VALUE / 2);
            // Self loops
            dist[i][i] = 0;
        }

        for (int[] e : edges) {
            // edge wt
            // bidirectional edges u-v and v-u
            int u = e[0], v = e[1], w = e[2];
            dist[u][v] = dist[v][u] = w;
        }
        // Floyd Warshall Algo
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    dist[i][j]
                        = Math.min(dist[i][j],
                                   dist[i][k] + dist[k][j]);
                }
            }
        }

        int maxCtr = Integer.MIN_VALUE;

        // Stores the count of total such nodes
        // having max neighbours with
        // atmax d distance
        int ansctr = 0;
        for (int node = 0; node < n; node++) {
            int ctr = 0;
            // How many neighbours within limit
            // for that node
            for (int adjNode = 0; adjNode < n; adjNode++) {
                if (dist[node][adjNode] <= d) {
                    ctr++;
                }
            }
            if (ctr > maxCtr) {
                // Update as we got a node that
                // has higher neighbours
                maxCtr = ctr;
                ansctr = 1;

            } // Equal to max, so increment counter
            // as of now this is the max
            else if (ctr == maxCtr) {
                ansctr++;
            }
        }
        return ansctr;
    }
    // Driver Code
    public static void main(String[] args)
    {
        // edges [u, v, wt]
        int[][] edges = { { 0, 1, 3 },
                          { 1, 2, 1 },
                          { 1, 3, 4 },
                          { 2, 3, 1 } };
        // Maximum distance to reach the neighbour
        int d = 4;

        // Number of nodes in graph
        int n = 4;
        int ans = findTheCity(n, edges, d);
        System.out.println(ans);
    }
}

Python3

from typing import List

def find_the_city(n: int, edges: List[List[int]], d: int) -> int:
    # Initialize a 2D list "dist" to store the distances between the nodes
    # Set all distances to 1e9 (a large value)
    dist = [[1e9] * n for _ in range(n)]

    # Set the distance between the nodes connected by an
    # edge to the weight of the edge
    for u, v, wt in edges:
        dist[u][v] = wt
        dist[v][u] = wt

    # Set the distance between a node and itself to 0
    for i in range(n):
        dist[i][i] = 0

    # Use the Floyd Warshall algorithm to find the shortest
    # distances between all pairs of nodes
    for k in range(n):
        for i in range(n):
            for j in range(n):
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

    # Initialize "max_ctr" to the minimum possible value
    max_ctr = float('-inf')

    # Initialize "ans_ctr" to 0
    ans_ctr = 0

    # Iterate through each node
    for node in range(n):

        # Initialize a counter "ctr" to count the number of
        # neighbors within the maximum distance "d"
        ctr = 0

        # Iterate through each adjacent node
        for adj_node in range(n):

            # If the distance between the current node and the
            # adjacent node is less than or equal to "d",
            # increment the counter
            if dist[node][adj_node] <= d:
                ctr += 1

        # If the number of neighbors is greater than the
        # current maximum number of neighbors,
        # update "max_ctr" and set "ans_ctr" to 1
        if ctr > max_ctr:
            max_ctr = ctr
            ans_ctr = 1

        # If the number of neighbors is equal to the
        # current maximum number of neighbors,
        # increment "ans_ctr"
        elif ctr == max_ctr:
            ans_ctr += 1

    # Return the final result
    return ans_ctr

# Driver Code
d=4
n=4
edges=[[0, 1, 3], [1, 2, 1], [1, 3, 4], [2, 3, 1]]
print(find_the_city(d,edges,n))
#This code is contributed by Potta Lokesh

Javascript

function findTheCity(n, edges, d) {

    // 2D matrix n*n
    let dist = new Array(n).fill(0).map(() => new Array(n).fill(Number.MAX_VALUE));
    for (let i = 0; i < edges.length; i++) {

        // nodes
        let u = edges[i][0];
        let v = edges[i][1];
        let w = edges[i][2];

