# Python3 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:

## Python3

 `# 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`

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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