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1930. Unique Length-3 Palindromic Subsequences

Description

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

  • class Solution {
    public int countPalindromicSubsequence(String s) {
        int ans = 0;
        for (char c = 'a'; c <= 'z'; ++c) {
            int l = s.indexOf(c), r = s.lastIndexOf(c);
            Set<Character> cs = new HashSet<>();
            for (int i = l + 1; i < r; ++i) {
                cs.add(s.charAt(i));
            }
            ans += cs.size();
        }
        return ans;
    }
}

  • class Solution {
public:
    int countPalindromicSubsequence(string s) {
        int ans = 0;
        for (char c = 'a'; c <= 'z'; ++c) {
            int l = s.find_first_of(c), r = s.find_last_of(c);
            unordered_set<char> cs;
            for (int i = l + 1; i < r; ++i) cs.insert(s[i]);
            ans += cs.size();
        }
        return ans;
    }
};

  • class Solution:
    def countPalindromicSubsequence(self, s: str) -> int:
        ans = 0
        for c in ascii_lowercase:
            l, r = s.find(c), s.rfind(c)
            if r - l > 1:
                ans += len(set(s[l + 1 : r]))
        return ans


  • func countPalindromicSubsequence(s string) (ans int) {
	for c := 'a'; c <= 'z'; c++ {
		l, r := strings.Index(s, string(c)), strings.LastIndex(s, string(c))
		cs := map[byte]struct{}{}
		for i := l + 1; i < r; i++ {
			cs[s[i]] = struct{}{}
		}
		ans += len(cs)
	}
	return
}

  • public class Solution {
    public int CountPalindromicSubsequence(string s) {
        int ans = 0;
        for (char c = 'a'; c <= 'z'; ++c) {
            int l = s.IndexOf(c), r = s.LastIndexOf(c);
            HashSet<char> cs = new HashSet<char>();
            for (int i = l + 1; i < r; ++i) {
                cs.Add(s[i]);
            }
            ans += cs.Count;
        }
        return ans;
    }
}

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