Check if the given Prufer sequence is valid or not
Last Updated :
30 Mar, 2023
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++
#include <bits/stdc++.h>
using namespace std;
bool isValidSeq(int a[], int n)
{
int nodes = n + 2;
for (int i = 0; i < n; i++) {
if (a[i] < 1 || a[i] > nodes)
return false;
}
return true;
}
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
import java.io.*;
class GFG
{
static boolean isValidSeq(int []a, int n)
{
int nodes = n + 2;
for (int i = 0; i < n; i++)
{
if (a[i] < 1 || a[i] > nodes)
return false;
}
return true;
}
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");
}
}
|
Python3
def isValidSeq(a, n) :
nodes = n + 2;
for i in range(n) :
if (a[i] < 1 or a[i] > nodes) :
return False;
return True;
if __name__ == "__main__" :
a = [ 4, 1, 3, 4 ];
n = len(a);
if (isValidSeq(a, n)) :
print("Valid");
else :
print("Invalid");
|
C#
using System;
class GFG
{
static Boolean isValidSeq(int []a, int n)
{
int nodes = n + 2;
for (int i = 0; i < n; i++)
{
if (a[i] < 1 || a[i] > nodes)
return false;
}
return true;
}
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");
}
}
|
Javascript
<script>
function isValidSeq( a, n)
{
var nodes = n + 2;
for (var i = 0; i < n; i++) {
if (a[i] < 1 || a[i] > nodes)
return false;
}
return true;
}
var a = [ 4, 1, 3, 4 ];
var n = a.length;
if (isValidSeq(a, n))
document.write( "Valid" );
else
document.write( "Invalid");
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
Approach 2: Using Maps:
This implementation uses a std::map<int, int> to count the frequency of each element in the sequence. It then iterates through the sequence and checks that each element is within the valid range (1 to n+2), occurs exactly once in the sequence, and corresponds to a leaf node in the tree (which is represented by the sequence itself). If all these conditions are met, the sequence is considered valid.
C++
#include <bits/stdc++.h>
using namespace std;
bool isValidSeq(const vector<int>& seq)
{
int n = seq.size();
int nodes = n + 2;
map<int, int> freq;
for (int i = 0; i < n; i++) {
freq[seq[i]]++;
}
for (int i = 0; i < n; i++) {
if (seq[i] < 1 || seq[i] > nodes
|| freq[seq[i]] != 1)
return false;
freq[nodes - i]--;
}
return true;
}
int main()
{
vector<int> seq = { 4, 1, 3, 4 };
if (isValidSeq(seq))
cout << "Inalid";
else
cout << "Valid";
return 0;
}
|
Java
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main {
public static boolean isValidSeq(int[] seq)
{
int n = seq.length;
int nodes = n + 2;
Map<Integer, Integer> freq
= new HashMap<Integer, Integer>();
for (int i = 0; i < n; i++) {
int key = seq[i];
if (freq.containsKey(key)) {
freq.put(key, freq.get(key) + 1);
}
else {
freq.put(key, 1);
}
}
for (int i = 0; i < n; i++) {
if (seq[i] < 1 || seq[i] > nodes
|| freq.get(seq[i]) != 1) {
return false;
}
freq.put(nodes - i, freq.get(nodes - i) - 1);
}
return true;
}
public static void main(String[] args)
{
int[] seq = { 4, 1, 3, 4 };
if (isValidSeq(seq))
System.out.println("Invalid");
else
System.out.println("Valid");
}
}
|
Python3
def is_valid_seq(seq):
n = len(seq)
nodes = n + 2
freq = {}
for i in range(n):
key = seq[i]
freq[key] = freq.get(key, 0) + 1
for i in range(n):
if seq[i] < 1 or seq[i] > nodes or freq[seq[i]] != 1:
return False
freq[nodes - i] = freq.get(nodes - i, 0) - 1
return True
seq = [4, 1, 3, 4]
if is_valid_seq(seq):
print("Inalid")
else:
print("Valid")
|
Javascript
function isValidSeq(seq) {
const n = seq.length;
const nodes = n + 2;
const freq = {};
for (let i = 0; i < n; i++) {
const key = seq[i];
freq[key] = (freq[key] || 0) + 1;
}
for (let i = 0; i < n; i++) {
if (seq[i] < 1 || seq[i] > nodes || freq[seq[i]] !== 1) {
return false;
}
freq[nodes - i] = (freq[nodes - i] || 0) - 1;
}
return true;
}
const seq = [4, 1, 3, 4];
if (isValidSeq(seq)) {
console.log("Invalid");
} else {
console.log("Valid");
}
|
C#
using System;
using System.Collections.Generic;
class Program {
static bool IsValidSeq(List<int> seq)
{
int n = seq.Count;
int nodes = n + 2;
Dictionary<int, int> freq
= new Dictionary<int, int>();
for (int i = 0; i < n; i++) {
if (!freq.ContainsKey(seq[i])) {
freq[seq[i]] = 0;
}
freq[seq[i]]++;
}
for (int i = 0; i < n; i++) {
if (seq[i] < 1 || seq[i] > nodes
|| freq[seq[i]] != 1) {
return false;
}
freq[nodes - i]--;
}
return true;
}
static void Main(string[] args)
{
List<int> seq = new List<int>() { 4, 1, 3, 4 };
if (IsValidSeq(seq)) {
Console.WriteLine("Invalid");
}
else {
Console.WriteLine("Valid");
}
}
}
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...