Python3 Program to Find Range sum queries for anticlockwise rotations of Array by K indices

Given an array arr consisting of N elements and Q queries of the following two types:

1 K: For this type of query, the array needs to be rotated by K indices anticlockwise from its current state.
2 L R: For this query, the sum of the array elements present in the indices [L, R] needs to be calculated.

Example:

Input: arr = { 1, 2, 3, 4, 5, 6 }, query = { {2, 1, 3}, {1, 3}, {2, 0, 3}, {1, 4}, {2, 3, 5} }
Output:
9
16
12
Explanation:
For the 1st query {2, 1, 3} -> Sum of the elements in the indices [1, 3] = 2 + 3 + 4 = 9.
For the 2nd query {1, 3} -> Modified array after anti-clockwise rotation by 3 places is { 4, 5, 6, 1, 2, 3 }
For the 3rd query {2, 0, 3} -> Sum of the elements in the indices [0, 3] = 4 + 5 + 6 + 1 = 16.
For the 4th query {1, 4} -> Modified array after anti-clockwise rotation by 4 places is { 2, 3, 4, 5, 6, 1 }
For the 5th query {2, 3, 5} -> Sum of the elements in the indices [3, 5] = 5 + 6 + 1 = 12.

Approach:

Create a prefix array which is double the size of the arr and copy the element at the i^th index of arr to i^th and N + i^th index of prefix for all i in [0, N).
Precompute the prefix sum for every index of that array and store in prefix.
Set the pointer start at 0 to denote the starting index of the initial array.
For query of type 1, shift start to

((start + K) % N)^th position

For query of type 2, calculate

prefix[start + R]
 - prefix[start + L- 1 ]

if start + L >= 1 ,then print the value of

prefix[start + R]

Below code is the implementation of the above approach:

Python3

# Python3 program to calculate range sum
# queries for anticlockwise
# rotations of the array by K
 
# Function to execute the queries
def rotatedSumQuery(arr, n, query, Q):
 
    # Construct a new array
    # of size 2*N to store
    # prefix sum of every index
    prefix = [0] * (2 * n)
 
    # Copy elements to the new array
    for i in range(n):
        prefix[i] = arr[i]
        prefix[i + n] = arr[i]
 
    # Calculate the prefix sum
    # for every index
    for i in range(1, 2 * n):
        prefix[i] += prefix[i - 1];
 
    # Set start pointer as 0
    start = 0;
 
    for q in range(Q):
 
        # Query to perform
        # anticlockwise rotation
        if (query[q][0] == 1):
            k = query[q][1]
            start = (start + k) % n;
 
        # Query to answer range sum
        elif (query[q][0] == 2):
            L = query[q][1]
            R = query[q][2]
 
            # If pointing to 1st index
            if (start + L == 0):
 
                # Display the sum upto start + R
                print(prefix[start + R])
 
            else:
 
                # Subtract sum upto start + L - 1
                # from sum upto start + R
                print(prefix[start + R]-
                      prefix[start + L - 1])
         
# Driver code
arr = [ 1, 2, 3, 4, 5, 6 ];
 
# Number of query
Q = 5
 
# Store all the queries
query= [ [ 2, 1, 3 ],
         [ 1, 3 ],
         [ 2, 0, 3 ],
         [ 1, 4 ],
         [ 2, 3, 5 ] ]
 
n = len(arr);
rotatedSumQuery(arr, n, query, Q);
 
# This code is contributed by ankitkumar34