# 3373. Maximize the Number of Target Nodes After Connecting Trees II from collections import deque """ there exists two undirect trees with 'n' and 'm' nodes, labeled from '[0, n - 1]' andc '[0, m - 1]' respectively. you are given two 2d integer arrays 'edges1' and 'edges2' of lengths 'n - 1' and 'm - 1' respectively. node 'u' is target to node 'v' if the number of edges on the path from 'u' to 'v' is even. note that a node is always target to itself. return an array of integers where 'ans[i]' is the maximum possible number of nodes that are target to node 'i' of the first tree if you had to connect one node from the first tree to another node in the second tree. note that queries are independent from each other. that is, for every query, you will remove the added edge before proceeding to the next query. """ class Solution(object): def maxTargetNodes(self, edges1, edges2): """ :type edges1: List[List[int]] :type edges2: List[List[int]] :rtype: List[int] """ def find_even(edges, n): graph = [[] for _ in range(n)] queue = deque([(0, -1, True)]) evens = [False] * n for u, v in edges: graph[u].append(v) graph[v].append(u) while queue: node, parent, curr = queue.popleft() evens[node] = curr for child in graph[node]: if child == parent: continue queue.append((child, node, not curr)) return evens n, m = len(edges1) + 1, len(edges2) + 1 e1, e2 = find_even(edges1, n), find_even(edges2, m) s1, s2 = sum(e1), sum(e2) x = max(s2, m - s2) return [x + (s1 if even else n - s1) for even in e1] if __name__ == "__main__": obj = Solution() print( obj.maxTargetNodes( edges1=[[0, 1], [0, 2], [2, 3], [2, 4]], edges2=[[0, 1], [0, 2], [0, 3], [2, 7], [1, 4], [4, 5], [4, 6]], ) ) print( obj.maxTargetNodes( edges1=[[0, 1], [0, 2], [0, 3], [0, 4]], edges2=[[0, 1], [1, 2], [2, 3]] ) )