LeetCode 686: Repeated String Match Solution

Master LeetCode problem 686 (Repeated String Match), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

686. Repeated String Match

Medium

String String Matching

Problem Explanation

Explanation

To solve this problem, we need to determine the minimum number of times we should repeat string a in order to make it long enough so that string b becomes a substring of it. We can achieve this by repeatedly concatenating a and checking if b is a substring of the concatenated string. We can optimize this process by calculating the maximum number of repetitions needed based on the lengths of strings a and b.

Calculate the maximum number of repetitions needed: repeat = ceil(b.length() / a.length()).
Generate a string by concatenating a repeat times.
Check if b is a substring of the generated string. If it is, return repeat. If not, check one more repetition and return -1 if b is still not a substring.

The time complexity of this approach is O(n * m) where n is the length of string a and m is the length of string b.

Solution Code

class Solution {
    public int repeatedStringMatch(String a, String b) {
        int repeat = (int) Math.ceil((double)b.length() / a.length());
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < repeat; i++) {
            sb.append(a);
        }
        if (sb.toString().contains(b)) {
            return repeat;
        }
        sb.append(a);
        if (sb.toString().contains(b)) {
            return repeat + 1;
        }
        return -1;
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 686 (Repeated String Match)?

The time complexity for LeetCode 686 (Repeated String Match) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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