<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:
        de = [0 for _ in range(n+1)]
        nei = [set() for _ in range(n+1)]
        for u, v in edges:
            de[u] += 1
            de[v] += 1
            nei[u].add(v)
            nei[v].add(u)
        
        odds = []
        for i, d in enumerate(de):
            if d & 1 == 1:
                odds.append(i)
        
        if not odds:
            return True
        if len(odds) > 4:
            return False
        if len(odds) in [1, 3]:
            return False
        # either 2 or 4
        if len(odds) == 2:
            if odds[0] not in nei[odds[1]]:
                return True
            a = nei[odds[1]]
            b = nei[odds[0]]
            for i in range(1, 1+n):
                if i not in a and i not in b:
                    return True
            return False
        if len(odds) == 4:
            
            a = odds[0]
            b = odds[1]
            c = odds[2]
            d = odds[3]
            an = [x for x in [b, c, d] if x in nei[a]]
            bn = [x for x in [a, c, d] if x in nei[b]]
            cn = [x for x in [a, b, d] if x in nei[c]]
            dn = [x for x in [a, b, c] if x in nei[d]]
            s = [
                [len(an), a, an],
                [len(bn), b, bn],
                [len(cn), c, cn],
                [len(dn), d, dn],
            ]
            s.sort()
            
            # case 1 all disconnected
            if s[-1][0] == 0:
                return True
            if [v[0] for v in s] == [0, 0, 1, 1]:
                return True
            if [v[0] for v in s] == [1, 1, 1, 1]:
                return True
            if [v[0] for v in s] == [1, 1, 2, 2]:
                return True
            if [v[0] for v in s] == [2, 2, 2, 2]:
                return True
            if [v[0] for v in s] == [0, 1, 1, 2]:
                return True
            if [v[0] for v in s] == [1, 1, 1, 3]:
                return False
        

        
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