Count ways to reach destination of given Graph in minimum time

Given an undirected graph with N nodes (numbered from 0 to N-1) such that any node can be reached from any other node and there are no parallel edges. Also, an integer array edges[] is given where each element is of the form {u, v, t} which represents an edge between u and v and it takes t time to move on the edge. The task is to find the number of ways to reach from node 0 to node N-1 in minimum time.

Examples:

Input: N = 7, M = 10, edges = [[0, 6, 7], [0, 1, 2], [1, 2, 3], [1, 3, 3], [6, 3, 3], [3, 5, 1], [6, 5, 1], [2, 5, 1], [0, 4, 5], [4, 6, 2]]
Output: 4
Explanation: The four ways to get there in 7 minutes are:
0->6
0->4->6
0->1->2->5->6
0->1->3->5->6

Input: N = 6, M = 8, edges = [[0, 5, 8], [0, 2, 2], [0, 1, 1], [1, 3, 3], [1, 2, 3], [2, 5, 6], [3, 4, 2], [4, 5, 2]]
Output: 3
Explanation: The three ways to get there in 8 minutes are:
0->5
0->2->5
0->1->3->4->5

Naive Approach: The problem can be solved using Dijkstra’s Algorithm based on the below idea:

For each node we can store the minimum time taken to reach from 0 and number of ways to reach the node. Say we keep an array time[], where time[i]  denotes the minimum time to travel from 0th node to ith node and another array ways[], where ways[i] denotes the number of way we can reach ith node from 0th node in minimum time. Consider time_taken(u, v) denotes time taken to go to v from u. Now there are two cases:

• If (time[u] + time_taken(u, v) < time[v])  , then ways[v] = ways[u] 
• If (time[u] + time_taken(u, v) == time[v]), then ways[v] = ways[v] + ways[u] 

Follow the below steps to solve the problem:

• First, we make an adjacency matrix from the roads array 
• Adjacency matrix is 2D matrix adj of size (V*V)  
• adj[u][v] stores the time it takes to go from u to 
• time[0] = 0  and ways[0] = 1
• Now we start from 0 node and start the Dijkstra algorithm
  • We find the node with minimum time to travel which is not visited yet, let’s take the node as a.
    • Run a loop through all the nodes which are not visited and keep track of the node with the smallest time to travel.
  • Then minimize the time of nodes that are adjacent to a:
    • we run a loop through all the nodes if there is a connection between a node and we check the following two conditions
  • If (time[u]+time_taken(u, v) < time[v])  , then ways[v] = ways[u] 
  • If (time[u] + time_taken(u, v) == time[v]), then ways[v] = ways[v] + ways[u]

Below is the implementation of the above approach: 

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to get the minimum position
int getmin(vector<long long int>& dist,
           vector<bool>& visited, int n)
{
    long long int x = LLONG_MAX, pos;
    for (int i = 0; i < n; i++) {
        if (!visited[i] && dist[i] < x) {
            x = dist[i];
            pos = i;
        }
    }
    return pos;
}

// Function to count number of paths
int countPaths(int n, vector<vector<int> >& roads)
{
    int mod = 1e9 + 7;
    long long int MMAX = 1e17;
    vector<vector<int> > adj(n, vector<int>(n, 0));
    for (auto road : roads) {
        int u = road[0];
        int v = road[1];
        int t = road[2];
        adj[u][v] = t;
        adj[v][u] = t;
    }
    vector<bool> visited(n, false);
    vector<long long int> time(n, MMAX);
    vector<long long int> ways(n, 0);
    time[0] = 0;
    ways[0] = 1;
    for (int i = 0; i < n - 1; i++) {

        // Find the intersection with minimum
        // time to travel
        int u = getmin(time, visited, n);
        visited[u] = true;
        for (int v = 0; v < n; v++) {
            if (!visited[v] && adj[u][v]
                && time[u] != MMAX) {
                if (time[u] + adj[u][v] < time[v]) {
                    time[v] = time[u] + adj[u][v];
                    ways[v] = ways[u];
                }
                else if (time[u] + adj[u][v] == time[v]) {
                    ways[v] = ways[v] + ways[u];
                    ways[v] %= mod;
                }
            }
        }
    }
    return ways[n - 1];
}

