Difference between the shortest and second shortest path in an Unweighted Bidirectional Graph

Given an unweighted bidirectional graph containing N nodes and M edges represented by an array arr[][2]. The task is to find the difference in length of the shortest and second shortest paths from node 1 to N. If the second shortest path does not exist, print 0.

Note: The graph is connected, does not contain multiple edges and self loops. (2<=N<=20)

Examples:

Input: N = 4, M = 4, arr[M][2]={{1, 2}, {2, 3}, {3, 4}, {1, 4}}
Output: 2
Explanation: The shortest path is 1->4 and the second shortest path is 1->2->3->4. Hence, the difference is 2.

Input: N = 6, M = 8, arr[M][2]={{1, 2}, {1, 3}, {2, 6}, {2, 3}, {2, 4}, {3, 4}, {3, 5}, {4, 6}}
Output:1

Approach: The idea is to Depth First Search to find all possible paths and store them in vector and sort the vector and find the difference between the shortest and the second shortest path. Follow the steps below to solve the problem:

• Define a function dfs(vector<vector<int> >& graph, int s, int e, vector<int> vis, int count, vector<int>& dp) and perform the following steps:
  • If s is equal to e, then it means that the current path is one of the possible ones, push the value of count in the vector dp[] and return.
  • Iterate over the range [0, graph[s]] using the variable i and performing the following steps:
    • If vis[i] is not equal to 1, then set the value of vis[i] to 1 and call the function dfs(graph, i, e, vis, count+1, dp) to find other possible paths and set the value of vis[0] again back to 0.
• Initialize a 2-D vector graph[][] with N number of rows to store the vertices connected from each vertex.
• Iterate over the range [0, M] using the variable i and perform the following steps:
  • Push the value of b-1 in the vector graph[][] in row a-1.
  • Push the value of a-1 in the vector graph[][] in row b-1.
• Initialize a vector vis[] of size N to keep track of visited nodes.
• Initialize a vector dp[] to store the length of all possible paths.
• Call the function dfs(graph, 0, N-1, vis, 0, dp) to find all possible paths and store them in the vector dp[].
• Sort the vector dp[] in ascending order.
• If the size of vector dp[] is greater than 1, then return the value dp[1]-dp[0] else return 0 as the answer.

Below is the implementation of the above approach.

C++

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

// DFS function to find all possible paths.
void dfs(vector<vector<int> >& graph, int s, int e,
         vector<int> v, int count, vector<int>& dp)
{
    if (s == e) {
        // Push the number of nodes required for
        // one of the possible paths
        dp.push_back(count);
        return;
    }
    for (auto i : graph[s]) {
        if (v[i] != 1) {

            // Mark the node as visited and
            // call the function to search for
            // possible paths and unmark the node.
            v[i] = 1;
            dfs(graph, i, e, v, count + 1, dp);
            v[i] = 0;
        }
    }
}

// Function to find the difference between the
// shortest and second shortest path
void findDifference(int n, int m, int arr[][2])
{
    // Construct the graph
    vector<vector<int> > graph(n, vector<int>());
    for (int i = 0; i < m; i++) {
        int a, b;
        a = arr[i][0];
        b = arr[i][1];
        graph[a - 1].push_back(b - 1);
        graph[b - 1].push_back(a - 1);
    }

    // Vector to mark the nodes as visited or not.
    vector<int> v(n, 0);

    // Vector to store the count of all possible paths.
    vector<int> dp;

    // Mark the starting node as visited.
    v[0] = 1;

    // Function to find all possible paths.
    dfs(graph, 0, n - 1, v, 0, dp);

    // Sort the vector
    sort(dp.begin(), dp.end());

    // Print the difference
    if (dp.size() != 1)
        cout << dp[1] - dp[0];
    else
        cout << 0;
}

// Driver Code
int main()
{
    int n, m;
    n = 6;
    m = 8;
    int arr[m][2]
        = { { 1, 2 }, { 1, 3 },
            { 2, 6 }, { 2, 3 },
            { 2, 4 }, { 3, 4 },
            { 3, 5 }, { 4, 6 } };

    findDifference(n, m, arr);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // DFS function to find all possible paths.
    static void dfs(Vector<Vector<Integer>> graph, int s,
                    int e, int[] v, int count, Vector<Integer> dp) {
      if (s == e)
      {

