C++ Program for 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, K): Left rotate the string by K characters.
(2, K): Print the K^th 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: 2^nd character is d
Query 3: str = “ghabcdef”
Query 4: 7^th 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 K^th character can be found by str[(ptr + K – 1) % N].
Below is the implementation of the above approach:

C++

// 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] << " ";

        }

    }
}
 
// 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;
}