// Driver Code

int main()
{

    int N = 7, M = 10;
    vector<vector<int> > edges
        = { { 0, 6, 7 }, { 0, 1, 2 }, { 1, 2, 3 }, { 1, 3, 3 }, { 6, 3, 3 }, { 3, 5, 1 }, { 6, 5, 1 }, { 2, 5, 1 }, { 0, 4, 5 }, { 4, 6, 2 } };

    // Function Call
    cout << countPaths(N, edges) << endl;

    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
import java.util.*;

public class Main {

  // Function to get the minimum position
  static int getmin(long[] dist, boolean[] visited, int n)
  {
    long x = Long.MAX_VALUE;
    int pos = 0;
    for (int i = 0; i < n; i++) {
      if (!visited[i] && dist[i] < x) {
        x = dist[i];
        pos = i;
      }
    }
    return pos;
  }

  // Function to count number of paths
  static int countPaths(int n, List<List<Integer> > roads)
  {
    int mod = (int)1e7;
    int MMAX = (int)1e17;
    int[][] adj = new int[n][n];
    for (List<Integer> road : roads) {
      int u = road.get(0);
      int v = road.get(1);
      int t = road.get(2);
      adj[u][v] = t;
      adj[v][u] = t;
    }
    boolean[] visited = new boolean[n];
    long[] time = new long[n];
    Arrays.fill(time, MMAX);
    long[] ways = new long[n];
    time[0] = 0;
    ways[0] = 1;
    for (int i = 0; i < n - 1; i++) {

      // Find the intersection with minimum
      // time to travel
      int u = getmin(time, visited, n);
      visited[u] = true;
      for (int v = 0; v < n; v++) {
        if (!visited[v] && adj[u][v]!=0
            && time[u] != MMAX) {
          if (time[u] + adj[u][v] < time[v]) {
            time[v] = time[u] + adj[u][v];
            ways[v] = ways[u];
          }
          else if (time[u] + adj[u][v]
                   == time[v]) {
            ways[v] = ways[v] + ways[u];
            ways[v] %= mod;
          }
        }
      }
    }
    return (int)ways[n - 1];
  }

  public static void main(String[] args)
  {
    int N = 7, M = 10;
    List<List<Integer> > edges = new ArrayList<>();
    edges.add(Arrays.asList(0, 6, 7));
    edges.add(Arrays.asList(0, 1, 2));
    edges.add(Arrays.asList(1, 2, 3));
    edges.add(Arrays.asList(1, 3, 3));
    edges.add(Arrays.asList(6, 3, 3));
    edges.add(Arrays.asList(3, 5, 1));
    edges.add(Arrays.asList(6, 5, 1));
    edges.add(Arrays.asList(2, 5, 1));
    edges.add(Arrays.asList(0, 4, 5));
    edges.add(Arrays.asList(4, 6, 2));

    // Function Call
    System.out.println(countPaths(N, edges));
  }
}

// This code is contributed by lokesh

Python3

# Python implementation
import sys

# Function to get the minimum position
def getmin(dist, visited, n):
    x = sys.maxsize
    pos = -1
    for i in range(n):
        if not visited[i] and dist[i] < x:
            x = dist[i]
            pos = i
    return pos

#  Function to count number of paths
def countPaths(n, roads):
    mod = 10**9 + 7
    MMAX = 10**17
    adj = [[0] * n for _ in range(n)]
    for road in roads:
        u, v, t = road[0], road[1], road[2]
        adj[u][v] = t
        adj[v][u] = t
    visited = [False] * n
    time = [MMAX] * n
    ways = [0] * n
    time[0] = 0
    ways[0] = 1
    for i in range(n - 1):

      # Find the intersection with minimum
      # time to travel
        u = getmin(time, visited, n)
        visited[u] = True
        for v in range(n):
            if not visited[v] and adj[u][v] and time[u] != MMAX:
                if time[u] + adj[u][v] < time[v]:
                    time[v] = time[u] + adj[u][v]
                    ways[v] = ways[u]
                elif time[u] + adj[u][v] == time[v]:
                    ways[v] = (ways[v] + ways[u]) % mod
    return ways[n - 1]

