<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 mx = 0;
int dp[100001][26]; 

vector<int> toposort(int n, vector<int> g[]) {
    vector<int> indeg(n, 0);
    for(int i = 0; i < n; i++) {
        for(auto& b: g[i]) {
            indeg[b]++;
        }
    }

    queue<int> q;
    for(int i = 0; i < n; i++) {
        if(indeg[i] == 0) q.push(i);
    }
    vector<int> topo;
    while(!q.empty()) {
        int node = q.front(); q.pop();
        topo.push_back(node);
        for(auto& b: g[node]) {
            indeg[b]--;
            if(indeg[b] == 0) q.push(b);
        }
    }
    return topo;
}

void dfs(int node, vector<int>& vis, string& col, vector<int> g[]) {
    vis[node] = 1;
    for(auto& b: g[node]) {
        if(!vis[b]) dfs(b, vis, col, g);
        for(int i = 0; i < 26; i++) dp[node][i] = max(dp[node][i], dp[b][i]);
    } 
    dp[node][col[node] - 'a']++;
    for(int i = 0; i < 26; i++) mx = max(mx, dp[node][i]);
}
    int largestPathValue(string col, vector<vector<int>>& e) {
        int n = col.size();
        // rev graph
        vector<int> g[n];
        for(auto& it: e) {
            int a = it[0], b = it[1];
            g[a].push_back(b);
        }
        vector<int> topo = toposort(n, g);
        if(topo.size() != n) return -1;
        vector<int> vis(n, 0);
        for(int i = 0; i < n; i++) {
            if(!vis[topo[i]]) dfs(topo[i], vis, col, g);
        }

        return mx;
    }
};
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