LeetCode 1930: Unique Length-3 Palindromic Subsequences Solution

Master LeetCode problem 1930 (Unique Length-3 Palindromic Subsequences), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1930. Unique Length-3 Palindromic Subsequences

Medium

Hash Table String Bit Manipulation Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through each character in the string s and check if there exists a valid palindromic subsequence of length 3 centered around the current character. We can do this by checking for characters that are the same as the current character but different from the previous and next characters. If such characters exist, we increment the count of valid palindromic subsequences.

Algorithmic Steps:

Initialize a variable count to store the count of valid palindromic subsequences.
Iterate through each character c in the string s.
For each character c, check if there exists two other characters a and b such that a = b = c and a is not equal to the previous character and b is not equal to the next character.
If the above condition is met, increment the count.
Finally, return the count.

Time Complexity

The time complexity of this solution is O(n), where n is the length of the input string s.

Space Complexity

The space complexity of this solution is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int countPalindromicSubsequences(String s) {
        int count = 0;
        for (int i = 1; i < s.length() - 1; i++) {
            char c = s.charAt(i);
            if (s.charAt(i - 1) == c) {
                for (char ch = 'a'; ch <= 'z'; ch++) {
                    if (ch != c && s.indexOf(ch, i + 1) != -1) {
                        count++;
                        break;
                    }
                }
            }
        }
        return count;
    }
}

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