<p>You are given an integer <code>n</code>, the number of nodes in a directed graph where the nodes are labeled from <code>0</code> to <code>n - 1</code>. Each edge is red or blue in this graph, and there could be self-edges and parallel edges.</p>

<p>You are given two arrays <code>redEdges</code> and <code>blueEdges</code> where:</p>

<ul>
	<li><code>redEdges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is a directed red edge from node <code>a<sub>i</sub></code> to node <code>b<sub>i</sub></code> in the graph, and</li>
	<li><code>blueEdges[j] = [u<sub>j</sub>, v<sub>j</sub>]</code> indicates that there is a directed blue edge from node <code>u<sub>j</sub></code> to node <code>v<sub>j</sub></code> in the graph.</li>
</ul>

<p>Return an array <code>answer</code> of length <code>n</code>, where each <code>answer[x]</code> is the length of the shortest path from node <code>0</code> to node <code>x</code> such that the edge colors alternate along the path, or <code>-1</code> if such a path does not exist.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, redEdges = [[0,1],[1,2]], blueEdges = []
<strong>Output:</strong> [0,1,-1]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]
<strong>Output:</strong> [0,1,-1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= redEdges.length,&nbsp;blueEdges.length &lt;= 400</code></li>
	<li><code>redEdges[i].length == blueEdges[j].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub>, u<sub>j</sub>, v<sub>j</sub> &lt; n</code></li>
</ul>

class Solution {
public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        // Using unordered_map as a replacement for Python's defaultdict
        unordered_map<int, vector<int>> red, blue;
        
        // Fill the red and blue edge adjacency lists
        for (auto& edge : redEdges) {
            red[edge[0]].push_back(edge[1]);
        }
        for (auto& edge : blueEdges) {
            blue[edge[0]].push_back(edge[1]);
        }
        
        // Initialize the answer vector
        vector<int> answer(n, -1);
        
        // Use a queue for BFS, storing tuples of (node, length, previous edge color)
        queue<tuple<int, int, int>> q;  // (node, length, edgeColor)
        q.push({0, 0, -1});  // Start at node 0, with length 0 and no previous color (-1)
        
        set<pair<int, int>> visit;  // To track visited nodes with their edge color
        visit.insert({0, -1});
        
        // BFS loop
        while (!q.empty()) {
            auto [node, length, edgeColor] = q.front();
            q.pop();
            
            if (answer[node] == -1) {
                answer[node] = length;  // Set the shortest path to the node
            }
            
            // Explore red edges if the previous edge was not red
            if (edgeColor != 0) {
                for (int nei : red[node]) {
                    if (visit.find({nei, 0}) == visit.end()) {
                        visit.insert({nei, 0});
                        q.push({nei, length + 1, 0});  // Push neighbor with red edge
                    }
                }
            }
            
            // Explore blue edges if the previous edge was not blue
            if (edgeColor != 1) {
                for (int nei : blue[node]) {
                    if (visit.find({nei, 1}) == visit.end()) {
                        visit.insert({nei, 1});
                        q.push({nei, length + 1, 1});  // Push neighbor with blue edge
                    }
                }
            }
        }
        
        return answer;
    }
};

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