<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:


    int dp[100005][26];

    bool recStack[100005];

    bool vis[100005];

    void find(int source , vector<int> adj[],string& colors,bool vis[]){


        vis[source] = true;


        for(int j = 0;j<adj[source].size();j++){

            int neighbour = adj[source][j];

            if(!vis[neighbour])
            find(neighbour,adj,colors,vis);

            for(int i = 0;i<26;i++){
                dp[source][i] = max(dp[source][i],dp[neighbour][i]);
            }
        }

        dp[source][colors[source]-'a']++;
    }

    bool hasCycle(vector<int> adj[],int source){


        vis[source] = true;
        recStack[source] = true;

        for(int i=0;i<adj[source].size();i++){
            int neighbour = adj[source][i];

            if(recStack[neighbour]) return true;

            if(!vis[neighbour])
            {
             
               bool check = hasCycle(adj , neighbour);

              if(check) return check;
            }
        }

        recStack[source] = false;
        return false;

    } 


    int largestPathValue(string colors, vector<vector<int>>& edges) {


        memset(dp,0,sizeof(dp));
        memset(vis,false,sizeof(vis));
        memset(recStack,false,sizeof(recStack));


        int n  = colors.size();

        vector<int> adj[n];


        for(int i =0;i<edges.size();i++){

            int from = edges[i][0];
            int to = edges[i][1];

            adj[from].push_back(to);
        }
         
         for(int i=0;i<n;i++){

            if(!vis[i]) {
                if(hasCycle(adj,i)) return -1;
            }
         }


         memset(vis,false,sizeof(vis));

         int ans = 0;

         for(int i=0;i<n;i++){

            if(!vis[i]){
                find(i,adj,colors,vis);
            }

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