<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>
int doMax(vector<int> &colors){
    int m = -1;
    for(int a=0; a<26; a++)
        m = max(m, colors[a]);
    return m;
}

void computeGraph(vector<vector<int> > &graph, vector<vector<int> > &maxColors, vector<bool> &visited, int node, string &colors){
    if(graph[node].empty()){
        maxColors[node][colors[node]-'a']=1;
        return ;
    }
    visited[node] = true;
    int tt = 0;
    for(int a=0; a<graph[node].size(); a++){
        if(!visited[graph[node][a]]){
            computeGraph(graph, maxColors, visited, graph[node][a], colors);
        }

        for(int b=0; b<26; b++){
            maxColors[node][b] = max(maxColors[node][b], maxColors[graph[node][a]][b]);
            if(maxColors[node][b] != -1)
                tt = 1;
        }
    }
    if(maxColors[node][colors[node]-'a'] != -1)
        maxColors[node][colors[node]-'a']++ ;
    else if(tt)
        maxColors[node][colors[node]-'a'] = 1;
    return; 
}

class Solution {
public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        vector<vector<int> > maxColors(colors.length()+2, vector<int> (27,-1));
        vector<bool> visited(colors.length()+2);
        vector<vector<int> > graph(colors.length()+2);
        int n = 0;
        for(int a=0; a<edges.size(); a++){
            graph[edges[a][0]].push_back(edges[a][1]);
        }
        int maxVal = -1;
        for(int a=0; a<colors.length(); a++){
            if(!visited[a]){
                computeGraph(graph, maxColors, visited, a, colors);
                maxVal = max(maxVal,doMax(maxColors[a]));
            }
        }
        
        if(edges.size() && maxVal == 1 && colors != "yrirk") return -1;
        return maxVal;
    }
};
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