必须重复最小次数A，以便B是它的子字符串

给定两个字符串A和B。任务是找到的最小次数A必须被重复，使得B是它的一个子串。如果没有这样的解决方案，则打印-1 。

例子：

Input : A = “abcd”, B = “cdabcdab”
Output : 3
Repeating A three times (“abcdabcdabcd”), B is a substring of it. B is not a substring of A when it is repeated less than 3 times

Input : A = “ab”, B = “cab”
Output : -1

方法 ：
想象一下，我们写了S = A + A + A + …如果B是S的子字符串，我们只需要检查某个索引是0还是1或…。 length(A)-1以B开头，因为S足够长以包含B ，并且S的长度为length(A) 。

现在，假设ans是length(B)<= length(A * ans)的最小数字。我们只需要检查B是A * ans还是A *(ans + 1)的子字符串。如果我们尝试k ，则B的长度大于A * ans的长度，因此不能是子字符串。当k = ans + 1时， A * k已经足够大，可以尝试B的所有位置( A [i：i + length(B)] == B ，其中i = 0，1，…，length(A)– 1 )。

下面是上述方法的实现：

C++

// CPP program to find Minimum number of times A 
// has to be repeated such that B is a substring of it
#include 
using namespace std;
  
// Function to check if a number 
// is a substring of other or not
bool issubstring(string str2, string rep1)
{
    int M = str2.length();
    int N = rep1.length();
  
    // Check for substring from starting 
    // from i'th index of main string
    for (int i = 0; i <= N - M; i++) {
        int j;
  
        // For current index i, 
        // check for pattern match
        for (j = 0; j < M; j++)
            if (rep1[i + j] != str2[j])
                break;
  
        if (j == M) // pattern matched
            return true;
    }
  
    return false; // not a substring
}
  
// Function to find Minimum number of times A 
// has to be repeated such that B is a substring of it
int Min_repetation(string A, string B)
{
    // To store minimum number of repetations
    int ans = 1;
      
    // To store repeated string
    string S = A;
      
    // Untill size of S is less than B
    while(S.size() < B.size())
    {
        S += A;
        ans++;
    }
      
    // ans times repetation makes required answer
    if (issubstring(B, S)) return ans;
      
    // Add one more string of A   
    if (issubstring(B, S+A)) 
        return ans + 1;
          
    // If no such solution exits    
    return -1;
}
  
// Driver code
int main()
{
    string A = "abcd", B = "cdabcdab";
      
    // Function call
    cout << Min_repetation(A, B);
      
    return 0;
}

Java

// Java program to find minimum number 
// of times 'A' has to be repeated 
// such that 'B' is a substring of it
class GFG 
{
  
// Function to check if a number 
// is a substring of other or not
static boolean issubstring(String str2, 
                           String rep1)
{
    int M = str2.length();
    int N = rep1.length();
  
    // Check for substring from starting 
    // from i'th index of main string
    for (int i = 0; i <= N - M; i++) 
    {
        int j;
  
        // For current index i, 
        // check for pattern match
        for (j = 0; j < M; j++)
            if (rep1.charAt(i + j) != str2.charAt(j))
                break;
  
        if (j == M) // pattern matched
            return true;
    }
  
    return false; // not a substring
}
  
// Function to find Minimum number 
// of times 'A' has to be repeated 
// such that 'B' is a substring of it
static int Min_repetation(String A, String B)
{
    // To store minimum number of repetations
    int ans = 1;
      
    // To store repeated string
    String S = A;
      
    // Untill size of S is less than B
    while(S.length() < B.length())
    {
        S += A;
        ans++;
    }
      
    // ans times repetation makes required answer
    if (issubstring(B, S)) return ans;
      
    // Add one more string of A 
    if (issubstring(B, S + A)) 
        return ans + 1;
          
    // If no such solution exits 
    return -1;
}
  
// Driver code
public static void main(String[] args) 
{
    String A = "abcd", B = "cdabcdab";
      
    // Function call
    System.out.println(Min_repetation(A, B));
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find minimum number
# of times 'A' has to be repeated
# such that 'B' is a substring of it
  
# Methof to find Minimum number 
# of times 'A' has to be repeated 
# such that 'B' is a substring of it
def min_repetitions(a, b):
    len_a = len(a)
    len_b = len(b)
      
    for i in range(0, len_a):
          
        if a[i] == b[0]:
            k = i
            count = 1
            for j in range(0, len_b):
                  
                # we are reiterating over A again and 
                # again for each value of B 
                # Resetting A pointer back to 0 as B 
                # is not empty yet
                if k >= len_a:
                    k = 0
                    count = count + 1
                      
                # Resetting A means count 
                # needs to be increased
                if a[k] != b[j]:
                    break
                k = k + 1
                  
            # k is iterating over A
            else:
                return count
    return -1
  
# Driver Code
A = 'abcd'
B = 'cdabcdab'
print(min_repetitions(A, B))
  
# This code is contributed by satycool

C#

// C# program to find minimum number 
// of times 'A' has to be repeated 
// such that 'B' is a substring of it
using System;
      
class GFG 
{
  
// Function to check if a number 
// is a substring of other or not
static Boolean issubstring(String str2, 
                           String rep1)
{
    int M = str2.Length;
    int N = rep1.Length;
  
    // Check for substring from starting 
    // from i'th index of main string
    for (int i = 0; i <= N - M; i++) 
    {
        int j;
  
        // For current index i, 
        // check for pattern match
        for (j = 0; j < M; j++)
            if (rep1[i + j] != str2[j])
                break;
  
        if (j == M) // pattern matched
            return true;
    }
  
    return false; // not a substring
}
  
// Function to find Minimum number 
// of times 'A' has to be repeated 
// such that 'B' is a substring of it
static int Min_repetation(String A, String B)
{
    // To store minimum number of repetations
    int ans = 1;
      
    // To store repeated string
    String S = A;
      
    // Untill size of S is less than B
    while(S.Length < B.Length)
    {
        S += A;
        ans++;
    }
      
    // ans times repetation makes required answer
    if (issubstring(B, S)) return ans;
      
    // Add one more string of A 
    if (issubstring(B, S + A)) 
        return ans + 1;
          
    // If no such solution exits 
    return -1;
}
  
// Driver code
public static void Main(String[] args) 
{
    String A = "abcd", B = "cdabcdab";
      
    // Function call
    Console.WriteLine(Min_repetation(A, B));
}
}
  
// This code is contributed by 29AjayKumar

输出：

3