# C++ Program for Queries for rotation and Kth character of the given string in constant time

• Last Updated : 09 Jun, 2022

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:

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:

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++

 `// C++ implementation of the approach``#include ``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] == 1) {` `            ``// Update the pointer pointing to the``            ``// starting character of the string``            ``ptr = (ptr + queries[i]) % n;``        ``}``        ``else` `{` `            ``int` `k = queries[i];` `            ``// 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);` `    ``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!

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