<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>

class Solution:
    def isPossible(self, n: int, edges: List[List[int]]) -> bool:
        
        degrees = defaultdict(int)
        adjList = defaultdict(list)
        n = 0 

        for start, end in edges:
            if start not in degrees:
                n += 1 

            if end not in degrees:
                n += 1 

            degrees[start] += 1 
            degrees[end] += 1 
            adjList[start].append(end)
            adjList[end].append(start)

        evenDegrees = 0 
        oddVertices = []
        evenVertices = []
        for i in degrees:
            if degrees[i] % 2 == 0:
                evenDegrees += 1 
                evenVertices.append(i)
            else:
                oddVertices.append(i)

        for i in range(1, n+1):
            if degrees[i] == 0:
                evenVertices.append(i)

        def checkValidThree(oddVertices, evenVertices):

            for i in evenVertices:
                if oddVertices[0] not in adjList[i] and oddVertices[1] not in adjList[i]:
                    return True

            return False



        def checkValidFour(oddVertices):

            first = oddVertices[0]
            second = oddVertices[1]
            third = oddVertices[2]
            fourth = oddVertices[3]

            if first not in adjList[second] and third not in adjList[fourth]:
                return True

            if first not in adjList[third] and second not in adjList[fourth]:
                return True

            if first not in adjList[fourth] and third not in adjList[second]:
                return True

            return False


        if evenDegrees == n:
            return True 

        elif evenDegrees == n - 2 and (oddVertices[0] not in adjList[oddVertices[1]] or checkValidThree(oddVertices, evenVertices)):
            return True 

        elif evenDegrees == n - 4 and checkValidFour(oddVertices):
            return True 

        return False
        

        
1302