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 or print the value of 

prefix[start + R]

• otherwise.

Below code is the implementation of the above approach: 

C++

// C++ Program to calculate range sum
// queries for anticlockwise
// rotations of array by K

#include <bits/stdc++.h>
using namespace std;

// Function to execute the queries
void rotatedSumQuery(
    int arr[], int n,
    vector<vector<int> >& query,
    int Q)
{
    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int prefix[2 * n];

    // Copy elements to the new array
    for (int i = 0; i < n; i++) {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }

    // Calculate the prefix sum
    // for every index
    for (int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];

    // Set start pointer as 0
    int start = 0;

    for (int q = 0; q < Q; q++) {

        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1) {
            int k = query[q][1];
            start = (start + k) % n;
        }

        // Query to answer range sum
        else if (query[q][0] == 2) {

            int L, R;
            L = query[q][1];
            R = query[q][2];

            // If pointing to 1st index
            if (start + L == 0)

                // Display the sum upto start + R
                cout << prefix[start + R] << endl;

            else

                // Subtract sum upto start + L - 1
                // from sum upto start + R
                cout << prefix[start + R]
                            - prefix[start + L - 1]
                     << endl;
        }
    }
}

// Driver code
int main()
{

    int arr[] = { 1, 2, 3, 4, 5, 6 };

    // Number of query
    int Q = 5;

    // Store all the queries
    vector<vector<int> > query
        = { { 2, 1, 3 },
            { 1, 3 },
            { 2, 0, 3 },
            { 1, 4 },
            { 2, 3, 5 } };

    int n = sizeof(arr) / sizeof(arr[0]);
    rotatedSumQuery(arr, n, query, Q);

    return 0;
}

Java

// Java program to calculate range sum
// queries for anticlockwise
// rotations of array by K
class GFG{

// Function to execute the queries
static void rotatedSumQuery(int arr[], int n,
                            int [][]query, int Q)
{

    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int []prefix = new int[2 * n];

    // Copy elements to the new array
    for(int i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }

    // Calculate the prefix sum
    // for every index
    for(int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];

    // Set start pointer as 0
    int start = 0;

    for(int q = 0; q < Q; q++)
    {

        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1)
        {
            int k = query[q][1];
            start = (start + k) % n;
        }

        // Query to answer range sum
        else if (query[q][0] == 2)
        {
            int L, R;
            L = query[q][1];
            R = query[q][2];

            // If pointing to 1st index
            if (start + L == 0)

                // Display the sum upto start + R
                System.out.print(prefix[start + R] + "\n");

            else

                // Subtract sum upto start + L - 1
                // from sum upto start + R
                System.out.print(prefix[start + R] -
                                 prefix[start + L - 1] +
                                 "\n");
        }
    }
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };

    // Number of query
    int Q = 5;

    // Store all the queries
    int [][]query = { { 2, 1, 3 },
                      { 1, 3 },
                      { 2, 0, 3 },
                      { 1, 4 },
                      { 2, 3, 5 } };

    int n = arr.length;
    rotatedSumQuery(arr, n, query, Q);
}
}

// This code is contributed by Rohit_ranjan

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

C#

// C# program to calculate range sum
// queries for anticlockwise
// rotations of array by K
using System;

class GFG{

// Function to execute the queries
static void rotatedSumQuery(int[] arr, int n,
                            int[,] query, int Q)
{

    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    int[] prefix = new int[2 * n];

    // Copy elements to the new array
    for(int i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }

    // Calculate the prefix sum
    // for every index
    for(int i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];

    // Set start pointer as 0
    int start = 0;

    for(int q = 0; q < Q; q++)
    {

        // Query to perform
        // anticlockwise rotation
        if (query[q, 0] == 1)
        {
            int k = query[q, 1];
            start = (start + k) % n;
        }

        // Query to answer range sum
        else if (query[q, 0] == 2)
        {
            int L, R;
            L = query[q, 1];
            R = query[q, 2];

            // If pointing to 1st index
            if (start + L == 0)

                // Display the sum upto start + R
                Console.Write(prefix[start + R] + "\n");

            else

                // Subtract sum upto start + L - 1
                // from sum upto start + R
                Console.Write(prefix[start + R] -
                              prefix[start + L - 1] +
                              "\n");
        }
    }
}

// Driver code
public static void Main()
{
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };

    // Number of query
    int Q = 5;

    // Store all the queries
    int[,] query = new int[,] { { 2, 1, 3 },
                                { 1, 3, 0 },
                                { 2, 0, 3 },
                                { 1, 4, 0 },
                                { 2, 3, 5 } };

    int n = arr.Length;
    rotatedSumQuery(arr, n, query, Q);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program to calculate range sum
// queries for anticlockwise
// rotations of array by K

// Function to execute the queries
function rotatedSumQuery(arr, n,
                            query, Q)
{

    // Construct a new array
    // of size 2*N to store
    // prefix sum of every index
    let prefix = [];

    // Copy elements to the new array
    for(let i = 0; i < n; i++)
    {
        prefix[i] = arr[i];
        prefix[i + n] = arr[i];
    }

    // Calculate the prefix sum
    // for every index
    for(let i = 1; i < 2 * n; i++)
        prefix[i] += prefix[i - 1];

    // Set start pointer as 0
    let start = 0;

    for(let q = 0; q < Q; q++)
    {

        // Query to perform
        // anticlockwise rotation
        if (query[q][0] == 1)
        {
            let k = query[q][1];
            start = (start + k) % n;
        }

        // Query to answer range sum
        else if (query[q][0] == 2)
        {
            let L, R;
            L = query[q][1];
            R = query[q][2];

            // If pointing to 1st index
            if (start + L == 0)

                // Display the sum upto start + R
                document.write(prefix[start + R] + "<br/>");

            else

                // Subtract sum upto start + L - 1
                // from sum upto start + R
                document.write(prefix[start + R] -
                                 prefix[start + L - 1] +
                                 "<br/>");
        }
    }
}

// Driver code

    let arr = [ 1, 2, 3, 4, 5, 6 ];

    // Number of query
    let Q = 5;

    // Store all the queries
    let query = [[ 2, 1, 3 ],
                      [ 1, 3 ],
                      [ 2, 0, 3 ],
                      [ 1, 4 ],
                      [ 2, 3, 5 ]];

    let n = arr.length;
    rotatedSumQuery(arr, n, query, Q);

    // This code is contributed by susmitakundugoaldanga.
</script>

Output:

9
16
12

Time Complexity: O(N+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(N+Q).

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