<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>




// get the shortest path from zero to each node with alternating color 
class Solution {
public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        // first build the graph 
        vector<vector<pair<int, int>>> graph(n, vector<pair<int, int>>(n));

        for (auto& edge : redEdges) {

            // zero represents color red
            graph[edge[0]].push_back({ edge[1], 0 });
        }

        for (auto& edge : blueEdges) {
            graph[edge[0]].push_back({ edge[1], 1 });

        }

        return BFS(0, n, graph); 
    }


    vector<int> BFS(int src, int n, vector<vector<pair<int, int>>>& graph) {
        
        // this holds the node the distance , the color
        queue<vector<int>> q; 

        q.push({ src, 0, 0 }); 
        q.push({ src, 0, 1 }); 

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


        // two vis one for red and one for blue 
        vector<vector<bool>> vis(n, vector<bool>(2));

        while (!q.empty()) {
            auto vc = q.front(); 
            q.pop(); 

            auto node = vc[0]; 
            auto nodeDis = vc[1]; 
            auto nodeColor = vc[2]; 

            vis[node][nodeColor] = true;

            for (auto& [neighbour, neighbourColor] : graph[node]) {

                if (!vis[neighbour][neighbourColor] && nodeColor != neighbourColor) {
                    if (dis[neighbour] == -1) {
                        dis[neighbour] = nodeDis + 1; 
                    }

                    q.push({ neighbour , nodeDis + 1 , neighbourColor }); 
                }
            }


        }

        return dis;
    }
};
26894