Count Paths That Can Form a Palindrome in a Tree — LeetCode 2791 Python Solution

The problem

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1.

Example

Input

parent = [-1,0,0,1,1,2], s = "acaabc"

Output

8

Explanation

The valid pairs are:

Python solution

class Solution:
    def countPalindromePaths(self, parent: List[int], s: str) -> int:
        def dfs(i: int, xor: int):
            nonlocal ans
            for j, v in g[i]:
                x = xor ^ v
                ans += cnt[x]
                for k in range(26):
                    ans += cnt[x ^ (1 << k)]
                cnt[x] += 1
                dfs(j, x)

        n = len(parent)
        g = defaultdict(list)
        for i in range(1, n):
            p = parent[i]
            g[p].append((i, 1 << (ord(s[i]) - ord('a'))))
        ans = 0
        cnt = Counter({0: 1})
        dfs(0, 0)
        return ans

Complexity

Pattern: Bit Manipulation

Use XOR, masks and the low-bit trick to replace whole data structures with an integer. LeetCode 2791. Count Paths That Can Form a Palindrome in a Tree is filed here because LeetCode tags it Bit Manipulation and Bitmask, which is the vocabulary this hub collects.

The bit manipulation guide has the Python template for the pattern and the 194 LeetCode problems that use it.

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