How do you solve Count Different Palindromic Subsequences on LeetCode?

Count Different Palindromic Subsequences (LeetCode #730) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Dynamic programming allows us to efficiently count palindromic subsequences by building on smaller solutions.

What is the time complexity of Count Different Palindromic Subsequences?

The optimal solution for Count Different Palindromic Subsequences runs in O(n²) time and uses O(n²) space. The time complexity is O(n²) due to the nested loops filling the DP table, while space complexity is also O(n²) for storing the DP results.

What is the brute force solution for Count Different Palindromic Subsequences?

The brute force approach for Count Different Palindromic Subsequences is Brute Force, with O(n² * 2^n) time complexity and O(n) space complexity. The brute force approach involves generating all possible subsequences of the string and checking each one to see if it's a palindrome. This method is straightforward but inefficient for larger strings due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Count Different Palindromic Subsequences?

Count Different Palindromic Subsequences uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion, Memoization.

What are the interview tips for Count Different Palindromic Subsequences?

Explain your thought process clearly while solving the problem; this shows your understanding. Consider edge cases, such as strings with all identical characters or no palindromes. Practice problems with similar patterns to build intuition for dynamic programming.

What are common mistakes when solving Count Different Palindromic Subsequences?

Failing to account for duplicate palindromic subsequences when characters at the ends are the same. Not initializing the DP table correctly, leading to incorrect counts.

#730

Count Different Palindromic Subsequences

Hard

StringDynamic ProgrammingDynamic ProgrammingRecursionMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * 2^n)	O(n²)
Space	O(n)	O(n²)

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to count palindromic subsequences efficiently. By breaking the problem into smaller subproblems and using previously computed results, we can avoid redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a 2D DP array where dp[i][j] represents the count of unique palindromic subsequences in the substring s[i:j].
2Step 2: Fill the DP table by considering substrings of increasing lengths.
3Step 3: For each substring, check the characters at the ends. If they are the same, include them in the count, adjusting for duplicates.
4Step 4: Return dp[0][n-1] as the result.

solution.py21 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def count_palindromic_subsequences(s):
5    n = len(s)
6    dp = [[0] * n for _ in range(n)]
7
8    for i in range(n):
9        dp[i][i] = 1  # Single character palindromes
10
11    for length in range(2, n + 1):  # Length of substring
12        for i in range(n - length + 1):
13            j = i + length - 1
14            if s[i] == s[j]:
15                dp[i][j] = dp[i + 1][j] + dp[i][j - 1] + 1
16            else:
17                dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]
18            dp[i][j] = (dp[i][j] + MOD) % MOD
19
20    return dp[0][n - 1]
21

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops filling the DP table, while space complexity is also O(n²) for storing the DP results.

1Dynamic programming allows us to efficiently count palindromic subsequences by building on smaller solutions.
2Understanding how to handle overlapping subproblems is crucial in dynamic programming.

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