<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:
    vector<int> find(vector<vector<int>> &g, int node, vector<bool> &vis, string &colors, vector<vector<int>> &dp){
        vector<int> res(26,0);
        for(int i =0;i<g[node].size();i++){
            int child = g[node][i];
            if(!vis[child]){
                vis[child]=1;
                vector<int> temp = find(g, child, vis, colors,dp);
                for(int i =0;i<26;i++){
                    res[i] = max(res[i], temp[i]);
                }
            }else{
               for(int i =0;i<26;i++){
                    res[i] = max(res[i], dp[child][i]);
                } 
            }
        }
        res[colors[node]-'a']++;
        return dp[node] = res;
    }
    bool dfs(vector<vector<int>> &g, int node, vector<bool> &vis,vector<bool> &dfs_vis){
        for(int i=0;i<g[node].size();i++){
            int child = g[node][i];
            if(!vis[child]){
                vis[child]=1;
                dfs_vis[child]=1; 
                if(dfs(g,child, vis, dfs_vis)) return 1;
                dfs_vis[child]=0;
            }else if(dfs_vis[child]){
                return 1;
            }
        }
        return 0;
    }
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        int n = colors.size();
        vector<vector<int>> g(n);
        for(int i=0;i<edges.size();i++){
            g[edges[i][0]].push_back(edges[i][1]);
        }
        
        vector<int> vis_color(26,0);
        vector<bool> vis(n,0), dfs_vis(n,0);
        int ans = -1;
        for(int i=0;i<n;i++){
            if(!vis[i]){
                vis[i]=1;
                dfs_vis[i]=1;
                if(dfs(g, i, vis, dfs_vis)) return -1;
                dfs_vis[i]=0;
            }
        }
        vector<vector<int>> dp(n, vector<int>(26,0));
        vis.assign(n,0);
        for(int i=0;i<n;i++){
            if(!vis[i]){
                vis[i]=1;
                vector<int> v = find(g, i, vis, colors,dp);
                for(int i =0;i<26;i++){
                    ans = max(ans , v[i]);
                }
            }
        }
        return ans;
    }
};
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