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author | Jack Sangdahl <[email protected]> | 2024-10-06 21:57:58 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2024-10-06 21:58:06 -0600 |
commit | b9c16a156eff8ebe6627a0869fdd6d360aec9bc5 (patch) | |
tree | a683212bc75ff3fba193586f76ad8a17d4772273 /count_paths_form_palindrome_tree.py | |
parent | e4e9e626e5b5a9659510d4b71a742f14e4580079 (diff) | |
download | leetcode-b9c16a156eff8ebe6627a0869fdd6d360aec9bc5.tar.gz leetcode-b9c16a156eff8ebe6627a0869fdd6d360aec9bc5.zip |
2791. count paths that can form a palindrome in a tree
Diffstat (limited to 'count_paths_form_palindrome_tree.py')
-rw-r--r-- | count_paths_form_palindrome_tree.py | 39 |
1 files changed, 39 insertions, 0 deletions
diff --git a/count_paths_form_palindrome_tree.py b/count_paths_form_palindrome_tree.py new file mode 100644 index 0000000..e4740d8 --- /dev/null +++ b/count_paths_form_palindrome_tree.py @@ -0,0 +1,39 @@ +# 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")) |