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>.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>.A valid <strong>path</strong> in the graph is a sequence of nodes <code>x<sub>1</sub> -> x<sub>2</sub> -> x<sub>3</sub> -> … -> 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 <= i < 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.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>.
class Solution: # Time: O(n + m) with a constant factor of 26 colors # Space: O(n) def largest_path_value(self, colors: str, edges: list[list[int]]) -> int: n = len(colors) adj: list[list[int]] = [[] for _ in range(n)] indegree = [0] * n for src, dst in edges: adj[src].append(dst) indegree[dst] += 1 counts = [[0] * 26 for _ in range(n)] for node in range(n): counts[node][ord(colors[node]) - 97] = 1 queue = [node for node in range(n) if indegree[node] == 0] processed = 0 best = 0 while queue: node = queue.pop() processed += 1 node_counts = counts[node] local_best = max(node_counts) if local_best > best: best = local_best for nxt in adj[node]: nxt_counts = counts[nxt] nxt_color = ord(colors[nxt]) - 97 for c in range(26): cand = node_counts[c] + (1 if c == nxt_color else 0) if cand > nxt_counts[c]: nxt_counts[c] = cand indegree[nxt] -= 1 if indegree[nxt] == 0: queue.append(nxt) return best if processed == n else -1
