How do you solve Repeated String Match on LeetCode?

Repeated String Match (LeetCode #686) 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: Understanding string lengths helps in determining how many times to repeat 'a'.

What is the brute force solution for Repeated String Match?

The brute force approach for Repeated String Match is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly concatenating string 'a' until its length is at least as long as string 'b'. We then check if 'b' is a substring of this concatenated string. This method 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 Repeated String Match?

Repeated String Match uses the following concepts: String, String Matching. The recommended patterns to study are: String Matching, Two Pointers.

What are the interview tips for Repeated String Match?

Always clarify the problem constraints and edge cases with the interviewer. Think aloud while solving to show your thought process and reasoning. Consider both brute-force and optimal solutions to demonstrate your understanding of efficiency.

What are common mistakes when solving Repeated String Match?

Not considering the case where 'b' is longer than 'a' and failing to repeat enough times. Confusing substring checks with string equality checks.

#686

Repeated String Match

Medium

StringString MatchingString MatchingTwo Pointers

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)

The optimal solution leverages the properties of string matching. By calculating how many times we need to repeat 'a' based on the lengths of 'a' and 'b', we can minimize unnecessary concatenations and directly check for substring presence.

⚙️

Algorithm

4 steps

1Step 1: Calculate the minimum number of times 'a' needs to be repeated to ensure its length is at least the length of 'b'.
2Step 2: Create a repeated string by repeating 'a' the calculated number of times.
3Step 3: Check if 'b' is a substring of this repeated string. If not, repeat 'a' one more time and check again.
4Step 4: Return the count if 'b' is found; otherwise, return -1.

solution.py7 lines

1def repeatedStringMatch(a, b):
2    repeat_count = -(-len(b) // len(a))  # Ceiling division
3    repeated = a * repeat_count
4    if b in repeated:
5        return repeat_count
6    repeated += a
7    return repeat_count + 1 if b in repeated else -1

ℹ

Complexity note: The time complexity is O(n) because we only concatenate 'a' a limited number of times based on the length of 'b'. The space complexity is O(n) due to the storage of the repeated string.

1Understanding string lengths helps in determining how many times to repeat 'a'.
2Checking for substrings efficiently can save unnecessary computations.

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