Given a string S consisting of N characters and positive integer K, the task is to check if there exist any (K + 1) strings i.e., A1, A2, A3, …, AK, A(K + 1) such that the concatenation of strings A1, A2, A3, …, AK, and A(K + 1) and the concatenation of the reverse of each strings AK, A(K – 1), A(K – 2), …, A1, and A0 is the string S. If found to be true, then print “Yes”. Otherwise, print “No”.
Examples:
Input: S = “qwqwq”, K = 1
Output: Yes
Explanation:
Consider the string A1 as “qw”, and A2 as “q”. Now the concatenation of A1, A2, reverse of A1 is “qwqwq”, which is the same as the given string S.Input: S = “qwqwa”, K = 2
Output: No
Approach: The given problem can be solved based on the observation that for a string S to satisfy the given condition, the first K characters must be equal to the last K characters of the given string. Follow the steps below to solve the problem:
- If the value of (2*K + 1) is greater than N, then print “No” and return from the function.
- Otherwise, store the prefix of size K i.e., S[0, …, K] in a string A, and the suffix of size K i.e., S[N – K, …, N – 1] in a string B.
- Reverse the string B and check if A is equal to B or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the string S// can be obtained by (K + 1) non-empty// substrings whose concatenation and// concatenation of the reverse// of these K stringsvoid checkString(string s, int k){ // Stores the size of the string int n = s.size(); // If n is less than 2*k+1 if (2 * k + 1 > n) { cout << "No"; return; } // Stores the first K characters string a = s.substr(0, k); // Stores the last K characters string b = s.substr(n - k, k); // Reverse the string reverse(b.begin(), b.end()); // If both the strings are equal if (a == b) cout << "Yes"; else cout << "No";}// Driver Codeint main(){ string S = "qwqwq"; int K = 1; checkString(S, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{ // Function to check if the string S // can be obtained by (K + 1) non-empty // substrings whose concatenation and // concatenation of the reverse // of these K strings static void checkString(String s, int k) { // Stores the size of the string int n = s.length(); // If n is less than 2*k+1 if (2 * k + 1 > n) { System.out.println("No"); return; } // Stores the first K characters String a = s.substring(0, k); // Stores the last K characters String b = s.substring(n - k, n); // Reverse the string StringBuffer str = new StringBuffer(b); // To reverse the string str.reverse(); b = str.toString(); // If both the strings are equal if (a.equals(b)) System.out.println("Yes"); else System.out.println("No"); } // Driver Code public static void main (String[] args) { String S = "qwqwq"; int K = 1; checkString(S, K); }}// This code is contributed by Dharanendra L V. |
Python3
# Python3 program for the above approach# Function to check if the S# can be obtained by (K + 1) non-empty# substrings whose concatenation and# concatenation of the reverse# of these K stringsdef checkString(s, k): # Stores the size of the string n = len(s) # If n is less than 2*k+1 if (2 * k + 1 > n): print("No") return # Stores the first K characters a = s[0:k] # Stores the last K characters b = s[n - k:n] # Reverse the string b = b[::-1] # If both the strings are equal if (a == b): print("Yes") else: print("No")# Driver Codeif __name__ == '__main__': S = "qwqwq" K = 1 checkString(S, K)# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;class GFG { // Function to check if the string S // can be obtained by (K + 1) non-empty // substrings whose concatenation and // concatenation of the reverse // of these K strings static void checkString(string s, int k) { // Stores the size of the string int n = s.Length; // If n is less than 2*k+1 if (2 * k + 1 > n) { Console.Write("No"); return; } // Stores the first K characters string a = s.Substring(0, k); // Stores the last K characters string b = s.Substring(n - k, k); // Reverse the string char[] arr = b.ToCharArray(); Array.Reverse(arr); b = new String(arr); // If both the strings are equal if (a == b) Console.Write("Yes"); else Console.Write("No"); } // Driver Code public static void Main() { string S = "qwqwq"; int K = 1; checkString(S, K); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript program for the above approach// Function to check if the string S// can be obtained by (K + 1) non-empty// substrings whose concatenation and// concatenation of the reverse// of these K stringsfunction checkString(s, k){ // Stores the size of the string let n = s.length; // If n is less than 2*k+1 if (2 * k + 1 > n) { document.write("No"); return; } // Stores the first K characters let a = s.substr(0, k); // Stores the last K characters let b = s.substr(n - k, k); // Reverse the string b.split("").reverse().join("") // If both the strings are equal if (a == b) document.write("Yes"); else document.write("No");}// Driver Code let S = "qwqwq"; let K = 1; checkString(S, K);// This code is contributed by gfgking.</script> |
Output:
Yes
Time complexity: O(N)
Auxiliary Space: O(N)
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