Queries for characters in a repeated string

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 = "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 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 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.
?>

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 = "geeksforgeeks";

    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 GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.