N = 7
M = 10
edges = [[0, 6, 7], [0, 1, 2], [1, 2, 3], [1, 3, 3], [6, 3, 3], [3, 5, 1], [6, 5, 1], [2, 5, 1], [0, 4, 5], [4, 6, 2]]

print(countPaths(N, edges))

# This code is contributed by ksam24000

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;

public class GFG {

  // Function to get the minimum position
  static int getmin(long[] dist, bool[] visited, int n)
  {
    long x = long.MaxValue;
    int pos = 0;
    for (int i = 0; i < n; i++) {
      if (!visited[i] && dist[i] < x) {
        x = dist[i];
        pos = i;
      }
    }
    return pos;
  }

  // Function to count number of paths
  static int countPaths(int n, List<List<int> > roads)
  {
    int mod = (int)1e7;
    long MMAX = (long)1e17;
    int[, ] adj = new int[n, n];
    foreach(List<int> road in roads)
    {
      int u = road[0];
      int v = road[1];
      int t = road[2];
      adj[u, v] = t;
      adj[v, u] = t;
    }
    bool[] visited = new bool[n];
    long[] time = new long[n];
    Array.Fill(time, MMAX);
    long[] ways = new long[n];
    time[0] = 0;
    ways[0] = 1;
    for (int i = 0; i < n - 1; i++) {

      // Find the intersection with minimum
      // time to travel
      int u = getmin(time, visited, n);
      visited[u] = true;
      for (int v = 0; v < n; v++) {
        if (!visited[v] && adj[u, v] != 0
            && time[u] != MMAX) {
          if (time[u] + adj[u, v] < time[v]) {
            time[v] = time[u] + adj[u, v];
            ways[v] = ways[u];
          }
          else if (time[u] + adj[u, v]
                   == time[v]) {
            ways[v] = ways[v] + ways[u];
            ways[v] %= mod;
          }
        }
      }
    }
    return (int)ways[n - 1];
  }

  static public void Main()
  {

    // Code
    int N = 7, M = 10;
    List<List<int> > edges = new List<List<int> >();
    edges.Add(new List<int>() { 0, 6, 7 });
    edges.Add(new List<int>() { 0, 1, 2 });
    edges.Add(new List<int>() { 1, 2, 3 });
    edges.Add(new List<int>() { 1, 3, 3 });
    edges.Add(new List<int>() { 6, 3, 3 });
    edges.Add(new List<int>() { 3, 5, 1 });
    edges.Add(new List<int>() { 6, 5, 1 });
    edges.Add(new List<int>() { 2, 5, 1 });
    edges.Add(new List<int>() { 0, 4, 5 });
    edges.Add(new List<int>() { 4, 6, 2 });

    // Function Call
    Console.WriteLine(countPaths(N, edges));
  }
}

// This code is contributed by lokeshmvs21.

Javascript

// JavaScript implementation of the approach

// Function to get the minimum position
function getmin(dist, visited) {
    let x = Number.MAX_SAFE_INTEGER, pos;
    for (let i = 0; i < dist.length; i++) {
        if (!visited[i] && dist[i] < x) {
            x = dist[i];
            pos = i;
        }
    }
    return pos;
}

// Function to count number of paths
function countPaths(n, roads) {
    const mod = 1e9 + 7;
    const MMAX = 1e17;
    let adj = new Array(n);
    for (let i = 0; i < n; i++) {
        adj[i] = new Array(n).fill(0);
    }
    for (let i = 0; i < roads.length; i++) {
        let u = roads[i][0];
        let v = roads[i][1];
        let t = roads[i][2];
        adj[u][v] = t;
        adj[v][u] = t;
    }
    let visited = new Array(n).fill(false);
    let time = new Array(n).fill(MMAX);
    let ways = new Array(n).fill(0);
    time[0] = 0;
    ways[0] = 1;
    for (let i = 0; i < n - 1; i++) {

