<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 Solution {
private:
    int clr;
    string color;
    
    unordered_map<char,int> cl;
    vector<bool> vis;
    vector<int> pathvis;
    // vector<vector<int>> graph;
    // vector<vector<int>> dp(n,vector<int>(clr,0));

    bool dfs(int parent,vector<vector<int>>& graph,vector<vector<int>>& dp){
        vis[parent]=1;
        pathvis[parent]=1;
        for(auto it: graph[parent]){
            if(!vis[it]){
                if(dfs(it,graph,dp)==false) return false;
            }
            else if(pathvis[it]) return false;
            for(int i=0;i<clr;i++){
                dp[parent][i]=max(dp[parent][i],dp[it][i]);
            }
        }
        dp[parent][cl[color[parent]]]+=1;
        pathvis[parent]=0;
        return true;
    }
public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n=colors.size();
        vector<vector<int>> graph(n);
        // graph(n);
        for(int i=0;i<edges.size();i++){
            graph[edges[i][0]].push_back(edges[i][1]);
        }
        for(int i=0;i<n;i++){
            cl[colors[i]]=1;
        }
        for(int i=0;i<n;i++){
            vis.push_back(0);
            pathvis.push_back(0);
        }
        clr=cl.size();
        int k=0;
        for(auto& it:cl){
            it.second=k;
            k++;
        }
        // vector<bool> vis(n,0);
        // vector<int> pathvis(n,0);
        // vis(n,0);
        color=colors;
        // pathvis(n,0);
        vector<vector<int>> col(n,vector<int>(clr,0));
        // col(n,vector<int>(clr,0));
        for(int i=0;i<n;i++){
            if(vis[i]==0){
                if(dfs(i,graph,col)==false) return -1;
            }
        }
        int maxi=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<clr;j++)
                maxi=max(maxi,col[i][j]);
        }
        return maxi; 
    }
};
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