检查一个字符串可以分为两个子字符串，以便一个子字符串是另一个子字符串

给定长度为N的字符串S ，任务是检查字符串可以分为两个子字符串，例如A和B ，使得B是A的子字符串。如果无法打印，请打印No。否则，打印是。

例子 ：

Input: S = “abcdab”
Output: Yes
Explanation: Considering the two splits to be A=”abcd” and B=”ab”, B is a substring of A.

Input: S = “abcd”
Output: No

天真的方法：解决该问题的最简单方法是在每个可能的索引处分割字符串S ，并检查右子字符串是否为左子字符串的子字符串。如果任何拆分均满足条件，则打印“是”。否则，打印“否”。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：为了优化上述方法，其思想是检查字符串S的最后一个字符是否存在于其余字符串。请按照以下步骤解决问题：

• 将S的最后一个字符存储在c中。
• 检查c是否存在于子字符串S [0，N-2]中。
• 如果发现是真的，则打印“是” 。否则打印“否” 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if a string can be
// divided into two substrings such that
// one substring is substring of the other
void splitString(string S, int N)
{
    // Store the last character of S
    char c = S[N - 1];
 
    int f = 0;
 
    // Traverse the characters at indices [0, N-2]
    for (int i = 0; i < N - 1; i++) {
 
        // Check if the current character is
        // equal to the last character
        if (S[i] == c) {
 
            // If true, set f = 1
            f = 1;
 
            // Break out of the loop
            break;
        }
    }
 
    if (f)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    // Given string, S
    string S = "abcdab";
 
    // Store the size of S
    int N = S.size();
 
    // Function Call
    splitString(S, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to check if a String can be
// divided into two subStrings such that
// one subString is subString of the other
static void splitString(String S, int N)
{
   
    // Store the last character of S
    char c = S.charAt(N - 1);
    int f = 0;
 
    // Traverse the characters at indices [0, N-2]
    for (int i = 0; i < N - 1; i++)
    {
 
        // Check if the current character is
        // equal to the last character
        if (S.charAt(i) == c)
        {
 
            // If true, set f = 1
            f = 1;
 
            // Break out of the loop
            break;
        }
    }
 
    if (f > 0)
        System.out.print("Yes");
    else
        System.out.print("No");
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given String, S
    String S = "abcdab";
 
    // Store the size of S
    int N = S.length();
 
    // Function Call
    splitString(S, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to check if a can be
# divided into two substrings such that
# one subis subof the other
def splitString(S, N):
     
    # Store the last character of S
    c = S[N - 1]
 
    f = 0
 
    # Traverse the characters at indices [0, N-2]
    for i in range(N - 1):
         
        # Check if the current character is
        # equal to the last character
        if (S[i] == c):
             
            # If true, set f = 1
            f = 1
             
            # Break out of the loop
            break
 
    if (f):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    # Given string, S
    S = "abcdab"
 
    # Store the size of S
    N = len(S)
 
    # Function Call
    splitString(S, N)
     
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach 
using System;
class GFG{
  
// Function to check if a string can be
// divided into two substrings such that
// one substring is substring of the other
static void splitString(string S, int N)
{
    // Store the last character of S
    char c = S[N - 1];
    int f = 0;
  
    // Traverse the characters at indices [0, N-2]
    for (int i = 0; i < N - 1; i++)
    {
  
        // Check if the current character is
        // equal to the last character
        if (S[i] == c)
        {
  
            // If true, set f = 1
            f = 1;
  
            // Break out of the loop
            break;
        }
    }
  
    if (f != 0)
        Console.Write("Yes");
    else
        Console.Write("No");
}
  
// Driver code
public static void Main()
{
    // Given string, S
    string S = "abcdab";
  
    // Store the size of S
    int N = S.Length;
  
    // Function Call
    splitString(S, N);
}
}
 
// This code is contributed by susmitakundugoaldanga

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)