1930. Unique Length-3 Palindromic Subsequences

Problem Link

Description

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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.

Solution

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Python3

class Solution:
    def countPalindromicSubsequence(self, s: str) -> int:
        N = len(s)
        first = [-1] * 26
        last = [-1] * 26
        res = 0
 
        for i, x in enumerate(s):
            k = ord(x) - ord('a')
            if first[k] == -1:
                first[k] = i
            
            last[k] = i
        
        for i in range(26):
            A = [0] * 26
 
            if first[i] != -1 and first[i] != last[i]:
                for j in range(first[i] + 1, last[i]):
                    A[ord(s[j]) - ord('a')] = 1
            
            res += sum(A)
 
        return res