<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:
        def checkValid(odds, connected):
            within_degrees = defaultdict(int)
            for node in odds:
                for other in odds:
                    if other != node and (other, node) in connected:
                        within_degrees[node] += 1
            ds = within_degrees.values()

            # nothing connecting the odd nodes, can do anything to even out
            if len(ds) == 0:
                return True

            # 2 scenarios: one node connects to all other odd nodes, not possible to even out
            #              three odd nodes form a clique, not possible to even out
            if max(ds) == 3:
                return False
            elif Counter(ds)[2] == 3:
                return False
            return True

        degrees = defaultdict(int)
        connected = defaultdict(bool)
        #adj_list = defaultdict(list)
        for edge in edges:
            s, t = edge
            degrees[s] += 1
            degrees[t] += 1
            connected[(s, t)] = True
            connected[(t, s)] = True
            #adj_list[s].append(t)
            #adj_list[t].append(s)
        
        # add in missing node so some odd nodes can connect to
        for i in range(1, n+1):
            if i not in degrees:
                degrees[i] = 0
        
        odds = [k for k, v in degrees.items() if v % 2 == 1]

        if len(odds) == 0:
            return True
        elif len(odds) == 2:
            s1, s2 = odds
            # case 1: no edge between s1 and s2, we can just make the connection
            if (s1, s2) not in connected:
                return True
            # if there is any even node that is not connecting to either odd nodes, then we can even out
            for node in degrees:
                if degrees[node] % 2 == 0 and (s1, node) not in connected and (s2, node) not in connected:
                    return True
            return False
        elif len(odds) == 4:
            # check validity for the subgraph
            if checkValid(odds, connected):
                return True
            return False
        else:
            return False
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