Package g2701_2800.s2791_count_paths_that_can_form_a_palindrome_in_a_tree

Class Solution

  • public class Solution
extends Object
    2791 - Count Paths That Can Form a Palindrome in a Tree.

    Hard

    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 all i >= 1
    • parent[0] == -1
    • parent represents a valid tree.
    • s consists of only lowercase English letters.

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • countPalindromePaths

        public long countPalindromePaths​(List<Integer> parent,
                                 String s)