Python3 Program to Find array sum using Bitwise OR after splitting given array in two halves after K circular shifts

Last Updated : 06 Sep, 2024

Given an array A[] of length N, where N is an even number, the task is to answer Q independent queries where each query consists of a positive integer K representing the number of circular shifts performed on the array and find the sum of elements by performing Bitwise OR operation on the divided array.
Note: Each query begins with the original array.
Examples:

Input: A[] = {12, 23, 4, 21, 22, 76}, Q = 1, K = 2
Output: 117
Explanation:
Since K is 2, modified array A[]={22, 76, 12, 23, 4, 21}.
Bitwise OR of first half of array = (22 | 76 | 12) = 94
Bitwise OR of second half of array = (21 | 23 | 4) = 23
Sum of OR values is 94 + 23 = 117
Input: A[] = {7, 44, 19, 86, 65, 39, 75, 101}, Q = 1, K = 4
Output: 238
Since K is 4, modified array A[]={65, 39, 75, 101, 7, 44, 19, 86}.
Bitwise OR of first half of array = 111
Bitwise OR of second half of array = 127
Sum of OR values is 111 + 127 = 238

Naive Approach:
To solve the problem mentioned above the simplest approach is to shift each element of the array by K % (N / 2) and then traverse the array to calculate the OR of the two halves for every query. But this method is not efficient and hence can be optimized further.
Efficient Approach:
To optimize the above mentioned approach we can take the help of Segment Tree data structure.

Observation:

We can observe that after exactly N / 2 right circular shifts the two halves of the array become the same as in the original array. This effectively reduces the number of rotations to K % (N / 2).
Performing a right circular shift is basically shifting the last element of the array to the front. So for any positive integer X performing X right circular shifts is equal to shifting the last X elements of the array to the front.

Following are the steps to solve the problem :

Construct a segment tree for the original array A[] and assign a variable let’s say i = K % (N / 2).
Then for each query we use the segment tree of find the bitwise OR; that is Bitwise OR of i elements from the end OR bitwise OR of the first (N / 2) – i – 1 elements.
Then calculate the bitwise OR of elements in range [(N / 2) – i, N – i – 1].
Add the two results to get the answer for the ith query.

Below is the implementation of the above approach:

Python3

# Python3 program to find Bitwise OR of two
# equal halves of an array after performing
# K right circular shifts
MAX = 100005

# Array for storing
# the segment tree
seg = [0] * (4 * MAX)

# Function to build the segment tree
def build(node, l, r, a):

    if (l == r):
        seg[node] = a[l]

    else:
        mid = (l + r) // 2

        build(2 * node, l, mid, a)
        build(2 * node + 1, mid + 1, r, a)
        
        seg[node] = (seg[2 * node] | 
                     seg[2 * node + 1])

# Function to return the OR
# of elements in the range [l, r]
def query(node, l, r, start, end, a):
    
    # Check for out of bound condition
    if (l > end or r < start):
        return 0

    if (start <= l and r <= end):
        return seg[node]

    # Find middle of the range
    mid = (l + r) // 2

    # Recurse for all the elements in array
    return ((query(2 * node, l, mid, 
                       start, end, a)) | 
            (query(2 * node + 1, mid + 1, 
                       r, start, end, a)))

# Function to find the OR sum
def orsum(a, n, q, k):

    # Function to build the segment Tree
    build(1, 0, n - 1, a)

    # Loop to handle q queries
    for j in range(q):
        
        # Effective number of
        # right circular shifts
        i = k[j] % (n // 2)

        # Calculating the OR of
        # the two halves of the
        # array from the segment tree

        # OR of second half of the
        # array [n/2-i, n-1-i]
        sec = query(1, 0, n - 1, n // 2 - i,
                          n - i - 1, a)

        # OR of first half of the array
        # [n-i, n-1]OR[0, n/2-1-i]
        first = (query(1, 0, n - 1, 0, 
                             n // 2 - 
                             1 - i, a) |
                 query(1, 0, n - 1, 
                             n - i, 
                             n - 1, a))

        temp = sec + first

        # Print final answer to the query
        print(temp)

# Driver Code
if __name__ == "__main__":

    a = [ 7, 44, 19, 86, 65, 39, 75, 101 ]
    n = len(a)
    
    q = 2
    k = [ 4, 2 ]
    
    orsum(a, n, q, k)

# This code is contributed by chitranayal