Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

Check if a string can be repeated to make another string

  • Difficulty Level : Medium
  • Last Updated : 13 May, 2021

Given two strings a and b, the task is to check how many times the string a can be repeated to generate the string b. If b cannot be generated by repeating a then print -1.
Examples: 
 

Input: a = “geeks”, b = “geeksgeeks” 
Output: 2 
“geeks” can be repeated twice to generate “geeksgeeks”
Input: a = “df”, b = “dfgrt” 
Output: -1 
 

 

Approach: 
 

  • If len(b) % len(a) != 0 then print -1 as b cannot be generated by repeating a.
  • Else set count = len(b) / len(a) and repeat a count number of times. 
    • If a = b then print count.
    • Else print -1.

Below is the implementation of the above approach: 
 

C++




// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to return the count of repetitions
// of string a to generate string b
int getCount(string a, string b)
{
 
    // If b cannot be generated by repeating a
    if(b.length() % a.length() != 0)
        return -1;
     
    int count = b.length() /a.length();
 
    // Repeat a count number of times
    string str = "";
    for(int i = 0; i < count; i++)
    {
        str = str + a;
    }
    if(str == b)
        return count;
     
    return -1;
}
 
// Driver code
int main()
{
    string a = "geeks";
    string b = "geeksgeeks";
    cout << (getCount(a, b));
    return 0;
}
 
// This code is contributed by
// Surendra_Gangwar
Java



// Java implementation of the approach
 
class GfG
{
 
    // Function to return the count of repetitions
    // of string a to generate string b
    static int getCount(String a, String b)
    {
        // If b cannot be generated by repeating a
        if(b.length() % a.length() != 0)
            return -1;
         
        int count = b.length() / a.length();
     
        // Repeat a count number of times
        String str = "";
        for(int i = 0; i < count; i++)
        {
            str = str + a;
        }
         
        if(str.equals(b))
            return count;
         
        return -1;
    }
 
    // Driver code
    public static void main(String []args)
    {
        String a = "geeks";
        String b = "geeksgeeks";
        System.out.println(getCount(a, b));
    }
}
     
// This code is contributed by Rituraj Jain
Python 3



# Python3 implementation of the approach
 
# Function to return the count of repetitions
# of string a to generate string b
def getCount(a, b):
     
    # If b cannot be generated by repeating a
    if(len(b) % len(a) != 0):
        return -1;
     
    count = int(len(b) / len(a))
     
    # Repeat a count number of times
    a = a * count
     
    if(a == b):
        return count
     
    return -1;
 
# Driver code
if __name__ == '__main__':
    a = 'geeks'
    b = 'geeksgeeks'
    print(getCount(a, b))
C#



// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to return the count of repetitions
    // of string a to generate string b
    static int getCount(String a, String b)
    {
        // If b cannot be generated by repeating a
        if(b.Length % a.Length != 0)
            return -1;
         
        int count = b.Length / a.Length;
     
        // Repeat a count number of times
        String str = "";
        for(int i = 0; i < count; i++)
        {
            str = str + a;
        }
         
        if(str.Equals(b))
            return count;
         
        return -1;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        String a = "geeks";
        String b = "geeksgeeks";
        Console.WriteLine(getCount(a, b));
    }
}
 
// This code contributed by Rajput-Ji
PHP



<?php
// PHP implementation of the approach
 
// Function to return the count of repetitions
// of string a to generate string b
function getCount($a, $b)
{
     
    // If b cannot be generated by repeating a
    if(strlen($b) % strlen($a) != 0)
        return -1;
     
    $count = floor(strlen($b) / strlen($a));
     
    // Repeat a count number of times
    // Repeat a count number of times
    $str = "";
    for($i = 0; $i < $count; $i++)
    {
        $str = $str . $a ;
    }
     
    if(strcmp($a,$b))
        return $count;
     
    return -1;
}
 
// Driver code
$a = 'geeks';
$b = 'geeksgeeks';
echo getCount($a, $b);
 
// This code is contributed by Ryuga
?>
Javascript



<script>
 
// Javascript implementation of the approach
 
 
    // Function to return the count of repetitions
    // of string a to generate string b
    function getCount( a,  b) {
        // If b cannot be generated by repeating a
        if (b.length % a.length != 0)
            return -1;
 
        var count = parseInt(b.length / a.length);
 
        // Repeat a count number of times
        var str = "";
        for (i = 0; i < count; i++) {
            str = str + a;
        }
 
        if (str == (b))
            return count;
 
        return -1;
    }
 
    // Driver code
     
        var a = "geeks";
        var b = "geeksgeeks";
        document.write(getCount(a, b));
 
// This code is contributed by todaysgaurav
 
</script>
Output: 
2
 

My Personal Notes
arrow_drop_up
Previous
access command in linux with examples
Next

Implementing Directory Management using Shell Script
Recommended Articles
Page :
Check if string S1 can be formed using repeated insertions of another string S2
20, Aug 21
Check if a string can be obtained by rotating another string 2 places
16, Mar 17
Check if a string can be converted to another string by replacing vowels and consonants
11, Feb 19
Check if a string can be obtained by rotating another string d places
10, Jul 19
Check if a string can be formed from another string by at most X circular clockwise shifts
10, May 20
Check if a string can be obtained by appending subsequences of another string
27, Oct 20
Check if a string can be made equal to another string by swapping or replacement of characters
24, Jan 21
Check if a string can be converted to another given string by removal of a substring
21, Dec 20
C++ Program to Check if a string can be formed from another string by at most X circular clockwise shifts
10, Dec 21
Java Program for Check if a string can be formed from another string by at most X circular clockwise shifts
10, Jan 22
Python3 Program to Check if a string can be formed from another string by at most X circular clockwise shifts
10, Jan 22
C++ Program to Check if a string can be obtained by rotating another string d places
10, Dec 21
Python3 Program for Check if a string can be obtained by rotating another string d places
10, Dec 21
Javascript Program to Check if a string can be obtained by rotating another string d places
10, Dec 21
C++ Program to Check if a string can be obtained by rotating another string 2 places
16, Nov 21
Java Program to Check if a string can be obtained by rotating another string 2 places
16, Nov 21
Python3 Program to Check if a string can be obtained by rotating another string 2 places
16, Oct 21
Javascript Program to Check if a string can be obtained by rotating another string 2 places
16, Nov 21
Javascript Program to Check if a string can be formed from another string by at most X circular clockwise shifts
10, Jan 22
Check if a string can be formed from another string using given constraints
21, Jun 18
Check if any permutation of string is a K times repeated string
08, Apr 20
Check if given string is a substring of string formed by repeated concatenation of z to a
22, Jul 20
Check if permutation of one string can break permutation of another
28, Jul 20
Check if one string can be converted to another
21, Jan 20
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
dkp1903
@dkp1903
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!