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Python3 Program for Queries for rotation and Kth character of the given string in constant time

Last Updated : 05 Sep, 2024
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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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