Given a string X. Form a string S by repeating string X multiple times i.e appending string X multiple times with itself. There are Q queries of form i and j. The task is to print “Yes” if the element at index i is same as the element at index j in S else print “No” for each query.
Examples :
Input : X = "geeksforgeeks", Q = 3. Query 1: 0 8 Query 2: 8 13 Query 3: 6 15 Output : Yes Yes No String S will be "geeksforgeeksgeeksforgeeks....". For Query 1, index 0 and index 8 have same element i.e 'g'. For Query 2, index 8 and index 13 have same element i.e 'g'. For Query 3, index 6 = 'o' and index 15 = 'e' which are not same.
Let length of string X be n. Observe that element at indexes 0, n, 2n, 3n,…. are same. Similarly for index i, position i, n+i, 2n+i, 3n+i,….. contain same element.
So, for each query, find (i%n) and (j%n) and if both are same for string X.
Below is the implementation of above idea :
C++
// Queries for same characters in a repeated // string #include<bits/stdc++.h> using namespace std; // Print whether index i and j have same // element or not. void query(char s[], int i, int j) { int n = strlen(s); // Finding relative position of index i,j. i %= n; j %= n; // Checking is element are same at index i, j. (s[i]==s[j])? (cout << "Yes" << endl): (cout << "No" << endl); } // Driven Program int main() { char X[] = "geeksforgeeks"; query(X, 0, 8); query(X, 8, 13); query(X, 6, 15); return 0; } |
Java
// Java Program to Queries for // same characters in a // repeated string import java.io.*; public class GFG{ // Print whether index i and j // have same element or not static void query(String s, int i, int j) { int n = s.length(); // Finding relative position // of index i,j i %= n; j %= n; // Checking is element are same // at index i, j if(s.charAt(i) == s.charAt(j)) System.out.println("Yes"); else System.out.println("No"); } // Driver Code static public void main (String[] args) { String X = "geeksforgeeks"; query(X, 0, 8); query(X, 8, 13); query(X, 6, 15); } } // This code is contributed by vt_m. |
Python3
# Queries for same characters in a repeated # string # Print whether index i and j have same # element or not. def query(s, i, j): n = len(s) # Finding relative position of index i,j. i %= n j %= n # Checking is element are same at index i, j. print("Yes") if s[i] == s[j] else print("No") # Driver code if __name__ == "__main__": X = "geeksforgeeks" query(X, 0, 8) query(X, 8, 13) query(X, 6, 15) # This code is contributed by # sanjeev2552 |
C#
// C# Program to Queries for // same characters in a // repeated string using System; public class GFG{ // Print whether index i and j // have same element or not static void query(string s, int i, int j) { int n = s.Length; // Finding relative position // of index i,j. i %= n; j %= n; // Checking is element are // same at index i, j if(s[i] == s[j]) Console.WriteLine("Yes"); else Console.WriteLine("No"); } // Driver Code static public void Main () { string X = "geeksforgeeks"; query(X, 0, 8); query(X, 8, 13); query(X, 6, 15); } } // This code is contributed by vt_m. |
PHP
<?php // Queries for same characters // in a repeated string // Print whether index i and j // have same element or not. function query($s, $i, $j) { $n = strlen($s); // Finding relative position // of index i,j. $i %= $n; $j %= $n; // Checking is element are // same at index i, j. if(($s[$i] == $s[$j])) echo "Yes\n" ; else echo "No" ; } // Driver Code $X = "geeksforgeeks"; query($X, 0, 8); query($X, 8, 13); query($X, 6, 15); // This code is contributed by nitin mittal. ?> |
Output :
Yes Yes No
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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