What is the brute force solution for Maximum Product of the Length of Two Palindromic Subsequences?

The brute force approach for Maximum Product of the Length of Two Palindromic Subsequences is Brute Force, with O(n² * 2^n) time complexity and O(2^n) space complexity. The brute-force approach involves generating all possible pairs of disjoint subsequences and checking if they are palindromic. This method is straightforward but inefficient for larger strings due to its combinatorial nature. The optimal Optimal Solution approach improves this to O(n^3).

What are the interview tips for Maximum Product of the Length of Two Palindromic Subsequences?

Clearly explain your thought process while coding. Discuss the trade-offs of different approaches before diving into coding. Practice generating subsequences and understanding palindromes beforehand.

What are common mistakes when solving Maximum Product of the Length of Two Palindromic Subsequences?

Failing to check for disjoint subsequences correctly. Not optimizing the palindrome check, leading to excessive time complexity.

#2002

Maximum Product of the Length of Two Palindromic Subsequences

Medium

StringDynamic ProgrammingBacktrackingBit ManipulationBitmaskDynamic ProgrammingBit ManipulationBacktracking

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * 2^n)	O(n^3)
Space	O(2^n)	O(n^2)

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Intuition

Time O(n^3)Space O(n^2)

The optimal solution uses dynamic programming to precompute the lengths of the longest palindromic subsequences for all possible pairs of disjoint character sets. This significantly reduces the number of checks needed and allows us to find the maximum product efficiently.

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Algorithm

3 steps

1Step 1: Create a function to compute the longest palindromic subsequence for a given string.
2Step 2: Use bit manipulation to represent subsets of characters in the string.
3Step 3: For each pair of subsets, calculate the product of their palindromic lengths and update the maximum product.

solution.py26 lines

1def longest_palindrome_length(s):
2    n = len(s)
3    dp = [[0] * n for _ in range(n)]
4    for i in range(n):
5        dp[i][i] = 1
6    for length in range(2, n + 1):
7        for i in range(n - length + 1):
8            j = i + length - 1
9            if s[i] == s[j]:
10                dp[i][j] = 2 + (dp[i + 1][j - 1] if length > 2 else 0)
11            else:
12                dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
13    return dp[0][n - 1]
14
15def max_product_palindromic_subsequences(s):
16    n = len(s)
17    max_product = 0
18    for mask1 in range(1 << n):
19        for mask2 in range(mask1 + 1, 1 << n):
20            if mask1 & mask2 == 0:
21                subseq1 = ''.join(s[i] for i in range(n) if mask1 & (1 << i))
22                subseq2 = ''.join(s[i] for i in range(n) if mask2 & (1 << i))
23                len1 = longest_palindrome_length(subseq1)
24                len2 = longest_palindrome_length(subseq2)
25                max_product = max(max_product, len1 * len2)
26    return max_product

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Complexity note: The time complexity is reduced to O(n^3) due to the dynamic programming approach for finding palindromic lengths, which is more efficient than checking all pairs of subsequences directly.

1Understanding palindromic subsequences is crucial for solving this problem effectively.
2Bit manipulation can help efficiently represent and check disjoint subsequences.

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