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
Difference between the shortest and second shortest path in an Unweighted Bidirectional Graph using DFS:
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.
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 pathvoid 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 Codeint 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 approachimport 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 pathdef 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 Coden = 6m = 8arr = [ [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 approachusing 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 pathfunction 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 Codelet 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)
Difference between the shortest and second shortest path in an Unweighted Bidirectional Graph using BFS:
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 <bits/stdc++.h>using namespace std;// Function to get all the edges in// the shortest pathvoid 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 v2void 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 arrayvoid 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 pathvoid 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 Codeint 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 code for the above approachimport java.util.*;class Main { // Function to get all the edges in // the shortest path static void getEdges(int s, List<Integer> edges, int[] p) { if (s == -1) { return; } getEdges(p[s], edges, p); edges.add(s); } // Calculate the shortest distance // after removing an edge between // v1 and v2 static void distHelper(List<List<Integer> > graph, int[] d, int v1, int v2, int n) { boolean[] visited = new boolean[n]; Queue<int[]> q = new LinkedList<>(); q.offer(new int[] { 0, 0 }); visited[0] = true; while (!q.isEmpty()) { int[] a = q.poll(); for (int i : graph.get(a[0])) { if ((i == v1 && a[0] == v2) || (i == v2 && a[0] == v1)) { continue; } if (!visited[i]) { d[i] = 1 + a[1]; visited[i] = true; q.offer(new int[] { i, d[i] }); } } } } // Calculates the shortest distances and // maintain a parent array static void dist(List<List<Integer> > graph, int[] d, int[] p, int n) { // Vector to mark the nodes visited boolean[] visited = new boolean[n]; // For BFS Queue<int[]> q = new LinkedList<>(); q.offer(new int[] { 0, 0 }); visited[0] = true; while (!q.isEmpty()) { int[] a = q.poll(); for (int i : graph.get(a[0])) { if (!visited[i]) { p[i] = a[0]; d[i] = 1 + a[1]; visited[i] = true; q.offer(new int[] { i, d[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<Integer> > graph = new ArrayList<>(); for (int i = 0; i < n; i++) { graph.add(new ArrayList<>()); } for (int i = 0; i < m; i++) { int a = arr[i][0] - 1; int b = arr[i][1] - 1; graph.get(a).add(b); graph.get(b).add(a); } // Initializing the arrays int[] p = new int[n]; Arrays.fill(p, -1); int[] d = new int[n]; Arrays.fill(d, Integer.MAX_VALUE); // Calculate the shortest path dist(graph, d, p, n); // Vector to store the lengths // of possible paths List<Integer> distances = new ArrayList<>(); distances.add(d[n - 1]); List<Integer> edges = new ArrayList<>(); // Get all the edges along the shortest path getEdges(n - 1, edges, p); // Iterate over the range for (int i = 0; i + 1 < edges.size(); i++) { // Calculate shortest distance after // removing the edge distHelper(graph, d, edges.get(i), edges.get(i + 1), n); distances.add(d[n - 1]); } // Sort the paths in ascending order Collections.sort(distances); if (distances.size() == 1) { System.out.println(0); } else { System.out.println(distances.get(1) - distances.get(0)); } } // Driver Code public static void main(String[] args) { int 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); }} |
Python3
# Python3 program for the above approachedges, d, p = [], [], []# Function to get all the edges in# the shortest pathdef 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 v2def 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 arraydef 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 pathdef 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 approachusing 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)
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