Queries for rotation and Kth character of the given string in constant time

Given a string str, the task is to perform the following type of queries on the given string:

  1. (1, K): Left rotate the string by K characters.
  2. (2, K): Print the Kth character of the string.

Examples:

Input: str = “abcdefgh”, q[][] = {{1, 2}, {2, 2}, {1, 4}, {2, 7}}
Output:
d
e
Query 1: str = “cdefghab”
Query 2: 2nd character is d
Query 3: str = “ghabcdef”
Query 4: 7th character is e



Input: str = “abc”, q[][] = {{1, 2}, {2, 2}}
Output:
a

Approach: The main observation here is that the string doesn’t need to be rotated in every query instead we can create a pointer ptr pointing to the first character of the string and which can be updated for every rotation as ptr = (ptr + K) % N where K the integer by which the string needs to be rotated and N is the length of the string. Now for every query of the second type, the Kth character can be found by str[(ptr + K – 1) % N].

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define size 2
  
// Function to perform the required
// queries on the given string
void performQueries(string str, int n,
                    int queries[][size], int q)
{
  
    // Pointer pointing to the current starting
    // character of the string
    int ptr = 0;
  
    // For every query
    for (int i = 0; i < q; i++) {
  
        // If the query is to rotate the string
        if (queries[i][0] == 1) {
  
            // Update the pointer pointing to the
            // starting character of the string
            ptr = (ptr + queries[i][1]) % n;
        }
        else {
  
            int k = queries[i][1];
  
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
  
            // Print the kth character
            cout << str[index] << "\n";
        }
    }
}
  
// Driver code
int main()
{
    string str = "abcdefgh";
    int n = str.length();
  
    int queries[][size] = { { 1, 2 }, { 2, 2 }, 
                            { 1, 4 }, { 2, 7 } };
    int q = sizeof(queries) / sizeof(queries[0]);
  
    performQueries(str, n, queries, q);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
import java.util.*;
class GFG 
{
static int size = 2;
  
// Function to perform the required
// queries on the given string
static void performQueries(String str, int n,
                           int queries[][], int q)
{
  
    // Pointer pointing to the current 
    // starting character of the string
    int ptr = 0;
  
    // For every query
    for (int i = 0; i < q; i++) 
    {
  
        // If the query is to rotate the string
        if (queries[i][0] == 1)
        {
  
            // Update the pointer pointing to the
            // starting character of the string
            ptr = (ptr + queries[i][1]) % n;
        }
        else 
        {
            int k = queries[i][1];
  
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
  
            // Print the kth character
            System.out.println(str.charAt(index));
        }
    }
}
  
// Driver code
public static void main(String[] args)
{
    String str = "abcdefgh";
    int n = str.length();
  
    int queries[][] = { { 1, 2 }, { 2, 2 }, 
                        { 1, 4 }, { 2, 7 } };
    int q = queries.length;
  
    performQueries(str, n, queries, q);
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
size = 2
  
# Function to perform the required 
# queries on the given string 
def performQueries(string, n, queries, q) : 
  
    # Pointer pointing to the current starting 
    # character of the string 
    ptr = 0
  
    # For every query 
    for i in range(q) :
  
        # If the query is to rotate the string 
        if (queries[i][0] == 1) : 
  
            # Update the pointer pointing to the 
            # starting character of the string 
            ptr = (ptr + queries[i][1]) % n; 
              
        else :
  
            k = queries[i][1]; 
  
            # Index of the kth character in the 
            # current rotation of the string 
            index = (ptr + k - 1) % n; 
  
            # Print the kth character 
            print(string[index]); 
  
# Driver code 
if __name__ == "__main__"
  
    string = "abcdefgh"
    n = len(string); 
  
    queries = [[ 1, 2 ], [ 2, 2 ], 
               [ 1, 4 ], [ 2, 7 ]]; 
    q = len(queries); 
  
    performQueries(string, n, queries, q); 
      
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
      
class GFG 
{
static int size = 2;
  
// Function to perform the required
// queries on the given string
static void performQueries(String str, int n,
                           int [,]queries, int q)
{
  
    // Pointer pointing to the current 
    // starting character of the string
    int ptr = 0;
  
    // For every query
    for (int i = 0; i < q; i++) 
    {
  
        // If the query is to rotate the string
        if (queries[i, 0] == 1)
        {
  
            // Update the pointer pointing to the
            // starting character of the string
            ptr = (ptr + queries[i, 1]) % n;
        }
        else
        {
            int k = queries[i, 1];
  
            // Index of the kth character in the
            // current rotation of the string
            int index = (ptr + k - 1) % n;
  
            // Print the kth character
            Console.WriteLine(str[index]);
        }
    }
}
  
// Driver code
public static void Main(String[] args)
{
    String str = "abcdefgh";
    int n = str.Length;
  
    int [,]queries = { { 1, 2 }, { 2, 2 }, 
                       { 1, 4 }, { 2, 7 } };
    int q = queries.GetLength(0);
  
    performQueries(str, n, queries, q);
}
}
  
// This code is contributed by PrinciRaj1992

chevron_right


Output:

d
e


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.