Given a sequence of non-negative integers arr[], the task is to check if there exists a simple graph corresponding to this degree sequence. Note that a simple graph is a graph with no self-loops and parallel edges.
Examples:
Input: arr[] = {3, 3, 3, 3}
Output: Yes
This is actually a complete graph(K4)Input: arr[] = {3, 2, 1, 0}
Output: No
A vertex has degree n-1 so it’s connected to all the other n-1 vertices.
But another vertex has degree 0 i.e. isolated. It’s a contradiction.
Approach: One way to check the existence of a simple graph is by Havel-Hakimi algorithm given below:
- Sort the sequence of non-negative integers in non-increasing order.
- Delete the first element(say V). Subtract 1 from the next V elements.
- Repeat 1 and 2 until one of the stopping conditions is met.
Stopping conditions:
- All the elements remaining are equal to 0 (Simple graph exists).
- Negative number encounter after subtraction (No simple graph exists).
- Not enough elements remaining for the subtraction step (No simple graph exists).
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function that returns true if // a simple graph exists bool graphExists(vector< int > &a, int n)
{ // Keep performing the operations until one
// of the stopping condition is met
while (1)
{
// Sort the list in non-decreasing order
sort(a.begin(), a.end(), greater<>());
// Check if all the elements are equal to 0
if (a[0] == 0)
return true ;
// Store the first element in a variable
// and delete it from the list
int v = a[0];
a.erase(a.begin() + 0);
// Check if enough elements
// are present in the list
if (v > a.size())
return false ;
// Subtract first element from next v elements
for ( int i = 0; i < v; i++)
{
a[i]--;
// Check if negative element is
// encountered after subtraction
if (a[i] < 0)
return false ;
}
}
} // Driver Code int main()
{ vector< int > a = {3, 3, 3, 3};
int n = a.size();
graphExists(a, n) ? cout << "Yes" :
cout << "NO" << endl;
return 0;
} // This code is contributed by // sanjeev2552 |
// Java implementation of the approach import java.util.*;
@SuppressWarnings ( "unchecked" )
class GFG{
// Function that returns true if // a simple graph exists static boolean graphExists(ArrayList a, int n)
{ // Keep performing the operations until one
// of the stopping condition is met
while ( true )
{
// Sort the list in non-decreasing order
Collections.sort(a, Collections.reverseOrder());
// Check if all the elements are equal to 0
if (( int )a.get( 0 ) == 0 )
return true ;
// Store the first element in a variable
// and delete it from the list
int v = ( int )a.get( 0 );
a.remove(a.get( 0 ));
// Check if enough elements
// are present in the list
if (v > a.size())
return false ;
// Subtract first element from
// next v elements
for ( int i = 0 ; i < v; i++)
{
a.set(i, ( int )a.get(i) - 1 );
// Check if negative element is
// encountered after subtraction
if (( int )a.get(i) < 0 )
return false ;
}
}
} // Driver Code public static void main(String[] args)
{ ArrayList a = new ArrayList();
a.add( 3 );
a.add( 3 );
a.add( 3 );
a.add( 3 );
int n = a.size();
if (graphExists(a, n))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "NO" );
}
} } // This code is contributed by pratham76 |
# Python3 implementation of the approach # Function that returns true if # a simple graph exists def graphExists(a):
# Keep performing the operations until one
# of the stopping condition is met
while True :
# Sort the list in non-decreasing order
a = sorted (a, reverse = True )
# Check if all the elements are equal to 0
if a[ 0 ] = = 0 and a[ len (a) - 1 ] = = 0 :
return True
# Store the first element in a variable
# and delete it from the list
v = a[ 0 ]
a = a[ 1 :]
# Check if enough elements
# are present in the list
if v> len (a):
return False
# Subtract first element from next v elements
for i in range (v):
a[i] - = 1
# Check if negative element is
# encountered after subtraction
if a[i]< 0 :
return False
# Driver code a = [ 3 , 3 , 3 , 3 ]
if (graphExists(a)):
print ( "Yes" )
else :
print ( "No" )
|
// C# implementation of the approach using System;
using System.Collections;
class GFG{
// Function that returns true if // a simple graph exists static bool graphExists(ArrayList a, int n)
{ // Keep performing the operations until one
// of the stopping condition is met
while ( true )
{
// Sort the list in non-decreasing order
a.Sort();
a.Reverse();
// Check if all the elements are equal to 0
if (( int )a[0] == 0)
return true ;
// Store the first element in a variable
// and delete it from the list
int v = ( int )a[0];
a.Remove(a[0]);
// Check if enough elements
// are present in the list
if (v > a.Count)
return false ;
// Subtract first element from
// next v elements
for ( int i = 0; i < v; i++)
{
a[i] = ( int )a[i] - 1;
// Check if negative element is
// encountered after subtraction
if (( int )a[i] < 0)
return false ;
}
}
} // Driver Code public static void Main( string [] args)
{ ArrayList a = new ArrayList(){ 3, 3, 3, 3 };
int n = a.Count;
if (graphExists(a, n))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "NO" );
}
} } // This code is contributed by rutvik_56 |
<script> // Javascript implementation of the approach // Function that returns true if // a simple graph exists function graphExists(a, n)
{ // Keep performing the operations until one
// of the stopping condition is met
while (1)
{
// Sort the list in non-decreasing order
a.sort((a, b) => b - a)
// Check if all the elements are equal to 0
if (a[0] == 0)
return true ;
// Store the first element in a variable
// and delete it from the list
var v = a[0];
a.shift();
// Check if enough elements
// are present in the list
if (v > a.length)
return false ;
// Subtract first element from next v elements
for ( var i = 0; i < v; i++)
{
a[i]--;
// Check if negative element is
// encountered after subtraction
if (a[i] < 0)
return false ;
}
}
} // Driver Code var a = [ 3, 3, 3, 3 ];
var n = a.length;
graphExists(a, n) ? document.write( "Yes" ):
document.write( "NO" );
// This code is contributed by rrrtnx </script> |
Output:
Yes
Time Complexity: O(
Auxiliary Space: O(1)