(Easy) Repeated String Match - LeetCode

Description:

Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1.

For example, with A = "abcd" and B = "cdabcdab".

Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").

Note:
The length of A and B will be between 1 and 10000.

Accepted

73,870

Submissions

233,180

 

Solution:

Initial Attempt:  Failed due to TIME OUT EXCEPTION

 

 

 

Apparantly , if String B is very long, and A is extremely short, the repeat times is almost the same length as String B. Thus, would cause timeout exception. 

 

 

class Solution {
    public int repeatedStringMatch(String A, String B) {
        
        if(A==null||A.length()==0){
            return -1;
        }
        
        int times = B.length() / A.length()+ 2; 
      
      // System.out.println(Repeat(A,3).indexOf(B));
        
        for(int i = 1; i<= times; i++){
            
            if(Repeat(A,i).indexOf(B)>-1){
                
                return i;
            }
        }
        
        return  -1;
    }
    
    static String Repeat (String a, int b){
        String result = a;
        
        for(int i = 1; i<b; i++){
            
            result = result + a; 
        }
        
        return result; 
    }
}

 

 Second Attempt

class Solution {
    public int repeatedStringMatch(String A, String B) {
        
        if(A==null||A.length()==0){
            return -1;
        }
        
        StringBuffer sb = new StringBuffer(A);
        
        int count = 1; 
        
        while(sb.indexOf(B)==-1){
            
            
            if(sb.length()>A.length()+B.length()){
                return -1;
            }
            sb = sb.insert(0,A);
            
            count++;
            
        }
         return count;
    }  
}