How do you solve Unique Length-3 Palindromic Subsequences on LeetCode?

Unique Length-3 Palindromic Subsequences (LeetCode #1930) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The maximum number of unique palindromic subsequences of length 3 is limited to the distinct characters in the string.

What is the brute force solution for Unique Length-3 Palindromic Subsequences?

The brute force approach for Unique Length-3 Palindromic Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences of length 3 and checking if they are palindromic. This is simple but inefficient, as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Unique Length-3 Palindromic Subsequences?

Unique Length-3 Palindromic Subsequences uses the following concepts: Hash Table, String, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Hash Map, Set, Dynamic Programming.

What are the interview tips for Unique Length-3 Palindromic Subsequences?

Always clarify the definition of a subsequence and a palindrome before diving into the solution. Think about how you can optimize brute-force solutions by identifying patterns or properties of the problem. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Unique Length-3 Palindromic Subsequences?

Failing to check for duplicates when adding palindromic subsequences, which can lead to incorrect counts. Not considering the order of characters when forming subsequences, which can lead to invalid palindromes.

#1930

Unique Length-3 Palindromic Subsequences

Medium

Hash TableStringBit ManipulationPrefix SumHash MapSetDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the properties of palindromes and uses a single pass through the string to track character occurrences. This allows us to efficiently count unique palindromic subsequences without generating all combinations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a set to store unique palindromic subsequences.
2Step 2: Use a loop to iterate through the string, keeping track of the last seen index of each character.
3Step 3: For each character, check if it has appeared before and if so, form palindromic subsequences with characters in between.

solution.py15 lines

1# Full working Python code
2
3def count_unique_palindromic_subsequences(s):
4    unique_palindromes = set()
5    last_seen = {}
6    n = len(s)
7    for i in range(n):
8        if s[i] in last_seen:
9            for j in last_seen[s[i]]:
10                unique_palindromes.add(s[j] + s[i] + s[j])
11        last_seen.setdefault(s[i], []).append(i)
12    return len(unique_palindromes)
13
14# Example usage
15print(count_unique_palindromic_subsequences('aabca'))  # Output: 3

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string, and the operations on the set and map are average O(1). The space complexity is O(n) due to the storage of indices in the map and unique palindromes in the set.

1The maximum number of unique palindromic subsequences of length 3 is limited to the distinct characters in the string.
2Using a set allows us to automatically handle duplicates, ensuring we only count unique palindromic subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.