<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[100001][26];
     vector<int> outgoing[100001];

    int dfs(int start, vector <int> &vis, string &s) {
        
        // if(vis[start])
        // return -1;

        if(dp[start][0] != -1)
        return 0;

        for(int i=0;i<outgoing[start].size();i++) {
            int ind = outgoing[start][i];

            if(vis[ind])
            return -1;

            vis[ind] = 1;

            cout<<start<<" "<<ind<<"\n";

            if(dp[ind][0] == -1) {
                
                int val = dfs(ind, vis, s);

                cout<<val<<" "<<ind<<"A\n";

                if(val == -1)
                return -1;
            }

            for(int j=0;j<26;j++) {
                if(j == s[start] - 'a') {
                    dp[start][j] = max(dp[start][j], 1+dp[ind][j]);
                    continue;
                }

                dp[start][j] = max(dp[start][j], dp[ind][j]);
            }

            vis[ind] = 0;
        }

        if(outgoing[start].size() == 0) {
            for(int i=0;i<26;i++)
            dp[start][i] = 0;

            dp[start][s[start] - 'a'] = 1;
        }

        return 0;
    }
    int largestPathValue(string colors, vector<vector<int>>& edges) {

        int n = colors.size();
        
        for(int i=0;i<edges.size();i++) {
            outgoing[edges[i][0]].push_back(edges[i][1]);

            //self edge
            if(edges[i][0] == edges[i][1])
            return -1;
        }

        vector<int> vis(n, 0);

        for(int i=0;i<=colors.size();i++)
        for(int j=0;j<26;j++)
        dp[i][j] = -1;

        for(int i=0;i<n;i++) {
            cout<<"start "<<i<<"\n";
            vis[i] = 1;
            int val = dfs(i, vis, colors);
            vis[i] = 0;

            if(val == -1)
            return -1;
        }

        int ans = 0;

        for(int i=0;i<=colors.size();i++)
        for(int j=0;j<26;j++)
        ans = max(ans, dp[i][j]);

        return ans;
    }
};
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