Last updated on October 5th, 2024 at 09:22 pm

Here, We see Repeated String Match LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Repeated String Match LeetCode Solution

Problem Statement

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 a after repeating it, return -1.

Notice: string “abc” repeated 0 times is “”, repeated 1 time is “abc” and repeated 2 times is “abcabc”.

Example 1:
Input: a = “abcd”, b = “cdabcdab”
Output: 3
Explanation: We return 3 because by repeating a three times “abcdabcdabcd”, b is a substring of it.

Example 2:
Input: a = “a”, b = “aa”
Output: 2

1. Repeated String Match Leetcode Solution C++

class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        string s="";
        int count = 0;
        while(s.size()<b.size())
        {
            s+=a;
            count++;
        }
        if(s.find(b)!=string::npos)
            return count;
        s+=a;
        count++;
        if(s.find(b)!=string::npos)
            return count;
        return -1;
    }
};

2. Repeated String Match Solution Java

class Solution {
    public int repeatedStringMatch(String a, String b) {
    StringBuilder gy=new StringBuilder();
        int I=0;
    for(I=1; gy.length()<=b.length(); I++){
        gy.append(a);
        if(gy.toString().contains(b))return I;
    }
        if(gy.append(a).toString().contains(b))return I;
        return -1;
    }
}

3. Repeated String Match Solution JavaScript

var repeatedStringMatch = function(a, b) {
    const count = Math.ceil(b.length / a.length)
    const str = a.repeat(count)
    return str.includes(b) ? count : (str + a).includes(b) ? count + 1 : -1     
};

4. Repeated String Match Solution Python

class Solution(object):
    def repeatedStringMatch(self, a, b):
    	if set(b).issubset(set(a)) == False: return -1
    	for i in range(1,int(len(b)/len(a))+3):
    		if b in a*i: return i
    	return -1