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Check if a string can be split into two substrings such that one substring is a substring of the other
  • Last Updated : 31 Dec, 2020

Given a string S of length N, the task is to check if a string can be split into two substrings, say A and B such that B is a substring of A. If not possible, print No. Otherwise, print Yes.

Examples :

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

Naive Approach: The simplest approach to solve the problem is to split the string S at every possible index, and check if the right substring is a substring of the left substring. If any split satisfies the condition, print “Yes“. Otherwise, print “No“. 



Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to check if the last character of the string S is present in the remaining string or not. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

Yes

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 

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