Repeated String Match LeetCode Solution

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.

List of all LeetCode Solution

Topics

Bit Manipulation, Hash Table

Companies

LinkedIn

Level of Question

Medium

Repeated String Match LeetCode Solution

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
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