Find if possible to visit every nodes in given Graph exactly once based on given conditions
Given a Graph having N+1 nodes with 2*N-1 edges and a boolean array arr[ ], the task is to find if one can visit every nodes exactly once and print any one possible path. It’s also known that N-1 edges go i-th to (i+1)-th node and rest edges go i-th to (n+1)-th node if arr[i] is 0 otherwise (n+1)-th to i-th.
Examples:
Input: N = 3, arr = {0, 1, 0}
Output: [1, 2, 3, 4]
Explanation:
from the given data, the edges will:
1 -> 2
2 -> 3
1 -> 4
4 -> 2
3 -> 4Therefore, one can follow the path: 1 -> 2 -> 3 -> 4
Input: N = 4, arr = {0, 1, 1, 1}
Output: [1, 4, 2, 3]
Approach: This problem can be solved using greedy approach with some observations. The observations are given below:
- If arr[1] is 1, then follow the path [N+1 -> 1 -> 2……..-> N].
- If arr[N] is 0, then follow the path [1 -> 2 ->……….-> N -> N+1].
- If arr[1] is 0 and arr[N] is 1, there must exists an integer i (1 ≤ i< N) where arr[i] =0 and arr[i+1] = 1, then the path [1 -> 2 -> ⋯ -> i -> N+1 -> i+1 -> i+2 -> ⋯N] is valid.
- Previous step proves that there always exists an Hamiltonian path in this graph.
Below is the implementation for the above approach:
C++
// C++ program for above approach.#include <bits/stdc++.h>using namespace std;// Function to find print pathvoid findpath(int N, int a[]){ // If a[0] is 1 if (a[0]) { // Printing path cout <<" "<< N + 1; for (int i = 1; i <= N; i++) cout <<" " << i; return; } // Seeking for a[i] = 0 and a[i+1] = 1 for (int i = 0; i < N - 1; i++) { if (!a[i] && a[i + 1]) { // Printing path for (int j = 1; j <= i; j++) cout <<" "<< j; cout <<" "<< N + 1; for (int j = i + 1; j <= N; j++) cout <<" "<< j; return; } } // If a[N-1] = 0 for (int i = 1; i <= N; i++) cout <<" "<< i; cout <<" "<< N + 1;}// Driver Codeint main(){ // Given Input int N = 3, arr[] = { 0, 1, 0 }; // Function Call findpath(N, arr);}// This code is contributed by shivanisinghss2110 |
C
// C++ program for above approach.#include <bits/stdc++.h>using namespace std;// Function to find print pathvoid findpath(int N, int a[]){ // If a[0] is 1 if (a[0]) { // Printing path printf("%d ", N + 1); for (int i = 1; i <= N; i++) printf("%d ", i); return; } // Seeking for a[i] = 0 and a[i+1] = 1 for (int i = 0; i < N - 1; i++) { if (!a[i] && a[i + 1]) { // Printing path for (int j = 1; j <= i; j++) printf("%d ", j); printf("%d ", N + 1); for (int j = i + 1; j <= N; j++) printf("%d ", j); return; } } // If a[N-1] = 0 for (int i = 1; i <= N; i++) printf("%d ", i); printf("%d ", N + 1);}// Driver Codeint main(){ // Given Input int N = 3, arr[] = { 0, 1, 0 }; // Function Call findpath(N, arr);} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to find print path static void findpath(int N, int a[]) { // If a[0] is 1 if (a[0] == 1) { // Printing path System.out.print((N + 1) + " "); for (int i = 1; i <= N; i++) System.out.print((i) + " "); return; } // Seeking for a[i] = 0 and a[i+1] = 1 for (int i = 0; i < N - 1; i++) { if (a[i] == 0 && a[i + 1] == 1) { // Printing path for (int j = 1; j <= i; j++) System.out.print((j) + " "); System.out.print((N + 1) + " "); for (int j = i + 1; j <= N; j++) System.out.print((j) + " "); return; } } // If a[N-1] = 0 for (int i = 1; i <= N; i++) System.out.print((i) + " "); System.out.print((N + 1) + " "); } // Driver Code public static void main(String[] args) { int N = 3, arr[] = { 0, 1, 0 }; // Function Call findpath(N, arr); }}// This code is contributed by dwivediyash |
Python3
# Python 3 program for above approach.# Function to find print pathdef findpath(N, a): # If a[0] is 1 if (a[0]): # Printing path print(N + 1) for i in range(1,N+1,1): print(i,end = " ") return # Seeking for a[i] = 0 and a[i+1] = 1 for i in range(N - 1): if (a[i]==0 and a[i + 1]): # Printing path for j in range(1,i+1,1): print(j,end = " ") print(N + 1,end = " "); for j in range(i + 1,N+1,1): print(j,end = " ") return # If a[N-1] = 0 for i in range(1,N+1,1): print(i,end = " ") print(N + 1,end = " ")# Driver Codeif __name__ == '__main__': # Given Input N = 3 arr = [0, 1, 0] # Function Call findpath(N, arr) # This code is contributed by ipg2016107. |
C#
// C# program for above approach.using System;class GFG{// Function to find print pathstatic void findpath(int N, int []a){ // If a[0] is 1 if (a[0]!=0) { // Printing path Console.Write(N + 1); for (int i = 1; i <= N; i++) Console.Write(i+ " "); return; } // Seeking for a[i] = 0 and a[i+1] = 1 for (int i = 0; i < N - 1; i++) { if (a[i]==0 && a[i + 1] !=0) { // Printing path for (int j = 1; j <= i; j++) Console.Write(j+" "); Console.Write(N + 1 + " "); for (int j = i + 1; j <= N; j++) Console.Write(j + " "); return; } } // If a[N-1] = 0 for (int i = 1; i <= N; i++) Console.Write(i+" "); Console.Write(N + 1 + " ");}// Driver Codepublic static void Main(){ // Given Input int N = 3; int []arr = { 0, 1, 0 }; // Function Call findpath(N, arr);}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find print path function findpath(N, a) { // If a[0] is 1 if (a[0]) { // Printing path document.write(N + 1); for (let i = 1; i <= N; i++) document.write(i); return; } // Seeking for a[i] = 0 and a[i+1] = 1 for (let i = 0; i < N - 1; i++) { if (!a[i] && a[i + 1]) { // Printing path for (let j = 1; j <= i; j++) document.write(j+" "); document.write(N + 1+" "); for (let j = i + 1; j <= N; j++) document.write(j+" "); return; } } // If a[N-1] = 0 for (let i = 1; i <= N; i++) document.write(i+" "); document.write(N + 1+" "); } // Driver Code // Given Input let N = 3, arr = [0, 1, 0]; // Function Call findpath(N, arr);// This code is contributed by Potta Lokesh </script> |
Output:
4 1 2 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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