        // Push the number of nodes required for
        // one of the possible paths
        dp.add(count);
        return;
      }
      for(int i : graph.get(s)) {
        if (v[i] != 1)
        {

          // Mark the node as visited and
          // call the function to search for
          // possible paths and unmark the node.
          v[i] = 1;
          dfs(graph, i, e, v, count + 1, dp);
          v[i] = 0;
        }
      }
    }

    // Function to find the difference between the
    // shortest and second shortest path
    static void findDifference(int n, int m, int[][] arr)
    {

      // Construct the graph
      Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
      for(int i = 0; i < n; i++)
      {
        graph.add(new Vector<Integer>());
      }

      for (int i = 0; i < m; i++) {
        int a, b;
        a = arr[i][0];
        b = arr[i][1];
        graph.get(a - 1).add(b - 1);
        graph.get(b - 1).add(a - 1);
      }

      // Vector to mark the nodes as visited or not.
      int[] v = new int[n];
      Arrays.fill(v, 0);

      // Vector to store the count of all possible paths.
      Vector<Integer> dp = new Vector<Integer>();

      // Mark the starting node as visited.
      v[0] = 1;

      // Function to find all possible paths.
      dfs(graph, 0, n - 1, v, 0, dp);

      // Sort the vector
      Collections.sort(dp);

      // Print the difference
      if (dp.size() != 1) System.out.print(dp.get(1) - dp.get(0));
      else System.out.print(0);
    }

    public static void main(String[] args) {
        int n, m;
        n = 6;
        m = 8;
        int[][] arr
            = { { 1, 2 }, { 1, 3 },
                { 2, 6 }, { 2, 3 },
                { 2, 4 }, { 3, 4 },
                { 3, 5 }, { 4, 6 } };

        findDifference(n, m, arr);
    }
}

// This code is contributed by mukesh07.

Python3

# Python3 program for the above approach

# DFS function to find all possible paths.
def dfs(graph, s, e, v, count, dp):
  if (s == e):

    # Push the number of nodes required for
    # one of the possible paths
    dp.append(count)
    return
  for i in graph[s]:
    if (v[i] != 1):

      # Mark the node as visited and
      # call the function to search for
      # possible paths and unmark the node.
      v[i] = 1
      dfs(graph, i, e, v, count + 1, dp)
      v[i] = 0

# Function to find the difference between the
# shortest and second shortest path
def findDifference(n, m, arr):
  # Construct the graph
  graph = []
  for i in range(n):
      graph.append([])

  for i in range(m):
    a = arr[i][0]
    b = arr[i][1]
    graph[a - 1].append(b - 1)
    graph[b - 1].append(a - 1)

  # Vector to mark the nodes as visited or not.
  v = [0]*(n)

  # Vector to store the count of all possible paths.
  dp = []

  # Mark the starting node as visited.
  v[0] = 1

  # Function to find all possible paths.
  dfs(graph, 0, n - 1, v, 0, dp)

  # Sort the vector
  dp.sort()

  # Print the difference
  if (len(dp) != 1):
    print(dp[1] - dp[0], end = "")
  else:
    print(0, end = "")

# Driver Code
n = 6
m = 8
arr = [
  [1, 2],
  [1, 3],
  [2, 6],
  [2, 3],
  [2, 4],
  [3, 4],
  [3, 5],
  [4, 6],
]

findDifference(n, m, arr)

# This code is contributed by divyesh072019.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // DFS function to find all possible paths.
    static void dfs(List<List<int>> graph, int s, int e, List<int> v, int count, List<int> dp) {
      if (s == e)
      {

        // Push the number of nodes required for
        // one of the possible paths
        dp.Add(count);
        return;
      }
      foreach(int i in graph[s]) {
        if (v[i] != 1)
        {

          // Mark the node as visited and
          // call the function to search for
          // possible paths and unmark the node.
          v[i] = 1;
          dfs(graph, i, e, v, count + 1, dp);
          v[i] = 0;
        }
      }
    }

    // Function to find the difference between the
    // shortest and second shortest path
    static void findDifference(int n, int m, int[,] arr)
    {

