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2059. Unique Length 3 Palindromic Subsequences

Hash Table String Bit Manipulation Prefix Sum

Problem - Unique Length 3 Palindromic Subsequences

Medium

Given a string s, return the number of unique palindromes of length three that are a subsequence of s.

Note that even if there are multiple ways to obtain the same subsequence, it is still only counted once.

A palindrome is a string that reads the same forwards and backwards.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

  • For example, "ace" is a subsequence of "abcde".

 

Example 1:

Input: s = "aabca"
Output: 3
Explanation: The 3 palindromic subsequences of length 3 are:
- "aba" (subsequence of "aabca")
- "aaa" (subsequence of "aabca")
- "aca" (subsequence of "aabca")

Example 2:

Input: s = "adc"
Output: 0
Explanation: There are no palindromic subsequences of length 3 in "adc".

Example 3:

Input: s = "bbcbaba"
Output: 4
Explanation: The 4 palindromic subsequences of length 3 are:
- "bbb" (subsequence of "bbcbaba")
- "bcb" (subsequence of "bbcbaba")
- "bab" (subsequence of "bbcbaba")
- "aba" (subsequence of "bbcbaba")

 

Constraints:

  • 3 <= s.length <= 105
  • s consists of only lowercase English letters.

Solutions

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class Solution:
    def countPalindromicSubsequence(self, s: str) -> int:
        result = 0
        for i in range(26):
            char = chr(ord('a') + i)
            left, right = s.find(char), s.rfind(char)
            if right - left > 1:
                result += len(set(s[left + 1 : right]))

        return result

Submission Stats:

  • Runtime: 67 ms (100.00%)
  • Memory: 18.3 MB (59.85%)