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:
- Store the last character of S in c.
- Check if c is present in the substring S[0, N-2].
- If found to be true, print “YES”. otherwise print “NO”.
Below is the implementation of the above approach:
C++
// 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 othervoid 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 Codeint 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 approachimport java.util.*;class GFG{// Function to check if a String can be// divided into two subStrings such that// one subString is subString of the otherstatic 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 Codepublic 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 otherdef 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 Codeif __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 otherstatic 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 codepublic 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 |
Javascript
<script>// JavaScript program to implement// the above approach// Function to check if a String can be// divided into two subStrings such that// one subString is subString of the otherfunction splitString(S , N){ // Store the last character of S var c = S.charAt(N - 1); var f = 0; // Traverse the characters at indices [0, N-2] for (var 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) document.write("Yes"); else document.write("No");}// Driver Code//Given String, Svar S = "abcdab";// Store the size of Svar N = S.length;// Function CallsplitString(S, N);// This code contributed by shikhasingrajput </script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