      // Construct the graph
      List<List<int>> graph = new List<List<int>>();
      for(int i = 0; i < n; i++)
      {
          graph.Add(new List<int>());
      }

      for (int i = 0; i < m; i++) {
        int a, b;
        a = arr[i,0];
        b = arr[i,1];
        graph[a - 1].Add(b - 1);
        graph[b - 1].Add(a - 1);
      }

      // Vector to mark the nodes as visited or not.
      List<int> v = new List<int>();
      for(int i = 0; i < n; i++)
      {
          v.Add(0);
      }

      // Vector to store the count of all possible paths.
      List<int> dp = new List<int>();

      // Mark the starting node as visited.
      v[0] = 1;

      // Function to find all possible paths.
      dfs(graph, 0, n - 1, v, 0, dp);

      // Sort the vector
      dp.Sort();

      // Print the difference
      if (dp.Count != 1) Console.Write(dp[1] - dp[0]);
      else Console.Write(0);
    }

  static void Main() {
    int n, m;
    n = 6;
    m = 8;
    int[,] arr
        = { { 1, 2 }, { 1, 3 },
            { 2, 6 }, { 2, 3 },
            { 2, 4 }, { 3, 4 },
            { 3, 5 }, { 4, 6 } };

    findDifference(n, m, arr);
  }
}

// This code is contributed by decode2207.

Javascript

<script>
// Javascript program for the above approach

// DFS function to find all possible paths.
function dfs(graph, s, e, v, count, dp) {
  if (s == e)
  {

    // Push the number of nodes required for
    // one of the possible paths
    dp.push(count);
    return;
  }
  for (let i of graph[s]) {
    if (v[i] != 1)
    {

      // Mark the node as visited and
      // call the function to search for
      // possible paths and unmark the node.
      v[i] = 1;
      dfs(graph, i, e, v, count + 1, dp);
      v[i] = 0;
    }
  }
}

// Function to find the difference between the
// shortest and second shortest path
function findDifference(n, m, arr)
{

  // Construct the graph
  let graph = new Array(n).fill(0).map(() => []);

  for (let i = 0; i < m; i++) {
    let a, b;
    a = arr[i][0];
    b = arr[i][1];
    graph[a - 1].push(b - 1);
    graph[b - 1].push(a - 1);
  }

  // Vector to mark the nodes as visited or not.
  let v = new Array(n).fill(0);

  // Vector to store the count of all possible paths.
  let dp = [];

  // Mark the starting node as visited.
  v[0] = 1;

  // Function to find all possible paths.
  dfs(graph, 0, n - 1, v, 0, dp);

  // Sort the vector
  dp.sort((a, b) => a - b);

  // Print the difference
  if (dp.length != 1) document.write(dp[1] - dp[0]);
  else document.write(0);
}

// Driver Code
let n, m;
n = 6;
m = 8;
let arr = [
  [1, 2],
  [1, 3],
  [2, 6],
  [2, 3],
  [2, 4],
  [3, 4],
  [3, 5],
  [4, 6],
];

findDifference(n, m, arr);

// This code is contributed by gfgking.
</script>

Output

1

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Efficient Approach: Using the fact that the second shortest path can not contain all the edges same as that in the shortest path. Remove each edge of the shortest path one at a time and keep finding the shortest path, then one of them has to be the required second shortest path. Use Breadth First Search to find the solution optimally. Follow the steps below to solve the problem:

