<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>

# 5 - 6 - 4
    # |   |
    # | - 1


class Solution:
    def isPossible(self, n: int, edges: List[List[int]]) -> bool:
        def canConnect(n1, n2):
            return n2 not in adj[n1]
        adj = defaultdict(set)
        ind = defaultdict(int)
        
        for a,b in edges:
            adj[a].add(b)
            adj[b].add(a)
            ind[a] += 1
            ind[b] += 1
        
        evenNodes = [] # even nodes that can accept more edges
        oddNodes = []
        for node in range(1, n + 1):
            if ind[node] % 2 == 1:
                if ind[node] == n-1:
                    return False
                
                oddNodes.append(node)
            
            elif ind[node] < n - 2:
                evenNodes.append(node)
        
        print(oddNodes)
        print(evenNodes)

        if len(oddNodes) > 4 or len(oddNodes) == 1 or len(oddNodes) == 3:
            return False

        if len(oddNodes) == 4:
            return (
                (canConnect(oddNodes[0], oddNodes[1]) and canConnect(oddNodes[2], oddNodes[3])) or 
                (canConnect(oddNodes[0], oddNodes[2]) and canConnect(oddNodes[1], oddNodes[3])) or 
                (canConnect(oddNodes[0], oddNodes[3]) and canConnect(oddNodes[1], oddNodes[2]))  )

        if len(oddNodes) == 2:
            if canConnect(oddNodes[0], oddNodes[1]):
                return True
            
            for node in evenNodes:
                if canConnect(node, oddNodes[0]) and canConnect(node, oddNodes[1]):
                    return True
            return False


        return True
                

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