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

• Last Updated : 26 Oct, 2021

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)

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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 ` `using` `namespace` `std;`   `// DFS function to find all possible paths.` `void` `dfs(vector >& 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 > 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> graph, ``int` `s, ` `                    ``int` `e, ``int``[] v, ``int` `count, Vector 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> graph = ``new` `Vector>();` `      ``for``(``int` `i = ``0``; i < n; i++)` `      ``{` `        ``graph.add(``new` `Vector());` `      ``}` `     `  `      ``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 dp = ``new` `Vector();` `     `  `      ``// 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> 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> graph = ``new` `List>();` `      ``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);` `  ``}` `}`   `// Ths code is contributed by decode2207.`

## Javascript

 ``

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:

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach` `#include ` `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 > graph, vector<``int``>& d,` `                 ``int` `v1, ``int` `v2, ``int` `n)` `{` `    ``// Vector to mark the nodes visited` `    ``vector<``int``> v(n, 0);`   `    ``// For BFS` `    ``queue > 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 > graph, vector<``int``>& d,` `          ``vector<``int``>& p, ``int` `n)` `{` `    ``// Vector to mark the nodes visited` `    ``vector<``int``> v(n, 0);`   `    ``// For BFS` `    ``queue > 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 > 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> 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> q = ``new` `List>();` `        ``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> 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> q = ``new` `List>();` `        ``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> graph = ``new` `List>();` `        ``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

 ``

Output

`1`

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

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