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

• Last Updated : 31 Mar, 2022

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 `

Output:

```238
230```

Time Complexity: O(N + Q*logN)

Auxiliary Space: O(4*MAX)

Please refer complete article on Find array sum using Bitwise OR after splitting given array in two halves after K circular shifts for more details!

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