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Check if the given Prufer sequence is valid or not

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  • Difficulty Level : Expert
  • Last Updated : 30 May, 2022

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++




// 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;
}

Java




// 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..

Python3




# 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

C#




// 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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true if
// given Prufer sequence is valid
function isValidSeq( a, n)
{
    var nodes = n + 2;
 
    // Iterate in the Prufer sequence
    for (var i = 0; i < n; i++) {
 
        // If out of range
        if (a[i] < 1 || a[i] > nodes)
            return false;
    }
 
    return true;
}
 
// Driver code
var a = [     4, 1, 3, 4 ];
var n = a.length;
if (isValidSeq(a, n))
    document.write( "Valid" );
else
    document.write( "Invalid");
 
// This code is contributed by itsok.
</script>

Output: 

Valid

 

Time Complexity: O(n)

Auxiliary Space: O(1)


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