🚀4. Minimum Repeats to Make Substring 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given two strings s1 and s2, return the minimum number of times s1 has to be repeated so that s2 becomes a substring of the repeated s1. If s2 cannot be a substring of any repeated version of s1, return -1.

🔍 Example Walkthrough:

Input:

s1 = "ww", s2 = "www"

Output:

2

Explanation: Repeating s1 two times ("wwww"), s2 is a substring of it.

Input:

s1 = "abcd", s2 = "cdabcdab"

Output:

3

Explanation: Repeating s1 three times ("abcdabcdabcd"), s2 is a substring. Less than 3 repeats would not contain s2.

Input:

s1 = "ab", s2 = "cab"

Output:

-1

Explanation: No matter how many times s1 is repeated, s2 can’t be a substring.

Constraints:

1 ≤ |s1|, |s2| ≤ 10^5

🎯 My Approach:

1. Repeated Concatenation Check:

   • Initialize a repeated version of s1 and keep appending s1 to it until its length meets or exceeds s2’s length.

   • After each append, check if s2 becomes a substring of this extended s1.

   • If it does, return the number of repetitions used.

2. Pattern Matching with KMP Algorithm:

   • Use the KMP (Knuth-Morris-Pratt) algorithm to efficiently search for s2 within the extended repeated string of s1.

   • The LPS (Longest Prefix Suffix) array is computed for s2 to enable efficient substring search.

3. Edge Cases:

   • If s1 is already a substring of s2, return the number of repetitions found.

   • If no valid repetition count satisfies the substring condition, return -1.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n + m), where n is the length of s1 and m is the length of s2. This is because the KMP algorithm requires linear time to find a substring.

• Expected Auxiliary Space Complexity: O(m), as we need additional space to store the LPS array for s2.

📝 Solution Code

Code (C++)

class Solution {
public:
    int minRepeats(string s1, string s2) {
        int repeat = 1;
        string repeatedStr = s1;

        while (repeatedStr.length() < s2.length()) {
            repeatedStr += s1;
            repeat++;
        }

        if (repeatedStr.find(s2) != string::npos)
            return repeat;

        repeatedStr += s1;
        repeat++;

        if (repeatedStr.find(s2) != string::npos)
            return repeat;

        return -1;
    }
};

Code (Java)

class Solution {

    static void computeLPSArray(String s, int[] lps) {
        int len = 0, idx = 1;
        lps[0] = 0;

        while (idx < s.length()) {
            if (s.charAt(idx) == s.charAt(len)) {
                len++;
                lps[idx] = len;
                idx++;
            } else {
                if (len == 0) {
                    lps[idx] = 0;
                    idx++;
                } else {
                    len = lps[len - 1];
                }
            }
        }
    }

    static boolean KMPSearch(String txt, String pat, int[] lps, int rep) {
        int n = txt.length(), m = pat.length();
        int i = 0, j = 0;

        while (i < n * rep) {
            if (txt.charAt(i % n) == pat.charAt(j)) {
                i++;
                j++;
                if (j == m) return true;
            } else {
                if (j != 0) j = lps[j - 1];
                else i++;
            }
        }
        return false;
    }

    static int minRepeats(String s1, String s2) {
        int n = s1.length();
        int m = s2.length();
        int[] lps = new int[m];
        computeLPSArray(s2, lps);

        int x = (m + n - 1) / n;
        if (KMPSearch(s1, s2, lps, x)) return x;
        if (KMPSearch(s1, s2, lps, x + 1)) return x + 1;

        return -1;
    }
}

Code (Python)

class Solution:
    def computeLPSArray(self, s):
        lps = [0] * len(s)
        len_ = 0
        idx = 1

        while idx < len(s):
            if s[idx] == s[len_]:
                len_ += 1
                lps[idx] = len_
                idx += 1
            else:
                if len_ == 0:
                    lps[idx] = 0
                    idx += 1
                else:
                    len_ = lps[len_ - 1]
        return lps

    def KMPSearch(self, txt, pat, lps, rep):
        n, m = len(txt), len(pat)
        i = j = 0

        while i < n * rep:
            if txt[i % n] == pat[j]:
                i += 1
                j += 1
                if j == m:
                    return True
            else:
                if j != 0:
                    j = lps[j - 1]
                else:
                    i += 1
        return False

    def minRepeats(self, s1, s2):
        n, m = len(s1), len(s2)
        lps = self.computeLPSArray(s2)
        x = (m + n - 1) // n
        if self.KMPSearch(s1, s2, lps, x):
            return x
        if self.KMPSearch(s1, s2, lps, x + 1):
            return x + 1
        return -1

🎯 Contribution and Support:

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