<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>
/*
Mistakes:
1. renaming of same variable and func
2. destructuring vector of size 2

a -> b
a -> c

0/1
dp(u):
    MAX(color[u] == current_color + dp(v)) for all v in children(u)

vis[], 0, 1, 2
dp[]

if vis[0]
 run the dp
 if vis[1]
    then its -1
if vis[2]
    return dp[u]

 for all v in children(u):
    dp(v)
    if -1
        return -1;

E + V;
for (color = 'a'; color <= 'z'; ++color)
    ans = max(ans, get_max_path_sum(color))



*/
class Solution {
private:
    int n;
    vector<vector<int> > adj;
    vector<int> vis;
    vector<int> mem;
    string colors;
    // 0 -> 1 -> 2 -> 3 -> 1
    int dp(int u, char color) {
        if (vis[u] == 2)
            return mem[u];
        if (vis[u] == 1)
            return -1;
        vis[u] = 1;
        int &ret = mem[u];

        for (int v : adj[u]) {
            int child_ans = dp(v, color);
            if (child_ans == -1)
                return -1;
            ret = max(ret, child_ans);
        }
        ret += colors[u] == color;
        vis[u] = 2;
        return ret;
    }

    int get_max_path_sum(char color) {
        vis.clear();
        vis.resize(n);
        mem.clear();
        mem.resize(n);

        int ans = 0;
        for (int u = 0; u < n; ++u) {
            int node_ans = dp(u, color);
            if (node_ans == -1)
                return -1;
            ans = max(ans, node_ans);
        }
        return ans;
    }
public:
    int largestPathValue(string colors, vector<vector<int>>& edges) {
        this->n = colors.size();
        this->colors = colors;

        adj.resize(n);
        for (auto &e: edges) {
            adj[e[0]].push_back(e[1]);
        }

        int ans = 0;
        for (char color = 'a'; color <= 'z'; ++color) {
            int color_ans = get_max_path_sum(color);
            if (color_ans == -1)
                return -1;
            ans = max(ans, color_ans);
        }
        return ans;
    }
};
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