algoadvance

Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of the repeated a, return -1.

Clarifying Questions

1. What constitutes a valid repetition?
   • A valid repetition is such that repeating the string a for a certain number of times contains b as a substring.
2. Are there constraints on the lengths of the strings a and b?
   • Generally, the constraints are such that the problem should be solvable within reasonable time and space complexity limits. For this problem, let’s assume typical input sizes that fit within memory constraints (e.g., lengths of a few thousand characters).
3. Can either a or b be empty?
   • If a is empty and b is not, it’s impossible to form b. If b is empty, 0 repeats would be a valid return value since an empty string is trivially a substring of any string.

Strategy

1. Determine Initial Length:
   • Calculate the minimum number of repetitions n of a such that it could potentially contain b. This can be approximated by the ceiling of the length of b divided by the length of a.
2. Check for Substring Inclusion:
   • Start with n repetitions of a and check if b is a substring. If not, incrementally add another repetition and check again.
3. Maximum Necessary Length:
   • Given the nature of the problem, you only need to check up to n + 2 repetitions. This is because increasing repetitions beyond this point without finding b will not suddenly make b fit.
4. Return Result:
   • If b is found within these checks, return the number of repetitions. If not, return -1.

Code

def repeatedStringMatch(a: str, b: str) -> int:
    import math

    # Get the minimum number of repetitions needed
    min_reps = math.ceil(len(b) / len(a))

    # Check if b is a substring in the possible range of repetitions
    for i in range(min_reps, min_reps + 3):
        if b in a * i:
            return i

    return -1

Time Complexity

• Checking if b is a substring of the repeated a involves a check that can run in O(m * n) time in the worst case, where m is the length of b and n is the length of a. However, generally, this will be much faster due to the nature of substring checking and early exits.

• The loop runs at most 3 times, making the overall time complexity about O(3 * m * n), which simplifies to O(m * n).

Space Complexity

• The space complexity is dominated by the space needed to store the repeated string, which may go up to a length of (min_reps + 2) * len(a). In the worst case, this can be space linear in the length of the repeated string. Thus, the space complexity is O((min_reps + 2) * n).

This approach ensures both efficiency and clarity in checking for the substring b within the repeated a.

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