How do you solve String Matching in an Array on LeetCode?

String Matching in an Array (LeetCode #1408) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + n*m) time complexity and O(n) space complexity. Key insight: Sorting helps reduce unnecessary comparisons.

What is the brute force solution for String Matching in an Array?

The brute force approach for String Matching in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every string against every other string to see if one is a substring of the other. This is straightforward but can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n + n*m).

What algorithm or data structure is used to solve String Matching in an Array?

String Matching in an Array uses the following concepts: Array, String, String Matching. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for String Matching in an Array?

Explain your choice of sorting and how it helps. Discuss the trade-offs between time and space complexity. Be prepared to handle edge cases and clarify constraints.

What are common mistakes when solving String Matching in an Array?

Not sorting the words before checking substrings. Assuming all strings are of similar length.

#1408

String Matching in an Array

Easy

ArrayStringString MatchingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + n*m)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + n*m)Space O(n)

An optimal approach involves sorting the words by length and then checking each word against only the longer words. This reduces unnecessary comparisons and improves efficiency.

⚙️

Algorithm

5 steps

1Step 1: Sort the words array by length.
2Step 2: Initialize an empty set to store results (to avoid duplicates).
3Step 3: Loop through each word and check if it is a substring of any longer word.
4Step 4: If a substring is found, add it to the set.
5Step 5: Convert the set to a list and return it.

solution.py9 lines

1def stringMatching(words):
2    words.sort(key=len)
3    result = set()
4    for i in range(len(words)):
5        for j in range(i + 1, len(words)):
6            if words[i] in words[j]:
7                result.add(words[i])
8                break
9    return list(result)

ℹ

Complexity note: The time complexity is O(n log n) for sorting the words and O(n*m) for checking substrings, where m is the average length of the strings. The space complexity is O(n) due to the storage of results in a set.

1Sorting helps reduce unnecessary comparisons.
2Using a set avoids duplicates in the result.

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