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

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  • 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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