<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 {
public:
   void dfs(int node, vector<int>graph[],vector<int>&vis,vector<vector<int>>&dp,string &colors, bool &val, int &ans){
       
       if(dp[node][0] != -1)return ;

       vector<int>v(26,0);

       for(int i =0 ; i<26 ; i++){
          if(i == (colors[node]-'a')){
             dp[node][i] = 1;
          }
          else{
            dp[node][i] = 0;
          }
       }

       vis[node] = 1;
    
       for(auto ch : graph[node]){
           if(vis[ch] == 0)dfs(ch,graph,vis,dp,colors,val,ans);
           else if(vis[ch] == 1){ val = false; return; }
           for(int i =0 ; i<26 ; i++){
              v[i] = max(v[i],dp[ch][i]);         
           }
       }
       for(int i =0 ; i<26 ; i++){
           dp[node][i] = v[i] + ((colors[node]-'a') == i);
       }
     
       vis[node] =  2;
       return ;

   }

//    bool check_cycle(vector<int>graph[],int &n,vector<int>&indegree){
//        vector<int>topo;
//        queue<int>q;
//        for(int i =0 ;i<n ;i++){
//         if(indegree[i] == 0){
//             q.push(i);
//          }
//        }

//        while(!q.empty()){
//           int x = q.front();
//           topo.push_back(x);
//           for(auto ch : graph[x]){
//                indegree[ch]--;
//                 if(indegree[ch] == 0){
//                     q.push(ch);
//                 }
//           }
//        }
       
//        if(topo.size()!= n){
//           return false;
//        }
//        return true ;

//    }
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n = colors.size();
        vector<int>graph[n];
        vector<int>indegree(n,0);
        for(int i =0 ; i<edges.size(); i++){
            int u = edges[i][0];
            int v = edges[i][1];
            indegree[v]++;
            graph[u].push_back(v);
        }
     
            bool val=true ;
            vector<int>vis(n,0);
            vector<vector<int>>dp(n,vector<int>(26,-1));
            int res = 0 ;
            for(int i =0 ; i<n ; i++){
                if(vis[i] == 0)dfs(i,graph,vis,dp,colors,val,res);
                if(val == false)return -1 ;
            }
            
            // for(int i =0 ;i<26; i++){
            //     cout<<dp[2][i]<<" ";
            // }
            // cout<<endl;

           int ans = 0 ;
            for(int i = 0 ; i<n ; i++){
                 for(int j =0 ; j<26 ; j++){
                    ans = max(ans,dp[i][j]);
                 }
            }
            return ans ;
     
    }
};
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