<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 {
    int find_path(int dst, vector<vector<int>>& blueAdj, vector<vector<int>>& redAdj, bool isBlue, int n){
        int path = 0;
        queue<int> q;
        vector<bool> blueVisted(n, false);
        vector<bool> redVisted(n, false);
        q.push(0);
        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; i++){
                int cur = q.front();
                if(cur == dst)
                    return path;
                q.pop();
                if(isBlue){
                    for(auto& next: blueAdj[cur]){
                        if(!blueVisted[next]){
                            blueVisted[next] = true;
                            q.push(next);
                        }
                    }
                }else{
                    for(auto& next: redAdj[cur]){
                        if(!redVisted[next]){
                            redVisted[next] = true;
                            q.push(next);
                        }
                    }
                }
            }
            isBlue = !isBlue;
            path++;
        }
        return -1;
    }
public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        vector<vector<int>> blueAdj(n, vector<int>());
        vector<vector<int>> redAdj(n, vector<int>());
        for(auto& edge:blueEdges){
            blueAdj[edge[0]].push_back(edge[1]);
        }
        for(auto& edge:redEdges){
            redAdj[edge[0]].push_back(edge[1]);
        }
        vector<int> res(n);
        for(int i = 1; i < n; i++){
            int bluePath = find_path(i, blueAdj, redAdj, true, n); // blue first
            int redPath = find_path(i, blueAdj, redAdj, false, n); // red first
            if(bluePath == -1){
                res[i] = redPath;
            }else if(redPath == -1){
                res[i] = bluePath;
            }else{
                res[i] = min(bluePath, redPath);
            }
        }
        return res;
    }
};
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