• Define a function get_edges(int s, vector<int>& edges, vector<int> p) and perform the following steps:
  • If s is equal to -1, then return.
  • Call the function get_edges(p[s], edges, p) to find all the edges along the shortest path.
  • Push the value of s in the vector edges[].
• Define a function dist_helper(vector<vector<int> > graph, vector<int>& d, int v1, int v2, int N) and perform the following steps:
  • Initialize the vector vis[] to keep track of visited nodes.
  • Initialize a queue of pairs q[] to do the Breadth First Search and find the path.
  • Enqueue the pair {0, 0} into the queue of pairs q[].
  • Mark the value of vis[0] as 1(visited).
  • Iterate in a while loop till the queue of pairs q[] is not empty.
    • Initialize a variable a as the front of the queue of pairs q[] and dequeue an element from the queue of pairs q[].
    • Iterate over the range [0, graph[a.first]] using the variable i and performing the following steps:
      • If i is equal to v1 and a.first is equal to v2 or i is equal to v2 and a.first is equal to v1, then continue.
      • If vis[i] is equal to 0, then set the value of d[i] as 1+a.second, p[i] as a.first and vis[i] as 1 and enqueue the pair {i, d[i]} into the queue of pairs q[].
• Define a function dist(vector<vector<int> > graph, vector<int>& d, vector<int> &p, int N) and perform the following steps:
  • Initialize the vector vis[] to keep track of visited nodes.
  • Initialize a queue of pairs q[] to do the Breadth First Search and find the path.
  • Enqueue the pair {0, 0} into the queue of pairs q[].
  • Mark the value of vis[0] as 1(visited).
  • Iterate in a while loop till the queue of pairs q[] is not empty.
    • Initialize the variable a as the front of the queue of pairs q[] and dequeue an element from the queue of pairs q[].
    • Iterate over the range [0, graph[a.first]] using the variable i and performing the following steps:
      • If vis[i] is equal to 0, then set the value of d[i] as 1+a.second and vis[i] as 1 and enqueue the pair {i, d[i]} into the queue of pairs q[].
• Initialize a 2-D vector graph[][] with N number of rows to store the vertices connected from each vertex.
• Iterate over the range [0, M] using the variable i and perform the following steps:
  • Push the value of b-1 in the vector graph[][] in row a-1.
  • Push the value of a-1 in the vector graph[][] in row b-1.
• Initialize the vectors p[] and d[] of size N to keep track of parent nodes and the length of the paths.
• Call the function dist(graph, d, p, N) to find the length of the shortest path.
• Initialize a vector distances[] to store the length of all possible paths.
• Push the value of d[N-1] in the vector distances[].
• Initialize a vector edges[] to get all the edges along the shortest path.
• Call the function get_edges(N-1, edges, p) to find all the edges along the shortest path.
• Iterate over the range [0, edges.size()-1] using the variable i and perform the following steps:
  • Call the function dist_helper(graph, d, edges[i], edges[i+1], N) to find all the edges along the shortest path.
  • Push the value of d[N-1] in the vector distances[].
• Sort the vector distances[] in ascending order.
• If the size of vector distances[] is greater than 1, then return the value distances[1]-distances[0] else return 0 as the answer.

Below is the implementation of the above approach.

C++

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

// Function to get all the edges in
// the shortest path
void get_edges(int s, vector<int>& edges, vector<int> p)
{
    if (s == -1)
        return;
    get_edges(p[s], edges, p);
    edges.push_back(s);
}

// Calculate the shortest distance
// after removing an edge between
// v1 and v2
void dist_helper(vector<vector<int> > graph, vector<int>& d,
                 int v1, int v2, int n)
{
    // Vector to mark the nodes visited
    vector<int> v(n, 0);

    // For BFS
    queue<pair<int, int> > q;
    q.push(make_pair(0, 0));
    v[0] = 1;

    // Iterate over the range
    while (!q.empty()) {
        auto a = q.front();
        q.pop();
        for (int i : graph[a.first]) {
            if ((i == v1 && a.first == v2)
                || (i == v2 && a.first == v1))
                continue;
            if (v[i] == 0) {
                d[i] = 1 + a.second;
                v[i] = 1;
                q.push(make_pair(i, d[i]));
            }
        }
    }
}

// Calculates the shortest distances and
// maintain a parent array
void dist(vector<vector<int> > graph, vector<int>& d,
          vector<int>& p, int n)
{
    // Vector to mark the nodes visited
    vector<int> v(n, 0);

    // For BFS
    queue<pair<int, int> > q;
    q.push(make_pair(0, 0));
    v[0] = 1;

    // Iterate over the range
    while (!q.empty()) {
        auto a = q.front();
        q.pop();
        for (int i : graph[a.first]) {
            if (v[i] == 0) {
                p[i] = a.first;
                d[i] = 1 + a.second;
                v[i] = 1;
                q.push(make_pair(i, d[i]));
            }
        }
    }
}

