<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:
    def largestPathValue(self, colors: str, edges: List[List[int]]) -> int:
        if not edges: return 1 # Edge Case -> no edge -> 1
        n = len(colors) # number of nodes
        roots = {i for i in range(n)} # to check who are root nodes
        graph = defaultdict(set) 
        indegree = defaultdict(int) 
        for a, b in edges:
            graph[a].add(b) # generate graph
            indegree[b] += 1 # save all nodes's indegree
            roots -= {b} # delete nodes who have parent
        
        Max = -1
        # if we use list[] to do dp, we'll use n*26 extra space, and exacly 26 times to update color for all nodes
        # so I decide use double hash to optimize TC & MC
        # dp define as -> max each color count when reach every node
        dp = defaultdict(dict) 
        for root in roots: # search from every root
            indegree[root] = 0 # set root's indegree as 0 (cause indegree hash didn't data for every root)
            dp[root][colors[root]] = 1 # set dp[root]'s color to 1
            q = [root]
            while q: # BFS
                cur = q.pop(0)
                for nxt in graph[cur]:
                    indegree[nxt] -= 1 # update indegree
                    # if we meet a node that indegree == -1, means we found a loop, just return -1
                    # (this situation won't happen when using Topological Sort on a graph without loop)
                    if indegree[nxt] == -1: return -1
                    # update next node's color count
                    for c in dp[cur]:
                        dp[nxt].setdefault(c, dp[cur][c])
                        dp[nxt][c] = max(dp[nxt][c], dp[cur][c])
                    # if we can finally visited next node
                    if indegree[nxt] == 0:
                        q.append(nxt)
                        cc = colors[nxt] # next node's color
                        # add 1 to next node's color count with it's own color
                        dp[nxt].setdefault(cc, 0)
                        dp[nxt][cc] += 1
                        Max = max(Max, dp[nxt][cc]) # update Max
        return Max
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