        // edge wt
        // bidirectional edges u-v and v-u
        dist[u][v] = w;
        dist[v][u] = w;
    }
    for (let i = 0; i < n; i++) {

        // Self loops
        dist[i][i] = 0;
    }

    // Floyd Warshall Algo
    for (let k = 0; k < n; k++) {
        for (let i = 0; i < n; i++) {
            for (let j = 0; j < n; j++) {
                dist[i][j] = Math.min(dist[i][j],
                                 dist[i][k] + dist[k][j]);
            }
        }
    }

    // To store the maximum neighbors
    // possible
    let maxCtr = Number.MIN_SAFE_INTEGER;

    // Stores the count of total such nodes
    // having max neighbours with
    // atmax d distance
    let ansctr = 0;

    for (let node = 0; node < n; node++) {

        // How many neighbours within limit
        // for that node
        let ctr = 0;
        for (let adjNode = 0; adjNode < n; adjNode++) {
            if (dist[node][adjNode] <= d) {
                ctr++;
            }
        }
        if (ctr > maxCtr) {

            // Update as we got a node that
            // has higher neighbours
            maxCtr = ctr;
            ansctr = 1;
        }

        // Equal to max, so increment counter
        // as of now this is the max
        else if (ctr === maxCtr)
            ansctr++;
    }
    return ansctr;
}

// Example usage:
// edges [u, v, wt]
let edges = [
    [0, 1, 3], [1, 2, 1], [1, 3, 4], [2, 3, 1]
];

// Maximum distance to reach the neighbour
let d = 4;

// Number of nodes in graph
let n = 4;

let ans = findTheCity(n, edges, d);
console.log(ans);

C#

using System;

public class MainClass
{
    static int FindTheCity(int n, int[][] edges, int d)
    {
        // 2D matrix n*n
        int[][] dist = new int[n][];
        for (int i = 0; i < n; i++) {
            dist[i] = new int[n];
            Array.Fill(dist[i], int.MaxValue / 2);
            // Self loops
            dist[i][i] = 0;
        }

        for (int i = 0; i < edges.Length; i++) {
            // edge wt
            // bidirectional edges u-v and v-u
            int u = edges[i][0], v = edges[i][1], w = edges[i][2];
            dist[u][v] = dist[v][u] = w;
        }
        // Floyd Warshall Algo
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    dist[i][j]
                        = Math.Min(dist[i][j],
                                   dist[i][k] + dist[k][j]);
                }
            }
        }

        int maxCtr = int.MinValue;

        // Stores the count of total such nodes
        // having max neighbours with
        // atmax d distance
        int ansctr = 0;
        for (int node = 0; node < n; node++) {
            int ctr = 0;
            // How many neighbours within limit
            // for that node
            for (int adjNode = 0; adjNode < n; adjNode++) {
                if (dist[node][adjNode] <= d) {
                    ctr++;
                }
            }
            if (ctr > maxCtr) {
                // Update as we got a node that
                // has higher neighbours
                maxCtr = ctr;
                ansctr = 1;

            } // Equal to max, so increment counter
            // as of now this is the max
            else if (ctr == maxCtr) {
                ansctr++;
            }
        }
        return ansctr;
    }

    public static void Main()
    {
        // edges [u, v, wt]
        int[][] edges = { new int[] { 0, 1, 3 },
                          new int[] { 1, 2, 1 },
                          new int[] { 1, 3, 4 },
                          new int[] { 2, 3, 1 } };
        // Maximum distance to reach the neighbour
        int d = 4;

        // Number of nodes in graph
        int n = 4;
        int ans = FindTheCity(n, edges, d);
        Console.WriteLine(ans);
    }
}

Output

2

Time Complexity: O(M) + O(N^3) + O(N^2) = O(N^3) Dominant factor

• O(M) = O(N^2) as at max edges can be N^2
• O(N^3) Floyd warshall algorithm takes
• O(N^2) To find out the count of such nodes.

Auxiliary Space Complexity: O(N+M) + O(N^2)

• O(N+M) the adjacency list takes as such
• O(N^2) the distance 2D Vector / Matrix