// Function to find the difference between the
// shortest and second shortest path
void findDifference(int n, int m, int arr[][2])
{

    // Initializing and constructing the graph
    vector<vector<int> > graph(n, vector<int>());
    for (int i = 0; i < m; i++) {
        int a, b;
        a = arr[i][0];
        b = arr[i][1];
        graph[a - 1].push_back(b - 1);
        graph[b - 1].push_back(a - 1);
    }

    // Initializing the arrays
    vector<int> p(n, -1);
    vector<int> d(n, 1e9);

    // Calculate the shortest path
    dist(graph, d, p, n);

    // Vector to store the lengths
    // of possible paths
    vector<int> distances;
    distances.push_back(d[n - 1]);

    vector<int> edges;

    // Get all the edges along the shortest path
    get_edges(n - 1, edges, p);

    // Iterate over the range
    for (int i = 0; i + 1 < edges.size(); i++) {
        // Calculate shortest distance after
        // removing the edge
        dist_helper(graph, d, edges[i], edges[i + 1], n);
        distances.push_back(d[n - 1]);
    }

    // Sort the paths in ascending order
    sort(distances.begin(), distances.end());
    if (distances.size() == 1)
        cout << 0 << endl;
    else
        cout << distances[1] - distances[0] << endl;
}

// Driver Code
int main()
{
    int n, m;
    n = 6;
    m = 8;
    int arr[m][2]
        = { { 1, 2 }, { 1, 3 },
            { 2, 6 }, { 2, 3 },
            { 2, 4 }, { 3, 4 },
            { 3, 5 }, { 4, 6 } };

    findDifference(n, m, arr);

    return 0;
}

Python3

# Python3 program for the above approach

edges, d, p = [], [], []

# Function to get all the edges in
# the shortest path
def get_edges(s):
    global edges, d, p
    if s == -1:
        return
    get_edges(p[s])
    edges.append(s)

# Calculate the shortest distance
# after removing an edge between
# v1 and v2
def dist_helper(graph, v1, v2, n):
    global edges, d, p
    # Vector to mark the nodes visited
    v = [0]*(n)

    # For BFS
    q = []
    q.append([0, 0])
    v[0] = 1

    # Iterate over the range
    while len(q) > 0:
        a = q[0]
        q.pop(0)
        for i in graph[a[0]]:
            if (i == v1 and a[0] == v2) or (i == v2 and a[0] == v1):
                continue
            if v[i] == 0:
                d[i] = 1 + a[1]
                v[i] = 1
                q.append([i, d[i]])

# Calculates the shortest distances and
# maintain a parent array
def dist(graph, n):
    global edges, d, p
    # Vector to mark the nodes visited
    v = [0]*(n)

    # For BFS
    q = []
    q.append([0, 0])
    v[0] = 1

    # Iterate over the range
    while len(q) > 0:
        a = q[0]
        q.pop(0)
        for i in graph[a[0]]:
            if v[i] == 0:
                p[i] = a[0]
                d[i] = 1 + a[1]
                v[i] = 1
                q.append([i, d[i]])

# Function to find the difference between the
# shortest and second shortest path
def findDifference(n, m, arr):
    global edges, d, p
    # Initializing and constructing the graph
    graph = []
    for i in range(n):
        graph.append([])
    for i in range(m):
        a = arr[i][0]
        b = arr[i][1]
        graph[a - 1].append(b - 1)
        graph[b - 1].append(a - 1)

    # Initializing the arrays
    p = [-1]*(n)
    d = [1e9]*(n)

    # Calculate the shortest path
    dist(graph, n)

    # Vector to store the lengths
    # of possible paths
    distances = []
    distances.append(d[n - 1])

    edges = []

    # Get all the edges along the shortest path
    get_edges(n - 1)

    # Iterate over the range
    i = 0
    while i + 1 < len(edges):

        # Calculate shortest distance after
        # removing the edge
        dist_helper(graph, edges[i], edges[i + 1], n)
        distances.append(d[n - 1])
        i+=1

