How do you solve Count Palindromic Subsequences on LeetCode?

Count Palindromic Subsequences (LeetCode #2484) 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: Focus on character matching to build palindromic subsequences.

What is the brute force solution for Count Palindromic Subsequences?

The brute force approach for Count Palindromic Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach generates all possible subsequences of length 5 from the string and checks each one to see if it is palindromic. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n²).

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

Count Palindromic Subsequences uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Combinatorial Counting.

What are the interview tips for Count Palindromic Subsequences?

Explain your thought process clearly. Discuss edge cases, like strings with all identical characters. Be prepared to optimize your solution if asked.

What are common mistakes when solving Count Palindromic Subsequences?

Not considering all combinations of characters. Overlooking the need to check for palindromic properties.

#2484

Count Palindromic Subsequences

Hard

StringDynamic ProgrammingDynamic ProgrammingCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n²)

The optimal solution leverages dynamic programming to efficiently count palindromic subsequences of length 5 by focusing on the first and last characters and using previously computed results to build up to the desired length.

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Algorithm

3 steps

1Step 1: Initialize a DP table to store counts of palindromic subsequences for different lengths.
2Step 2: Iterate through the string, updating the DP table based on matching characters and previously computed values.
3Step 3: Return the count of palindromic subsequences of length 5 from the DP table.

solution.py17 lines

1def countPalindromicSubsequences(s):
2    mod = 10**9 + 7
3    n = len(s)
4    dp = [[0] * n for _ in range(n)]
5
6    for i in range(n):
7        dp[i][i] = 1
8
9    for length in range(2, n + 1):
10        for left in range(n - length + 1):
11            right = left + length - 1
12            if s[left] == s[right]:
13                dp[left][right] = dp[left + 1][right] + dp[left][right - 1] + 1
14            else:
15                dp[left][right] = dp[left + 1][right] + dp[left][right - 1] - dp[left + 1][right - 1]
16
17    return sum(dp[i][j] for i in range(n) for j in range(n) if j - i + 1 == 5) % mod

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops for filling the DP table. The space complexity is also O(n²) because we store results for all pairs of indices in the DP table.

1Focus on character matching to build palindromic subsequences.
2Dynamic programming can optimize counting by reusing results.

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