<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:

        G=defaultdict(set)
        uneven = set()

        def add_edge(a,b):
            G[a].add(b)
            G[b].add(a)

            for x in (a,b):
                if x in uneven:
                    uneven.discard(x)
                else:
                    uneven.add(x)
        
        def del_edge(a,b):
            G[a].discard(b)
            G[b].discard(a)
            for x in (a,b):
                if x in uneven:
                    uneven.discard(x)
                else:
                    uneven.add(x)

        for a,b in edges:
            add_edge(a,b)

        def backtrack(edges_used):
            uneven_count = len(uneven)

            if uneven_count == 0:
                return True
            
            elif uneven_count == 2:
                a,b = list(uneven)
                if not (b in G[a] and a in G[b]):
                    return True
                
                if edges_used != 0:
                    return False
                
                # find some node, not already connected to neither a or b
                for i in range(1, n+1):
                    if i not in (a,b) and not (i in G[a] or i in G[b]):
                        return True
                
                return False
            
            elif uneven_count == 4:
                for a,b in itertools.combinations(uneven, 2):
                    if (b in G[a] and a in G[b]):
                        continue
                    add_edge(a,b)
                    success = backtrack(1)
                    del_edge(a,b)
                    if success:
                        return True
                return False
                    
            return False
        
        return backtrack(0)

        
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