        // Find the intersection with minimum
        // time to travel
        let u = getmin(time, visited);
        visited[u] = true;
        for (let v = 0; v < n; v++) {
            if (!visited[v] && adj[u][v] && time[u] !== MMAX) {
                if (time[u] + adj[u][v] < time[v]) {
                    time[v] = time[u] + adj[u][v];
                    ways[v] = ways[u];
                }
                else if (time[u] + adj[u][v] === time[v]) {
                    ways[v] = (ways[v] + ways[u]) % mod;
                }
            }
        }
    }
    return ways[n - 1];
}

// Example usage
let N = 7, M = 10;
let edges = [ [0, 6, 7], [0, 1, 2], [1, 2, 3], [1, 3, 3],
[6, 3, 3], [3, 5, 1], [6, 5, 1], [2, 5, 1], [0, 4, 5], [4, 6, 2] ];
console.log(countPaths(N, edges));

Output

4

Time complexity: As we check the node with minimum time to travel each time it takes O(n) time and we have to run this algorithm (v-1) times to find the minimum time of every nodes. So the total time complexity will be O(n*(n – 1)) ? O(n²).
Auxiliary Space: As we keep track of three extra arrays(time[], ways[], visited[]) and an adjacency matrix the space complexity is O(v²+3*v).  

Efficient approach: To solve the problem follow the below idea:

As discussed in previous approach to get the intersection with smallest travel time we can now get that efficiently by using a min heap . As we start minimizing the travel time we add those intersections in a priority queue and take the intersection with minimum time to travel.

Follow the below steps to solve the problem:

• First, we make an adjacency list from the edges array
• Now we create a min heap which stores the {minimum_time_to_travel_intersection, intersection_number}
• Now every time we takeout the intersection with minimum_time_to_travel and minimize its adjacent intersections time 
• If (time[u]+time_taken(u, v) < time[v])  , then ways[v] = ways[u]
• If (time[u] + time_taken(u, v) == time[v]), then ways[v] = ways[v] + ways[u] 

Below is the implementation of the above approach: 

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to count number of paths
int countPaths(int n, vector<vector<int> >& roads)
{
    int mod = 1e9 + 7;
    vector<vector<pair<int, int> > > graph(n);
    for (auto& road : roads) {
        graph[road[0]].push_back({ road[1], road[2] });
        graph[road[1]].push_back({ road[0], road[2] });
    }

    vector<long long> distance(n, LONG_MAX);
    vector<int> path(n, 0);

    priority_queue<pair<long long, int>,
                   vector<pair<long long, int> >,
                   greater<pair<long long, int> > >
        pq;
    pq.push({ 0, 0 });
    distance[0] = 0;
    path[0] = 1;

    while (!pq.empty()) {
        pair<long long, int> t = pq.top();
        pq.pop();

        for (auto& nbr : graph[t.second]) {
            long long vert = nbr.first;
            long long edge = nbr.second;

            if (distance[vert]
                > distance[t.second] + edge) {
                distance[vert] = distance[t.second] + edge;
                pq.push({ distance[vert], vert });
                path[vert] = path[t.second] % mod;
            }
            else if (distance[vert] == t.first + edge) {
                path[vert] += path[t.second];
                path[vert] %= mod;
            }
        }
    }

    return path[n - 1];
}

// Driver Code
int main()
{

    int n = 7, m = 10;

    vector<vector<int> > roads
        = { { 0, 6, 7 }, { 0, 1, 2 }, { 1, 2, 3 }, { 1, 3, 3 }, { 6, 3, 3 }, { 3, 5, 1 }, { 6, 5, 1 }, { 2, 5, 1 }, { 0, 4, 5 }, { 4, 6, 2 } };

    // Function call
    cout << countPaths(n, roads) << endl;

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

    static final int mod = (int)(1e9 + 7);

