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# 2791. Count Paths That Can Form a Palindrome in a Tree
from collections import Counter
"""
you are given a tree (ie. a connected, undirected graph that has no cycles)
rooted at node 0 consisting of 'n' nodes numbered from 0 to 'n - 1'. the tree
is represented by a 0-indexed array 'parent' of size 'n' where 'parent[i]' is
the parent of node 'i'. since node 0 is the root, 'parent[0] == -1'. you are
also given a string 's' of length 'n' where 's[i]' is the character assigned
to the edge between 'i' and 'parent[i]'. 's[0]' can be ignored. return the
number of pairs of nodes '(u, v)' such that 'u < v' and the characters
assigned to edges on the path from 'u' to 'v' can rearranged to form a
palindrome
"""
class Solution(object):
def countPalindromePaths(self, parent, s):
"""
:type parent: List[int]
:type s: str
:rtype: int
"""
def f(i):
return f(parent[i]) ^ (1 << (ord(s[i]) - ord("a"))) if i else 0
cnt, ans = Counter(), 0
for i in range(len(parent)):
mask = f(i)
ans += cnt[mask] + sum(cnt[mask ^ (1 << j)] for j in range(26))
cnt[mask] += 1
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.countPalindromePaths(parent=[-1, 0, 0, 1, 1, 2], s="acaabc"))
print(obj.countPalindromePaths(parent=[-1, 0, 0, 0, 0], s="aaaaa"))
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