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

• Last Updated : 31 May, 2022

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:

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 ith index of arr to ith and N + ith 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 or print the value of

`prefix[start + R]`
• otherwise.

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`

Output:

```9
16
12```

Time Complexity: O(Q), where Q is the number of queries, and as each query will cost O (1) time for Q queries time complexity would be O(Q).

Auxiliary Space: O(N), as we are using  extra space for prefix.

Please refer complete article on Range sum queries for anticlockwise rotations of Array by K indices for more details!

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