    # Sort the paths in ascending order
    distances.sort()
    if len(distances) == 1:
        print(0)
    else:
        print(distances[1] - distances[0])

n = 6;
m = 8;
arr = [ [ 1, 2 ], [ 1, 3 ], [ 2, 6 ], [ 2, 3 ], [ 2, 4 ], [ 3, 4 ], [ 3, 5 ], [ 4, 6 ] ]

findDifference(n, m, arr)

# This code is contributed by suresh07.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

   static List<int> edges = new List<int>();
   static List<int> d = new List<int>();
   static List<int> p = new List<int>();

    // Function to get all the edges in
    // the shortest path
    static void get_edges(int s)
    {
        if (s == -1)
            return;
        get_edges(p[s]);
        edges.Add(s);
    }

    // Calculate the shortest distance
    // after removing an edge between
    // v1 and v2
    static void dist_helper(List<List<int>> graph, int v1, int v2, int n)
    {
        // Vector to mark the nodes visited
        List<int> v = new List<int>();
        for(int i = 0; i < n; i++)
        {
            v.Add(0);
        }

        // For BFS
        List<Tuple<int,int>> q = new List<Tuple<int,int>>();
        q.Add(new Tuple<int,int>(0, 0));
        v[0] = 1;

        // Iterate over the range
        while (q.Count > 0) {
            Tuple<int,int> a = q[0];
            q.RemoveAt(0);
            for (int i = 0; i < graph[a.Item1].Count; i++) {
                if ((graph[a.Item1][i] == v1 && a.Item1 == v2)
                    || (graph[a.Item1][i] == v2 && a.Item1 == v1))
                    continue;
                if (v[graph[a.Item1][i]] == 0) {
                    d[graph[a.Item1][i]] = 1 + a.Item2;
                    v[graph[a.Item1][i]] = 1;
                    q.Add(new Tuple<int,int>(graph[a.Item1][i], d[graph[a.Item1][i]]));
                }
            }
        }
    }

    // Calculates the shortest distances and
    // maintain a parent array
    static void dist(List<List<int>> graph, int n)
    {
        // Vector to mark the nodes visited
        List<int> v = new List<int>();
        for(int i = 0; i < n; i++)
        {
            v.Add(0);
        }

        // For BFS
        List<Tuple<int,int>> q = new List<Tuple<int,int>>();
        q.Add(new Tuple<int,int>(0, 0));
        v[0] = 1;

        // Iterate over the range
        while (q.Count > 0) {
            Tuple<int,int> a = q[0];
            q.RemoveAt(0);
            for (int i = 0; i < graph[a.Item1].Count; i++) {
                if (v[graph[a.Item1][i]] == 0) {
                    p[graph[a.Item1][i]] = a.Item1;
                    d[graph[a.Item1][i]] = 1 + a.Item2;
                    v[graph[a.Item1][i]] = 1;
                    q.Add(new Tuple<int,int>(graph[a.Item1][i], d[graph[a.Item1][i]]));
                }
            }
        }
    }

    // Function to find the difference between the
    // shortest and second shortest path
    static void findDifference(int n, int m, int[,] arr)
    {

        // Initializing and constructing the graph
        List<List<int>> graph = new List<List<int>>();
        for(int i = 0; i < n; i++)
        {
            graph.Add(new List<int>());
        }
        for (int i = 0; i < m; i++) {
            int a, b;
            a = arr[i,0];
            b = arr[i,1];
            graph[a - 1].Add(b - 1);
            graph[b - 1].Add(a - 1);
        }

        // Initializing the arrays
        for(int i = 0; i < n; i++)
        {
            p.Add(-1);
            d.Add(1000000000);
        }

        // Calculate the shortest path
        dist(graph, n);

        // Vector to store the lengths
        // of possible paths
        List<int> distances = new List<int>();
        distances.Add(d[n - 1]);

        // Get all the edges along the shortest path
        get_edges(n - 1);

        // Iterate over the range
        for (int i = 0; i + 1 < edges.Count; i++) {
            // Calculate shortest distance after
            // removing the edge
            dist_helper(graph, edges[i], edges[i + 1], n);
            distances.Add(d[n - 1]);
        }

        // Sort the paths in ascending order
        distances.Sort();
        if (distances.Count == 1)
        {
            Console.WriteLine(0);
        }
        else
        {
            Console.WriteLine((distances[1] - distances[0]));
        }
    }

