<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:
bool dfs(vector<vector<int>>& adj,int n,int cur,vector<int>&vis,vector<int>&temp)
{
    vis[cur]=1;
    temp[cur]=1;
    for(int i=0;i<adj[cur].size();i++)
    {
        int node=adj[cur][i];
        if(temp[node]==1)return true;
        if(vis[node]==1)continue;
        else{
            if(dfs(adj,n,node,vis,temp))return true;
        }
    }
    temp[cur]=0;
    return false;
}
bool cycle(vector<vector<int>>& adj,int n)
{
    vector<int> vis(n,0);
    for(int i=0;i<n;i++)
    {
        if(vis[i]==0)
        {
            vector<int> temp(n,0);
            if(dfs(adj,n,i,vis,temp))return true;
        }
    }
    return false;
}

void solve(vector<vector<int>>& adj,int n,string& colors,int node,int &ans,vector<vector<int>>& dp)
{
if(dp[node][26]!=-1)return;
int c=(colors[node]-'a');
for(int i=0;i<adj[node].size();i++)
{
    int x=adj[node][i];
    solve(adj,n,colors,x,ans,dp);
    for(int j=0;j<27;j++)
    {
        dp[node][j]=max(dp[node][j],dp[x][j]);
    }
}
dp[node][c]++;
dp[node][26]=max(dp[node][26],dp[node][c]);
ans=max(ans,dp[node][26]);
return;
}


     int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n=colors.length();
        vector<vector<int>> adj(n);
        int m=edges.size();
        for(int i=0;i<m;i++)
        {
            adj[edges[i][0]].push_back(edges[i][1]);
        }
        if(cycle(adj,n))return -1;
        int ans=0;
        vector<vector<int>> dp(n,vector<int>(27,0));
        for(int i=0;i<n;i++)
        {
            dp[i][26]=-1;
        }
        for(int i=0;i<n;i++)
        {
        solve(adj,n,colors,i,ans,dp);
        }
        return  ans;
    }
};
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