Check if the given Prufer sequence is valid or not

Given a Prufer sequence of N integers, the task is to check if the given sequence is a valid Prufer sequence or not.

Examples:

Input: arr[] = {4, 1, 3, 4} 
Output: Valid 
The tree is:
2----4----3----1----5
     |
     6                 

Input: arr[] = {4, 1, 7, 4} 
Output: Invalid 

Approach: Since we know the Prufer sequence is of length N – 2 where N is the number of vertices. Hence we need to check if the Prufer sequence consists of elements which are in the range [1, N].



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if
// given Prufer sequence is valid
bool isValidSeq(int a[], int n)
{
    int nodes = n + 2;
  
    // Iterate in the Prufer sequence
    for (int i = 0; i < n; i++) {
  
        // If out of range
        if (a[i] < 1 || a[i] > nodes)
            return false;
    }
  
    return true;
}
  
// Driver code
int main()
{
    int a[] = { 4, 1, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    if (isValidSeq(a, n))
        cout << "Valid";
    else
        cout << "Invalid";
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
  
// Function that returns true if
// given Prufer sequence is valid
static boolean isValidSeq(int []a, int n)
{
    int nodes = n + 2;
  
    // Iterate in the Prufer sequence
    for (int i = 0; i < n; i++) 
    {
  
        // If out of range
        if (a[i] < 1 || a[i] > nodes)
            return false;
    }
  
    return true;
}
  
// Driver code
public static void main (String[] args) 
{
    int a[] = { 4, 1, 3, 4 };
    int n = a.length;
    if (isValidSeq(a, n))
        System.out.println( "Valid");
    else
        System.out.print( "Invalid");
}
}
  
// This code is contributed by anuj_67..

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function that returns true if 
# given Prufer sequence is valid 
def isValidSeq(a, n) : 
  
    nodes = n + 2
  
    # Iterate in the Prufer sequence 
    for i in range(n) :
  
        # If out of range 
        if (a[i] < 1 or a[i] > nodes) :
            return False
      
    return True
  
# Driver code 
if __name__ == "__main__"
  
    a = [ 4, 1, 3, 4 ]; 
      
    n = len(a); 
      
    if (isValidSeq(a, n)) :
        print("Valid"); 
    else :
        print("Invalid"); 
          
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
      
class GFG 
{
  
  
// Function that returns true if
// given Prufer sequence is valid
static Boolean isValidSeq(int []a, int n)
{
    int nodes = n + 2;
  
    // Iterate in the Prufer sequence
    for (int i = 0; i < n; i++) 
    {
  
        // If out of range
        if (a[i] < 1 || a[i] > nodes)
            return false;
    }
  
    return true;
}
  
// Driver code
public static void Main (String[] args) 
{
    int []a = { 4, 1, 3, 4 };
    int n = a.Length;
    if (isValidSeq(a, n))
        Console.WriteLine( "Valid");
    else
    Console.WriteLine( "Invalid");
}
}
  
// This code has been contributed by 29AjayKumar

chevron_right


Output:

Valid


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m, 29AjayKumar, AnkitRai01



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.