<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:
    void bfs(unordered_map<int, vector<pair<int, int>>> &adj, vector<vector<int>> &visited, int source, vector<int> &distance) {

        queue<vector<int>> q;
        visited[source][0] = 1;
        visited[source][1] = 1;
        q.push({source, 0, -1});

        while(!q.empty()) {

            int q_size = q.size();

            while(q_size--) {

                auto front = q.front();
                q.pop();

                int s = front[0];
                int d = front[1];
                int c = front[2];

                for(auto x: adj[s]) {
                    
                    int nbr = x.first;
                    char nbr_color = x.second;

                    if(!visited[nbr][nbr_color] && nbr_color != c) {
                        visited[nbr][nbr_color] = 1;
                        q.push({nbr, d + 1, nbr_color});
                        if(distance[nbr] == -1)
                            distance[nbr] = d + 1;
                    }

                }

            }

        }

    }
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        // we have to create a 2d visited array and mark for blue and red edges 
        // if red edge is goint from node mark it visited for red, if blue edge is going mark is visited for blue
        unordered_map<int, vector<pair<int, int>>> adj;

        for(auto x: redEdges) {
            int u = x[0];
            int v = x[1];

            adj[u].push_back({v, 0});
        }

        for(auto x: blueEdges) {
            int u = x[0];
            int v = x[1];

            adj[u].push_back({v, 1});
        }

        vector<vector<int>> visited(n, vector<int>(2, 0));

        vector<int> ans(n, -1);
        ans[0] = 0;

        for(int i = 0; i < n; i++) {

            bfs(adj, visited, 0, ans);

        }

        return ans;
    }
};
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