# Find if a degree sequence can form a simple graph | Havel-Hakimi Algorithm

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( )

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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 and a[a.size() - 1] == 0) ` `            ``return` `true``; ` ` `  `        ``// Store the first element in a variable ` `        ``// and delete it from the list ` `        ``int` `v = a; ` `        ``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 `

## Python3

 `# Python 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"``) `

Output:

```Yes
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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 : sanjeev2552