What is the time complexity of Longest Palindrome After Substring Concatenation I?

The optimal solution for Longest Palindrome After Substring Concatenation I runs in O(n) time and uses O(n) space. Counting characters is linear, O(n), and we only store counts, leading to O(n) space complexity.

What is the brute force solution for Longest Palindrome After Substring Concatenation I?

The brute force approach for Longest Palindrome After Substring Concatenation I is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible substrings from both strings and check each concatenation for palindromes. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Palindrome After Substring Concatenation I?

Longest Palindrome After Substring Concatenation I uses the following concepts: Two Pointers, String, Dynamic Programming, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Palindrome After Substring Concatenation I?

Explain your thought process clearly. Discuss edge cases like empty strings. Practice counting character frequencies.

What are common mistakes when solving Longest Palindrome After Substring Concatenation I?

Overlooking empty substrings. Not considering odd character counts for the center.

#3503

Longest Palindrome After Substring Concatenation I

Medium

Two PointersStringDynamic ProgrammingEnumerationHash MapArray

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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)

Count the frequency of characters in both strings. The longest palindrome can be formed by using pairs of characters from both strings, plus one odd character if available.

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Algorithm

3 steps

1Step 1: Count character frequencies in both strings.
2Step 2: Calculate the total length of pairs from both strings.
3Step 3: Add 1 if any character has an odd count for the center of the palindrome.

solution.py6 lines

1from collections import Counter
2
3def longestPalindrome(s, t):
4    count = Counter(s) + Counter(t)
5    length = sum(v // 2 * 2 for v in count.values())
6    return length + (1 if any(v % 2 for v in count.values()) else 0)

ℹ

Complexity note: Counting characters is linear, O(n), and we only store counts, leading to O(n) space complexity.

1Character pairs form the basis of palindromes.
2Odd counts can contribute a single character to the center.

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