    // Function to count number of paths
    static int countPaths(int n, int[][] roads)
    {
        List<List<int[]> > graph = new ArrayList<>();
        for (int i = 0; i < n; i++)
            graph.add(new ArrayList<>());
        for (int[] road : roads) {
            graph.get(road[0]).add(
                new int[] { road[1], road[2] });
            graph.get(road[1]).add(
                new int[] { road[0], road[2] });
        }

        long[] distance = new long[n];
        Arrays.fill(distance, Long.MAX_VALUE);
        int[] path = new int[n];
        PriorityQueue<long[]> pq = new PriorityQueue<>(
            (a, b) -> (int)(a[0] - b[0]));
        pq.offer(new long[] { 0, 0 });
        distance[0] = 0;
        path[0] = 1;

        while (!pq.isEmpty()) {
            long[] t = pq.poll();
            for (int[] nbr : graph.get((int)t[1])) {
                int vert = nbr[0];
                int edge = nbr[1];

                if (distance[vert]
                    > distance[(int)t[1]] + edge) {
                    distance[vert]
                        = distance[(int)t[1]] + edge;
                    pq.offer(new long[] { distance[vert],
                                          vert });
                    path[vert] = path[(int)t[1]] % mod;
                }
                else if (distance[vert] == t[0] + edge) {
                    path[vert]
                        = (path[vert] + path[(int)t[1]])
                          % mod;
                }
            }
        }

        return path[n - 1];
    }

    public static void main(String[] args)
    {
        int n = 7, m = 10;

        int[][] roads
            = { { 0, 6, 7 }, { 0, 1, 2 }, { 1, 2, 3 },
                { 1, 3, 3 }, { 6, 3, 3 }, { 3, 5, 1 },
                { 6, 5, 1 }, { 2, 5, 1 }, { 0, 4, 5 },
                { 4, 6, 2 } };

        // Function call
        System.out.println(countPaths(n, roads));
    }
}

// This code is contributed by karthik.

Python3

import heapq

# Function to count number of paths
def countPaths(n, roads):
    mod = int(1e9 + 7)
    graph = [[] for i in range(n)]
    for road in roads:
        graph[road[0]].append((road[1], road[2]))
        graph[road[1]].append((road[0], road[2]))

    distance = [float('inf') for i in range(n)]
    path = [0 for i in range(n)]

    heap = [(0, 0)]
    heapq.heapify(heap)
    distance[0] = 0
    path[0] = 1

    while heap:
        t = heapq.heappop(heap)

        for nbr in graph[t[1]]:
            vert = nbr[0]
            edge = nbr[1]

            if distance[vert] > distance[t[1]] + edge:
                distance[vert] = distance[t[1]] + edge
                heapq.heappush(heap, (distance[vert], vert))
                path[vert] = path[t[1]] % mod
            elif distance[vert] == t[0] + edge:
                path[vert] += path[t[1]]
                path[vert] %= mod

    return path[n - 1]

# Driver Code
if __name__ == "__main__":

    n = 7
    m = 10
    roads = [[0, 6, 7], [0, 1, 2], [1, 2, 3], [1, 3, 3], [6, 3, 3], [3, 5, 1], [6, 5, 1], [2, 5, 1], [0, 4, 5], [4, 6, 2]]

    # Function call
    print(countPaths(n, roads))

# This code is contributed by Vikram_Shirsat

C#

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;

// C# code to implement the approach
class HelloWorld {

  // custom sort function.
  class GFG : IComparer<KeyValuePair<int,int>>
  {
    public int Compare(KeyValuePair<int,int> x, KeyValuePair<int,int> y)
    {
      if (x.Key == y.Key)
      {
        return x.Value.CompareTo(y.Value);
      }

      // CompareTo() method
      return x.Key.CompareTo(y.Key);

    }
  }

  // Function to count number of paths
  static int countPaths(int n, int m, int[][] road)
  {
    int mod = 1000000007;
    List<KeyValuePair<int,int>>[] graph = new List<KeyValuePair<int,int>>[n];
    for(int i = 0; i < n; i++){
      graph[i] = new List<KeyValuePair<int, int>>();
    }

    for(int i = 0; i < m; i++){
      graph[road[i][0]].Add(new KeyValuePair<int,int> (road[i][1], road[i][2]));
      graph[road[i][1]].Add(new KeyValuePair<int,int> (road[i][0], road[i][2]));
    }

    int[] distance = new int[n];
    for(int i = 0; i < n; i++){
      distance[i] = 500000;
    }

