<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 find_degree(n: int, edges: List[List[int]]):
            nodes_degree = {}
            for i in range(1, n+1):
                nodes_degree[i] = 0

            for edge in edges:
                nodes_degree[edge[0]] += 1
                nodes_degree[edge[1]] += 1

            return nodes_degree
        
        def get_odd_degree_nodes(nodes_degree):
            odd_nodes = []
            for key, val in nodes_degree.items():
                if val % 2 != 0:
                    odd_nodes.append(key)
            return odd_nodes

        def find_non_connected_neighbors(n, edges):
            node_to_neighbors = {}
            for i in range(1, n+1):
                node_to_neighbors[i] = []
            
            for edge in edges:
                node_to_neighbors[edge[0]].append(edge[1])
                node_to_neighbors[edge[1]].append(edge[0])
            
            return node_to_neighbors

        nodes_degree = find_degree(n, edges)
        odd_degree_nodes = get_odd_degree_nodes(nodes_degree)
        node_to_neighbors = find_non_connected_neighbors(n, edges)

        if len(odd_degree_nodes) == 0:
            return True

        if len(odd_degree_nodes) > 4:
            return False
        print (len(odd_degree_nodes))
        if len(odd_degree_nodes) == 4:
            if ((odd_degree_nodes[0] not in node_to_neighbors[odd_degree_nodes[1]]) \
            and (odd_degree_nodes[2] not in node_to_neighbors[odd_degree_nodes[3]])) \
                or \
                ((odd_degree_nodes[0] not in node_to_neighbors[odd_degree_nodes[2]]) \
                and (odd_degree_nodes[1] not in node_to_neighbors[odd_degree_nodes[3]])) \
                or \
                ((odd_degree_nodes[0] not in node_to_neighbors[odd_degree_nodes[3]]) \
                and (odd_degree_nodes[1] not in node_to_neighbors[odd_degree_nodes[2]])):
                return True
        
        if len(odd_degree_nodes) == 2:
            if (odd_degree_nodes[0] not in node_to_neighbors[odd_degree_nodes[1]]):
                return True
            nodes = [i for i in range(1, n+1)]
            blacklist = list(set(node_to_neighbors[odd_degree_nodes[0]] + node_to_neighbors[odd_degree_nodes[1]]))
            whitelist = []
            for node in nodes:
                if (node not in blacklist) and (node not in [odd_degree_nodes[0], odd_degree_nodes[1]]):
                    whitelist.append(node)
            if len(whitelist) > 0:
                return True

        if len(odd_degree_nodes) == 1:
            return False

        if len(odd_degree_nodes) == 3:
            return False


        
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