检查一个字符串可以重复以制作另一个字符串

给定两个字符串a和b ，任务是检查字符串a可以重复多少次以生成字符串b 。如果不能通过重复a生成b ，则打印-1 。

例子：

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

Input: a = “df”, b = “dfgrt”
Output: -1

方法：

• 如果len(b)％len(a)！= 0，则不能通过重复a来将-1打印为b 。
• 否则设置count = len(b)/ len(a)并重复一次计数次数。
  • 如果a = b，则打印计数。
  • 否则打印-1 。

下面是上述方法的实现：

C++

// CPP implementation of the approach
#include
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

输出：

2