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. (1, K): Left rotate the string by K characters.
2. (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;
}

Output

d e 

Time Complexity: O(Q), Where Q is the number of queries.
Auxiliary Space: O(1)

Please refer complete article on Queries for rotation and Kth character of the given string in constant time for more details!