  // Driver code
  static void Main() {
    int n, m;
    n = 6;
    m = 8;
    int[,] arr
        = { { 1, 2 }, { 1, 3 },
            { 2, 6 }, { 2, 3 },
            { 2, 4 }, { 3, 4 },
            { 3, 5 }, { 4, 6 } };

    findDifference(n, m, arr);
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>
    // Javascript program for the above approach

    let edges = [], d = [], p = [];

    // Function to get all the edges in
    // the shortest path
    function get_edges(s)
    {
        if (s == -1)
            return;
        get_edges(p[s]);
        edges.push(s);
    }

    // Calculate the shortest distance
    // after removing an edge between
    // v1 and v2
    function dist_helper(graph, v1, v2, n)
    {
        // Vector to mark the nodes visited
        let v = [];
        for(let i = 0; i < n; i++)
        {
            v.push(0);
        }

        // For BFS
        let q = [];
        q.push([0, 0]);
        v[0] = 1;

        // Iterate over the range
        while (q.length > 0) {
            let a = q[0];
            q.shift();
            for (let i = 0; i < graph[a[0]].length; i++) {
                if ((graph[a[0]][i] == v1 && a[0] == v2)
                    || (graph[a[0]][i] == v2 && a[0] == v1))
                    continue;
                if (v[graph[a[0]][i]] == 0) {
                    d[graph[a[0]][i]] = 1 + a[1];
                    v[graph[a[0]][i]] = 1;
                    q.push([graph[a[0]][i], d[graph[a[0]][i]]]);
                }
            }
        }
    }

    // Calculates the shortest distances and
    // maintain a parent array
    function dist(graph, n)
    {
        // Vector to mark the nodes visited
        let v = [];
        for(let i = 0; i < n; i++)
        {
            v.push(0);
        }

        // For BFS
        let q = [];
        q.push([0, 0]);
        v[0] = 1;

        // Iterate over the range
        while (q.length > 0) {
            let a = q[0];
            q.shift();
            for (let i = 0; i < graph[a[0]].length; i++) {
                if (v[graph[a[0]][i]] == 0) {
                    p[graph[a[0]][i]] = a[0];
                    d[graph[a[0]][i]] = 1 + a[1];
                    v[graph[a[0]][i]] = 1;
                    q.push([graph[a[0]][i], d[graph[a[0]][i]]]);
                }
            }
        }
    }

    // Function to find the difference between the
    // shortest and second shortest path
    function findDifference(n, m, arr)
    {

        // Initializing and constructing the graph
        let graph = [];
        for(let i = 0; i < n; i++)
        {
            graph.push([]);
        }
        for (let i = 0; i < m; i++) {
            let a, b;
            a = arr[i][0];
            b = arr[i][1];
            graph[a - 1].push(b - 1);
            graph[b - 1].push(a - 1);
        }

        // Initializing the arrays
        for(let i = 0; i < n; i++)
        {
            p.push(-1);
            d.push(1e9);
        }

        // Calculate the shortest path
        dist(graph, n);

        // Vector to store the lengths
        // of possible paths
        let distances = [];
        distances.push(d[n - 1]);

        // Get all the edges along the shortest path
        get_edges(n - 1);

        // Iterate over the range
        for (let i = 0; i + 1 < edges.length; i++) {
            // Calculate shortest distance after
            // removing the edge
            dist_helper(graph, edges[i], edges[i + 1], n);
            distances.push(d[n - 1]);
        }

        // Sort the paths in ascending order
        distances.sort(function(a, b){return a - b});
        if (distances.length == 1)
        {
            document.write(0 + "</br>");
        }
        else
        {
            document.write((distances[1] - distances[0]) + "</br>");
        }
    }

    let n, m;
    n = 6;
    m = 8;
    let arr
        = [ [ 1, 2 ], [ 1, 3 ],
            [ 2, 6 ], [ 2, 3 ],
            [ 2, 4 ], [ 3, 4 ],
            [ 3, 5 ], [ 4, 6 ] ];

    findDifference(n, m, arr);

// This code is contributed by rameshtravel07.
</script>

Output

1

Time Complexity: O(N*M)
Auxiliary Space: O(N)