<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 u, vector<vector<int>> &adj, vector<vector<int>> &dp, string &colors){
        if(dp[u][0] != -1)return;

        for(int i=0;i<26;i++){
            dp[u][i] = 0;
        }

        // dp[u][colors[u] - 'a']++;
        for(auto ne : adj[u]){
            if(dp[ne][0] == -1){
                dfs(ne, adj, dp, colors);
            }

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

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

    bool isCycle(vector<vector<int>> &adj, vector<int> &in, int n){
        queue<int> q;
        for(int i=0;i<n;i++){
            if(in[i] == 0){
                q.push(i);
            }
        }
        int cnt = 0;
        while(!q.empty()){
            // ans.push_back(q.front());
            int u = q.front();
            cnt++;
            q.pop();
            for(auto ne : adj[u]){
                in[ne]--;
                if(in[ne] == 0){
                    q.push(ne);
                }
            }
        }

        return cnt != n;
    }

    int largestPathValue(string colors, vector<vector<int>>& edges) {
        // dp[u][c] -> path starting at u and color = c ka count
        // dp[u][c] = max(dp[ne][c]) + 1 whichever c is that node
        int n = colors.size();

        vector<vector<int>> adj(n);
        vector<int> in(n, 0);

        for(int i=0;i<edges.size();i++){
            int u = edges[i][0];
            int v = edges[i][1];

            adj[u].push_back(v);
            in[v]++;
        }

        if(isCycle(adj, in, n))return -1;
        in.clear();
        int ans = 0;

        vector<vector<int>> dp(n , vector<int> (26, -1));

        for(int i=0;i<n;i++){
            if(dp[i][0] == -1){
                dfs(i, adj, dp, colors);
            }
            // if(dp[i][0] == INT_MAX)return -1;
            for(int j=0;j<26;j++){
                if(dp[i][j] < 0)return -1;
                ans = max(ans, dp[i][j]);
            }
        }

        return ans;
    }
};
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