<p>There is an <strong>undirected</strong> graph consisting of <code>n</code> nodes numbered from <code>1</code> to <code>n</code>. You are given the integer <code>n</code> and a <strong>2D</strong> array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. The graph can be disconnected.</p>

<p>You can add <strong>at most</strong> two additional edges (possibly none) to this graph so that there are no repeated edges and no self-loops.</p>

<p>Return <code>true</code><em> if it is possible to make the degree of each node in the graph even, otherwise return </em><code>false</code><em>.</em></p>

<p>The degree of a node is the number of edges connected to it.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/26/agraphdrawio.png" style="width: 500px; height: 190px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[1,2],[2,3],[3,4],[4,2],[1,4],[2,5]]
<strong>Output:</strong> true
<strong>Explanation:</strong> The above diagram shows a valid way of adding an edge.
Every node in the resulting graph is connected to an even number of edges.
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/26/aagraphdrawio.png" style="width: 400px; height: 120px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[1,2],[3,4]]
<strong>Output:</strong> true
<strong>Explanation:</strong> The above diagram shows a valid way of adding two edges.</pre>

<p><strong>Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/26/aaagraphdrawio.png" style="width: 150px; height: 158px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[1,2],[1,3],[1,4]]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is not possible to obtain a valid graph with adding at most 2 edges.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>

from collections import defaultdict

class Solution:
    def isPossible(self, n: int, edges: List[List[int]]) -> bool:

        '''
            an edge will always increase the degree by 2.
            can't have more than 4 vertices with odd edges
            b/w the 4 vertices can't have an edge between them
        '''

        graph = defaultdict(lambda: defaultdict(int))
        degrees = [0] * (n+1)

        for edge in edges:
            degrees[edge[0]] += 1
            degrees[edge[1]] += 1
            graph[edge[0]][edge[1]] = 1
            graph[edge[1]][edge[0]] = 1

        odd_degrees, even_degrees = [], []
        for i in range(1, n+1):
            vertex, degree = i, degrees[i]
            if degree % 2:
                odd_degrees.append((vertex, degree))
            else:
                even_degrees.append((vertex, degree))
        
        print(odd_degrees)
        print(even_degrees)
        if len(odd_degrees) > 4 or len(odd_degrees) % 2:
            return False
        
        if len(odd_degrees) == 2:
            v_1, v_2 = odd_degrees[0][0], odd_degrees[1][0]
            even_works = False
            for v, d in even_degrees:
                even_works = even_works or (not graph[v_1][v] and not graph[v_2][v])
            return not graph[v_1][v_2] or even_works
        
        if len(odd_degrees) == 4:
            vertices = [v for v, d in odd_degrees]

            v0, v1, v2, v3 = vertices[0], vertices[1], vertices[2], vertices[3]

            c1 = not graph[v0][v1] and not graph[v2][v3]
            c2 = not graph[v0][v2] and not graph[v1][v3]
            c3 = not graph[v0][v3] and not graph[v1][v2]

            return c1 or c2 or c3
        
        return True
            



        

        
        return True







    
        
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