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 forms i and j. The task is to print “Yes” if the element at index i is the same as the element at index j in S else print “No” for each query.
Examples :
Input : X = "neveropen", Q = 3. Query 1: 0 8 Query 2: 8 13 Query 3: 6 15 Output : Yes Yes No String S will be "neveropenneveropen....". 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 the length of string X be n. Observe that elements at indexes 0, n, 2n, 3n,…. are the 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 the same for string X.
Below is the implementation of the 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 Programint main(){ char X[] = "neveropen"; 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 stringimport java.io.*;public class GFG{ // Print whether index i and j// have same element or notstatic 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 = "neveropen"; 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 codeif __name__ == "__main__": X = "neveropen" 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 stringusing System;public class GFG{ // Print whether index i and j// have same element or notstatic 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 = "neveropen"; 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$X = "neveropen";query($X, 0, 8);query($X, 8, 13);query($X, 6, 15);// This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript Program to Queries for // same characters in a // repeated string // Print whether index i and j // have same element or not function query(s, i, j) { let 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]) document.write("Yes" + "</br>"); else document.write("No" + "</br>"); } let X = "neveropen"; query(X, 0, 8); query(X, 8, 13); query(X, 6, 15); </script> |
Output
Yes Yes No
Time complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Anuj Chauhan. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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