Leetcode #2791: Count Paths That Can Form a Palindrome in a Tree
In this guide, we solve Leetcode #2791 Count Paths That Can Form a Palindrome in a Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Tree, Depth-First Search, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: parent = [-1,0,0,1,1,2], s = "acaabc"
Output: 8
Explanation: The valid pairs are:
- All the pairs (0,1), (0,2), (1,3), (1,4) and (2,5) result in one character which is always a palindrome.
- The pair (2,3) result in the string "aca" which is a palindrome.
- The pair (1,5) result in the string "cac" which is a palindrome.
- The pair (3,5) result in the string "acac" which can be rearranged into the palindrome "acca".
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
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.