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Find if possible to visit every nodes in given Graph exactly once based on given conditions

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  • Last Updated : 13 Sep, 2021
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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 -> 4

Therefore, 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 path
void 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 Code
int 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 path
void 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 Code
int main()
{
 
    // Given Input
    int N = 3, arr[] = { 0, 1, 0 };
 
    // Function Call
    findpath(N, arr);
}


Java




// Java program for the above approach
import 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 path
def 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 Code
if __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 path
static 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 Code
public 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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