<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[100005][26]= {}, vis[100005]={},realvis[100005] = {};
    vector<int> grp[100005];
    bool dfs(int s, string& color){
       
        bool f = false ;
        vis[s] = 1;
        realvis[s] = 1;
        for(auto x : grp[s]){
            if(vis[x] == 1){
                return true;
            } 
            if(realvis[x] == 0) {
                f |= dfs(x, color);
            }
            for(int i = 0; i < 26 ; i++)
                dp[s][i] = max(dp[s][i], dp[x][i]);
        }
        dp[s][color[s] - 'a'] += 1;
        vis[s] = 0;
        return f;
    }
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        const int n = colors.size(), m = edges.size();
        vector<int> roots;
        vector<int> a(n, 0);
        for(int i = 0; i < m ; i++){
            a[edges[i][1]] += 1;
            if(edges[i][0] == edges[i][1]) return -1;
            grp[edges[i][0]].push_back(edges[i][1]);
        }
        for(int i = 0; i < n; i++){                         // path will be from root to any leaf node which contains the largest number of same colors 
            if(a[i] == 0){
                roots.push_back(i);
            }
        }
        if(roots.size() == 0) return -1 ;
        int ans = 0;
        for(auto x : roots){
            bool f = dfs(x, colors);
            if(f) return -1;
            for(int i = 0; i < 26; i++){
                ans = max(ans, dp[x][i]);
            }
        }
        for(int i = 0; i < n; i++){
            if(realvis[i] == 0) return -1;             // if any of the node is still unvisited the there must be a cycle.
        }
        return ans ;
    }
};
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