What is the brute force solution for Longest Palindrome by Concatenating Two Letter Words?

The brute force approach for Longest Palindrome by Concatenating Two Letter Words is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of the given words and checking if they can form a palindrome. This method is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Palindrome by Concatenating Two Letter Words?

Longest Palindrome by Concatenating Two Letter Words uses the following concepts: Array, Hash Table, String, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Palindrome by Concatenating Two Letter Words?

Always think about how to use data structures like HashMaps for counting occurrences. Practice generating combinations and permutations, but also think about how to optimize those processes. Clarify the requirements of the problem, especially around the definition of a palindrome.

What are common mistakes when solving Longest Palindrome by Concatenating Two Letter Words?

Not considering pairs of words that are reverses of each other. Failing to check for a center word that can be used once.

#2131

Longest Palindrome by Concatenating Two Letter Words

Medium

ArrayHash TableStringGreedyCountingHash 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)

The optimal solution uses a frequency count of the pairs of words and their reverses. We can form pairs that mirror each other to build the palindrome, and we can use one unpaired word in the center if available.

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Algorithm

4 steps

1Step 1: Count occurrences of each word and its reverse using a HashMap.
2Step 2: For each unique word, calculate the maximum pairs that can be formed.
3Step 3: Check if there is any word that can be used as a center (i.e., a word that is its own reverse).
4Step 4: Sum the lengths of all pairs and add 2 if a center is used.

solution.py15 lines

1from collections import Counter
2
3def longestPalindrome(words):
4    count = Counter(words)
5    length = 0
6    center_added = False
7    for word, freq in count.items():
8        reverse_word = word[::-1]
9        if word == reverse_word:
10            length += (freq // 2) * 2
11            if freq % 2 == 1:
12                center_added = True
13        elif reverse_word in count:
14            length += min(freq, count[reverse_word]) * 2
15    return length + (2 if center_added else 0)

ℹ

Complexity note: This complexity is linear because we only traverse the list of words a couple of times: once for counting and once for calculating the palindrome length. The space complexity is linear due to the HashMap storing the counts.

1A palindrome can be formed by pairing words that are reverses of each other.
2Words that are the same can contribute to the center of the palindrome.

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