How do you solve Palindrome Pairs on LeetCode?

Palindrome Pairs (LeetCode #336) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k²) time complexity and O(n) space complexity. Key insight: Using a hash map allows for O(1) average time complexity lookups, significantly speeding up the process.

What is the brute force solution for Palindrome Pairs?

The brute force approach for Palindrome Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of strings to see if their concatenation forms a palindrome. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * k²).

What algorithm or data structure is used to solve Palindrome Pairs?

Palindrome Pairs uses the following concepts: Array, Hash Table, String, Trie, Hash Function. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Palindrome Pairs?

Always clarify the problem constraints and edge cases before jumping into coding. Think about how you can reduce the number of checks needed — look for patterns. Practice explaining your thought process out loud as you code; it helps in interviews.

What are common mistakes when solving Palindrome Pairs?

Not considering edge cases like empty strings or single-character strings. Failing to optimize the search for palindrome pairs, leading to timeouts.

#336

Palindrome Pairs

Hard

ArrayHash TableStringTrieHash FunctionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k²)
Space	O(1)	O(n)

💡

Intuition

Time O(n * k²)Space O(n)

The optimal solution uses a hash map to store the words and checks for possible palindromic combinations more efficiently. This reduces unnecessary checks and leverages the properties of palindromes.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store each word and its index.
2Step 2: For each word, check all possible prefixes and suffixes to see if they can form a palindrome when combined with another word.
3Step 3: If a valid palindrome pair is found, add it to the results.

solution.py13 lines

1def is_palindrome(s): return s == s[::-1]
2
3def palindrome_pairs(words):
4    result = []
5    word_map = {word: i for i, word in enumerate(words)}
6    for i, word in enumerate(words):
7        for j in range(len(word) + 1):
8            prefix, suffix = word[:j], word[j:]
9            if is_palindrome(suffix) and prefix[::-1] in word_map:
10                result.append([word_map[prefix[::-1]], i])
11            if j != len(word) and is_palindrome(prefix) and suffix[::-1] in word_map:
12                result.append([i, word_map[suffix[::-1]]])
13    return result

ℹ

Complexity note: This complexity arises because we iterate through each word and check all its prefixes and suffixes, where k is the average length of the words.

1Using a hash map allows for O(1) average time complexity lookups, significantly speeding up the process.
2Recognizing the properties of palindromes helps in reducing unnecessary checks.

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