<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:
        count = [0 for _ in range(n)]
        adj = {}
        for e in edges:
            count[e[0]-1] += 1
            count[e[1]-1] += 1
            if e[0]-1 not in adj:
                adj[e[0]-1] = {}
            adj[e[0]-1][e[1]-1] = 1

            if e[1]-1 not in adj:
                adj[e[1]-1] = {}
            adj[e[1]-1][e[0]-1] = 1
        
        odd_nodes = []
        numOdd = 0
        for i in range(n):
            if count[i] % 2 != 0:
                numOdd += 1
                odd_nodes.append(i)
        if numOdd == 0:
            return True
        print(odd_nodes)
        if numOdd == 2:
            print(adj[odd_nodes[1]])
            if odd_nodes[0] not in adj[odd_nodes[1]]:
                return True
            for i in range(n):
                k = i
                if k in adj:
                    v = adj[k]
                else:
                    v = {}
                if odd_nodes[1] in v or odd_nodes[0] in v:
                    continue
                if len(v) < n-1:
                    return True
        elif numOdd == 4:
            # for i in odd_nodes:
            #     print(i, adj[i])
            #     c = 0
            #     for j in odd_nodes:
            #         if j in adj[i]:
            #             c += 1
            #     if c == 3:
            #         return False
            # return True
            for x, y in [((0, 1), (2, 3)), ((0, 2), (1, 3)), ((0, 3), (1, 2))]:
                if odd_nodes[x[0]] in adj[odd_nodes[x[1]]]:
                    continue
                if odd_nodes[y[0]] in adj[odd_nodes[y[1]]]:
                    continue
                return True
            return False
        return False
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