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


    List<KeyValuePair<int,int>> pq = new List<KeyValuePair<int,int>>();

    pq.Add(new KeyValuePair<int, int>(0, 0));
    distance[0] = 0;
    path[0] = 1;
    GFG gg = new GFG();

    while (pq.Count > 0) {
      KeyValuePair<int, int> t = pq[0];
      pq.RemoveAt(0);

      for(int i = 0; i < graph[t.Value].Count; i++){

        KeyValuePair<int,int> nbr = graph[t.Value][i];
        int vert = nbr.Key;
        int edge = nbr.Value;

        if (distance[vert] > distance[t.Value] + edge) {
          distance[vert] = distance[t.Value] + edge;
          pq.Add(new KeyValuePair<int, int> (distance[vert], vert ));
          pq.Sort(gg);
          path[vert] = path[t.Value] % mod;
        }
        else if (distance[vert] == t.Key + edge) {
          path[vert] += path[t.Value];
          path[vert] %= mod;
        }
      }
    }

    return path[n - 1];
  }

  static void Main() {

    int n = 7;
    int m = 10;

    int[][] roads = new int[][] {
      new int[] { 0, 6, 7 },
      new int[] { 0, 1, 2 },
      new int[] { 1, 2, 3 },
      new int[] { 1, 3, 3 },
      new int[] { 6, 3, 3 },
      new int[] { 3, 5, 1 },
      new int[] { 6, 5, 1 },
      new int[] {2, 5, 1 },
      new int[] {0, 4, 5},
      new int[] {4, 6, 2}
    };

    // Function call
    Console.WriteLine(countPaths(n, m, roads));
  }
}

// The code is contributed by Nidhi goel.

Javascript

// javascript code to implement the approach

// Function to count number of paths
function countPaths(n, roads)
{
    let mod = 1e9 + 7;
    let graph = new Array(n);
    for(let i = 0; i < n; i++){
        graph[i] = new Array();
    }
    for (let i = 0; i < roads.length; i++) {
        let road = roads[i];

        graph[road[0]].push([road[1], road[2]]);
        graph[road[1]].push([road[0], road[2]]);
    }

    let distance = new Array(n).fill(2000);
    let path = new Array(n).fill(0);

    let pq = [];

    pq.push([0,0]);
    distance[0] = 0;
    path[0] = 1;

    while (pq.length > 0) {
        let t = pq[0];
        pq.shift();

        for(let i = 0; i < graph[t[1]].length; i++){
            let nbr = graph[t[1]][i];
            let vert = nbr[0];
            let edge = nbr[1];

            if (distance[vert] > distance[t[1]] + edge) {
                distance[vert] = distance[t[1]] + edge;
                pq.push([distance[vert], vert]);
                pq.sort(function(x, y){
                        if(x[0] == y[0]){
                            return x[1] - y[1];
                        }

                        return x[0] - y[0];
                    }
                )
                path[vert] = path[t[1]] % mod;
            }
            else if (distance[vert] == t[0] + edge) {
                path[vert] += path[t[1]];
                path[vert] %= mod;
            }
        }
    }

    return path[n - 1];
}


// Driver Code
let n = 7, m = 10;
let roads =  [ [ 0, 6, 7 ], [ 0, 1, 2 ], [ 1, 2, 3 ], [ 1, 3, 3 ], [ 6, 3, 3 ], [ 3, 5, 1 ], [ 6, 5, 1 ], [ 2, 5, 1 ], [ 0, 4, 5 ], [ 4, 6, 2 ] ];

// Function call
console.log(countPaths(n, roads));

// The code is contributed by Nidhi goel.

Output

4

Time complexity: As we will traverse every edge and use min heap to store intersections, the Time complexity is O(M+NlogN) where M is the number of edges and N is the number of intersections.
Auxiliary Space: We will need an adjacency list that takes O(M+N) space and three arrays with length N. So we can tell that the space complexity will be O(M+N).