<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:
    int cal(int s , int dest , vector<int> badj[], vector<int> radj[],int n)
    {
        vector<int>rvis(n ,0) , bvis(n , 0);

        queue<pair<pair<int,int>,int>> q;

        q.push({{ 0, 0 },-1 });
        rvis[0] = 1;
        bvis[0] = 1;
        while(!q.empty())
        {
            auto ele = q.front() ;q.pop();
            int u =ele.first.first;
            int dist =ele.first.second;
            int color = ele.second;

            if(u == dest)   return dist;

            if(color!=0)
            {
                for(auto it:radj[u])
                {
                    if(!rvis[it])
                    {
                        q.push({{it ,dist+1} , 0});
                        rvis[it] = 1;
                    }
                }
            }
            if(color!=1)
            {

                for(auto it:badj[u])
                {
                    if(!bvis[it])
                    {
                        q.push({{it ,dist+1} , 1});
                        bvis[it] = 1;
                    }
                }
            }
        }

        return -1;
    }
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        vector<int> badj[n];
        vector<int> radj[n];

        vector<int> ans(n ,-1);

        for(int i = 0 ; i < redEdges.size() ;i++)
        {
            int u = redEdges[i][0];
            int v = redEdges[i][1];

            radj[u].push_back(v);
        }
        for(int i = 0 ; i < blueEdges.size() ;i++)
        {
            int u = blueEdges[i][0];
            int v = blueEdges[i][1];

            badj[u].push_back(v);
        }

        for(int i = 0 ; i<n ;i++)
        {
            ans[i] = cal(0 , i , badj , radj ,n);
        }

        return ans;


    }
};
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