- Difficulty: Hard
- Tags: LeetCode, Hard, Bit Manipulation, Tree, Depth-First Search, Dynamic Programming, Bitmask, leetcode-2791, O(n), DFS, Freq Table
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
. 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 be rearranged to form a palindrome.
A string is a palindrome when it reads the same backwards as forwards.
Example 1:
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".
Example 2:
Input: parent = [-1,0,0,0,0], s = "aaaaa" Output: 10 Explanation: Any pair of nodes (u,v) where u < v is valid.
Constraints:
n == parent.length == s.length
1 <= n <= 105
0 <= parent[i] <= n - 1
for alli >= 1
parent[0] == -1
parent
represents a valid tree.s
consists of only lowercase English letters.