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This approach involves repeating the string a incrementally and checking if b becomes a substring. The minimum length of repeated a needed should be at least equal to b's length. Thus, start by repeating a just enough times to get a string longer than b and check if b is a substring. If not, repeat a one more time to cover scenarios where b spans two a's boundary.
Time Complexity: O((m + n) * n), where m is the length of a and n is the length of b. Space Complexity: O(m + n).
1def repeatedStringMatch(a: str, b: str) -> int:
2 m, n = len(a), len(b)
3 repeat = (n + m - 1) // m
4 for i in range(repeat + 2):
5 if b in a * i:
6 return i
7 return -1
8
9# Example
10print(repeatedStringMatch("abcd", "cdabcdab"))In this Python solution, `a` is repeatedly concatenated to make a longer string until either `b` is a substring or the repetition exceeds the possible size necessary to include `b`.
This approach relies on two repeated concatenations of string a. As per the pigeonhole principle, if b can fit within the repeated strings, it should certainly fit within two full concatenated repetitions and its prefix or suffix match.
Time Complexity: O(m * n). Space Complexity: O(m).
1#include <iostream>
#include <string>
int repeatedStringMatch(std::string a, std::string b) {
int m = a.size(), n = b.size();
int repeat = (n + m - 1) / m;
std::string doubled = a + a;
for (int i = 0; i < repeat; ++i) {
if (doubled.substr(i) + a).find(b) != std::string::npos) return i + 1;
}
return -1;
}
int main() {
std::string a = "abcd";
std::string b = "cdabcdab";
std::cout << repeatedStringMatch(a, b) << std::endl;
return 0;
}The solution effectively utilizes the doubled string to check for presence of b, reducing the calculations involved.