<p>There is a <strong>directed graph</strong> of <code>n</code> colored nodes and <code>m</code> edges. The nodes are numbered from <code>0</code> to <code>n - 1</code>.</p>

<p>You are given a string <code>colors</code> where <code>colors[i]</code> is a lowercase English letter representing the <strong>color</strong> of the <code>i<sup>th</sup></code> node in this graph (<strong>0-indexed</strong>). You are also given a 2D array <code>edges</code> where <code>edges[j] = [a<sub>j</sub>, b<sub>j</sub>]</code> indicates that there is a <strong>directed edge</strong> from node <code>a<sub>j</sub></code> to node <code>b<sub>j</sub></code>.</p>

<p>A valid <strong>path</strong> in the graph is a sequence of nodes <code>x<sub>1</sub> -&gt; x<sub>2</sub> -&gt; x<sub>3</sub> -&gt; ... -&gt; x<sub>k</sub></code> such that there is a directed edge from <code>x<sub>i</sub></code> to <code>x<sub>i+1</sub></code> for every <code>1 &lt;= i &lt; k</code>. The <strong>color value</strong> of the path is the number of nodes that are colored the <strong>most frequently</strong> occurring color along that path.</p>

<p>Return <em>the <strong>largest color value</strong> of any valid path in the given graph, or </em><code>-1</code><em> if the graph contains a cycle</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet1.png" style="width: 400px; height: 182px;" /></p>

<pre>
<strong>Input:</strong> colors = &quot;abaca&quot;, edges = [[0,1],[0,2],[2,3],[3,4]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The path 0 -&gt; 2 -&gt; 3 -&gt; 4 contains 3 nodes that are colored <code>&quot;a&quot; (red in the above image)</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/21/leet2.png" style="width: 85px; height: 85px;" /></p>

<pre>
<strong>Input:</strong> colors = &quot;a&quot;, edges = [[0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is a cycle from 0 to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == colors.length</code></li>
	<li><code>m == edges.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= m &lt;= 10<sup>5</sup></code></li>
	<li><code>colors</code> consists of lowercase English letters.</li>
	<li><code>0 &lt;= a<sub>j</sub>, b<sub>j</sub>&nbsp;&lt; n</code></li>
</ul>
class Graph
{
    unordered_map<int, vector<int>> adj;
    unordered_map<int, int> indegreeTracker;
    vector<vector<int>> nodeToColorFreq;
    string colors;
public:
    Graph(string &colors)
    {
        nodeToColorFreq.resize(colors.size(), vector<int>(26, 0));
        this->colors = colors;
        for(int color = 0; color < colors.size(); color++)
            indegreeTracker[color] = 0, nodeToColorFreq[color][colors[color] - 'a'] = 1;
    }
    void addDirectedEdge(int from, int to)
    {   
        indegreeTracker[to]++;
        adj[from].push_back(to);
    }

    int getLargestColorValue()
    {
        queue<int> bfsQ;
        
        //start the bfs with the vertices having indegree 0.
        for(auto [vertex, indegree] : indegreeTracker)
        {
            if(indegree == 0)
                bfsQ.push(vertex);
        }

        if(bfsQ.empty())
            return -1;
        
        int elementsCovered = 0;
        int maxLen = 0;
        while(!bfsQ.empty())
        {
            auto front = bfsQ.front();
            bfsQ.pop();
            elementsCovered += 1;
            maxLen = max(maxLen, 
            *max_element(nodeToColorFreq[front].begin(), nodeToColorFreq[front].end()));

            for(int neighbour : adj[front])
            {
                indegreeTracker[neighbour] -= 1;

                for(int color = 0; color < 26; color++)
                    nodeToColorFreq[neighbour][color] = max(nodeToColorFreq[neighbour][color],
                    nodeToColorFreq[front][color] + (color == (colors[neighbour] - 'a'))); 
                
                if(indegreeTracker[neighbour] == 0)
                    bfsQ.push(neighbour);
            }
        }

        return (elementsCovered != colors.size()) ? -1 : maxLen;
    }
};

class Solution {
    Graph buildGraph(vector<vector<int>>& edges, string &colors)
    {   
        Graph graph(colors);

        for(auto edge : edges)
            graph.addDirectedEdge(edge[0], edge[1]);

        return graph;
    }

public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        return buildGraph(edges, colors).getLargestColorValue();
    }
};
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