<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:
        connections = defaultdict(int)
        visited = set()

        for x,y in edges:
            connections[x] += 1
            connections[y] += 1
            visited.add((x,y))

        odd_nodes = []
        

        for n in connections:
            if connections[n] % 2 == 1:
                odd_nodes.append(n)
        
        print(odd_nodes)

        if len(odd_nodes) > 4 or len(odd_nodes) % 2 == 1:
            return False
        if len(odd_nodes) == 0:
            return True
        if len(odd_nodes) == 2:
            if (odd_nodes[0], odd_nodes[1]) not in visited and (odd_nodes[1], odd_nodes[0]) not in visited:
                return True
            
            for n in range(1, n + 1):
                if connections[n] % 2 == 0:
                    if (odd_nodes[0], n) not in visited and (n, odd_nodes[0]) not in visited and (odd_nodes[1], n) not in visited and (n, odd_nodes[1]) not in visited:
                        return True
            return False

        if len(odd_nodes) == 4:
            for i in range(1,4):
                first = [0, i]
                second = [i for i in range(4) if i not in first]
                if (odd_nodes[first[0]], odd_nodes[first[1]]) not in visited and (odd_nodes[first[1]], odd_nodes[first[0]]) not in visited and (odd_nodes[second[0]], odd_nodes[second[1]]) not in visited and (odd_nodes[second[1]], odd_nodes[second[0]]) not in visited:
                    return True
            return False
        
        
        
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