<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 {
private:
    int fun(int dest, vector<int> adjBlue[], vector<int> adjRed[],int n) {
        vector<int>distRed(n,INT_MAX);
        vector<int>distBlue(n,INT_MAX);
        distRed[0]=distBlue[0]=0;
        priority_queue<pair<int, pair<int, int>>,
                       vector<pair<int, pair<int, int>>>,
                       greater<pair<int, pair<int, int>>>>
            pq;
        pq.push({0, {0, -1}});
        while (!pq.empty()) {
            // first one is distance second one is node and the third one is
            // prev -1 for none 0 for blue 1 for red
            auto it = pq.top();
            int dist = it.first;
            int node = it.second.first;
            int prev = it.second.second;
            pq.pop();
            if (node == dest)
                return dist;
            if (prev == -1 || prev == 1) {
                for (auto neighbor : adjBlue[node]) {
                    if (distBlue[neighbor] > dist + 1) {
                        distBlue[neighbor] = dist + 1;
                        pq.push({dist + 1, {neighbor, 0}});
                    }
                }
            }
            if (prev == -1 || prev == 0) {
                for (auto neighbor : adjRed[node]) {
                    if (distRed[neighbor] > dist + 1) {
                        distRed[neighbor] = dist + 1;
                        pq.push({dist + 1, {neighbor, 1}});
                    }
                }
            }
        }
        return -1;
    }

public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges,
                                         vector<vector<int>>& blueEdges) {

        vector<int> adjRed[n];
        for (int i = 0; i < redEdges.size(); i++) {
            int u = redEdges[i][0];
            int v = redEdges[i][1];
            adjRed[u].push_back(v);
        }
        vector<int> adjBlue[n];
        for (int i = 0; i < blueEdges.size(); i++) {
            int u = blueEdges[i][0];
            int v = blueEdges[i][1];
            adjBlue[u].push_back(v);
        }
        vector<int> ans(n, 0);
        ans[0]=0;
        for (int i = 1; i < n; i++) {
            int value = fun(i, adjBlue, adjRed,n);
            ans[i]=value;
        }
        return ans;
    }
};
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