<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 maxi = 0;
    // vector<int>topoSort(int n,string&col,vector<int>&indeg,vector<int>adj[],map<int,vector<int>>&mp){
    //     queue<int>q; // node,colorCount
    //     vector<int>ans;
    //     for(int i=0;i<n;i++){
    //         if(indeg[i]==0){
    //             ans.push_back(i);
    //             mp[i][col[i]-'a']++;
    //             q.push(i);
    //         }
    //     }

    //     while(!q.empty()){
    //         int node = q.front();
    //         q.pop();
    //         maxi = max(maxi,mp[node][col[node]-'a']);
    //         for(auto it:adj[node]){
    //             indeg[it]--;
                
    //             for(int i=0;i<26;i++){
    //                 mp[it][i] = max(mp[it][i],mp[node][i]+ (col[it]-'a'==i));
    //             }
                

    //             if(indeg[it]==0){
    //                 ans.push_back(it);
    //                 q.push(it);
    //             }
    //         }
    //     }

    //     return ans;
    // }


    int largestPathValue(string col, vector<vector<int>>& edges) {
        int n = col.size();
        vector<int>adj[n];
        int ans =0;
        map<int,vector<int>>mp;
        for(int i=0;i<n;i++){
            vector<int>temp(26,0);
            mp[i] = temp;
        }
        
        vector<int>indeg(n,0);
        for(auto it:edges){
            adj[it[0]].push_back(it[1]);
            indeg[it[1]]++;
        }

        queue<int>q; // node,
        for(int i=0;i<n;i++){
            if(indeg[i]==0){
                mp[i][col[i]-'a']++;
                q.push(i);
            }
        }
        int count = 0;
        while(!q.empty()){
            int node = q.front();
            q.pop();
            count++;
            maxi = max(maxi,mp[node][col[node]-'a']);
            for(auto it:adj[node]){
                indeg[it]--;
                for(int i=0;i<26;i++){
                    mp[it][i] = max(mp[it][i],mp[node][i]+ (col[it]-'a'==i));
                }
                if(indeg[it]==0){
                    q.push(it);
                }
            }
        }

        if(count != n){
            return -1;
        }
        

        return maxi;
    }
};
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