How do you solve Maximize Palindrome Length From Subsequences on LeetCode?

Maximize Palindrome Length From Subsequences (LeetCode #1771) 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: Subsequences can be formed by deleting characters without changing order.

What is the time complexity of Maximize Palindrome Length From Subsequences?

The optimal solution for Maximize Palindrome Length From 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. The space complexity is also O(n²) because we store results in a 2D array.

What is the brute force solution for Maximize Palindrome Length From Subsequences?

The brute force approach for Maximize Palindrome Length From Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible non-empty subsequences from both strings and checking which concatenated subsequences can form a palindrome. Although this method is straightforward, it can be very slow 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 Maximize Palindrome Length From Subsequences?

Maximize Palindrome Length From Subsequences uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Two Pointers.

What are the interview tips for Maximize Palindrome Length From Subsequences?

Understand the definitions of subsequences and palindromes clearly. Practice generating subsequences and checking for palindromes. Be prepared to explain your thought process while coding.

What are common mistakes when solving Maximize Palindrome Length From Subsequences?

Not considering all combinations of subsequences. Forgetting to check both ends of the palindrome during DP.

#1771

Maximize Palindrome Length From Subsequences

Hard

StringDynamic ProgrammingDynamic ProgrammingTwo Pointers

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 approach uses dynamic programming to find the longest palindromic subsequence in the concatenated string of word1 and word2. This method is efficient because it systematically builds solutions to subproblems, avoiding redundant calculations.

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Algorithm

4 steps

1Step 1: Concatenate word1 and word2 into a new string.
2Step 2: Create a 2D DP array where dp[i][j] represents the length of the longest palindromic subsequence in the substring from index i to j.
3Step 3: Fill the DP table using the relationships: if characters match, extend the palindrome; if not, take the maximum from the adjacent subproblems.
4Step 4: The result will be in dp[0][length-1] where length is the length of the concatenated string.

solution.py16 lines

1# Full working Python code
2
3def max_palindrome_length(word1, word2):
4    combined = word1 + word2
5    n = len(combined)
6    dp = [[0] * n for _ in range(n)]
7    for i in range(n):
8        dp[i][i] = 1
9    for length in range(2, n + 1):
10        for i in range(n - length + 1):
11            j = i + length - 1
12            if combined[i] == combined[j]:
13                dp[i][j] = dp[i + 1][j - 1] + 2
14            else:
15                dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
16    return dp[0][n - 1]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops filling the DP table. The space complexity is also O(n²) because we store results in a 2D array.

1Subsequences can be formed by deleting characters without changing order.
2Palindromes can be efficiently found using dynamic programming.

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