How do you solve Find Maximum Number of String Pairs on LeetCode?

Find Maximum Number of String Pairs (LeetCode #2744) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each string can only be paired once.

What is the brute force solution for Find Maximum Number of String Pairs?

The brute force approach for Find Maximum Number of String Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of strings to see if one is the reverse of the other. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Maximum Number of String Pairs?

Find Maximum Number of String Pairs uses the following concepts: Array, Hash Table, String, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Maximum Number of String Pairs?

Think about how to reduce the number of comparisons. Use data structures like hash maps effectively. Always consider edge cases, such as strings that can't be paired.

What are common mistakes when solving Find Maximum Number of String Pairs?

Not checking for already paired strings. Confusing string reversal logic.

#2744

Find Maximum Number of String Pairs

Easy

ArrayHash TableStringSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can use a hash map to store the reversed strings and check for matches in a single pass, which is much more efficient than comparing every pair.

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Algorithm

3 steps

1Step 1: Create a hash map to store the reversed strings and their counts.
2Step 2: Iterate through the words and for each word, check if its reverse exists in the hash map.
3Step 3: If it exists, increment the pair count and remove both from the map to ensure they are not reused.

solution.py16 lines

1# Full working Python code
2words = ["cd", "ac", "dc", "ca", "zz"]
3
4def maxPairs(words):
5    count = 0
6    word_map = {}
7    for word in words:
8        rev = word[::-1]
9        if rev in word_map:
10            count += 1
11            del word_map[rev]
12        else:
13            word_map[word] = True
14    return count
15
16print(maxPairs(words))

ℹ

Complexity note: The time complexity is O(n) because we only pass through the list of words once. The space complexity is O(n) due to the hash map storing the words.

1Each string can only be paired once.
2Reversing a string is a key operation in this problem.

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