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

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:

## C++

 `// C++ Program to find Bitwise OR of two``// equal halves of an array after performing``// K right circular shifts``#include ``const` `int` `MAX = 100005;``using` `namespace` `std;` `// Array for storing``// the segment tree``int` `seg[4 * MAX];` `// Function to build the segment tree``void` `build(``int` `node, ``int` `l, ``int` `r, ``int` `a[])``{``    ``if` `(l == r)``        ``seg[node] = a[l];` `    ``else` `{``        ``int` `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]``int` `query(``int` `node, ``int` `l, ``int` `r,``          ``int` `start, ``int` `end, ``int` `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``    ``int` `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``void` `orsum(``int` `a[], ``int` `n, ``int` `q, ``int` `k[])``{``    ``// Function to build the segment Tree``    ``build(1, 0, n - 1, a);` `    ``// Loop to handle q queries``    ``for` `(``int` `j = 0; j < q; j++) {``        ``// Effective number of``        ``// right circular shifts``        ``int` `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]``        ``int` `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]``        ``int` `first = (query(1, 0, n - 1, 0,``                           ``n / 2 - 1 - i, a)``                     ``| query(1, 0, n - 1,``                             ``n - i, n - 1, a));` `        ``int` `temp = sec + first;` `        ``// Print final answer to the query``        ``cout << temp << endl;``    ``}``}` `// Driver Code``int` `main()``{` `    ``int` `a[] = { 7, 44, 19, 86, 65, 39, 75, 101 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``int` `q = 2;` `    ``int` `k[q] = { 4, 2 };` `    ``orsum(a, n, q, k);` `    ``return` `0;``}`

## Java

 `// Java program to find Bitwise OR of two``// equal halves of an array after performing``// K right circular shifts``import` `java.util.*;` `class` `GFG{``    ` `static` `int` `MAX = ``100005``; ` `// Array for storing``// the segment tree``static` `int` `[]seg = ``new` `int``[``4` `* MAX];` `// Function to build the segment tree``static` `void` `build(``int` `node, ``int` `l, ``                  ``int` `r, ``int` `a[])``{``    ``if` `(l == r)``        ``seg[node] = a[l];` `    ``else``    ``{``        ``int` `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]``static` `int` `query(``int` `node, ``int` `l, ``int` `r,``                 ``int` `start, ``int` `end, ``int` `a[])``{``    ` `    ``// Check for out of bound condition``    ``if` `(l > end || r < start)``        ``return` `0``;` `    ``if` `(start <= l && r <= end)``        ``return` `seg[node];` `    ``// Find middle of the range``    ``int` `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``static` `void` `orsum(``int` `a[], ``int` `n, ``                  ``int` `q, ``int` `k[])``{``    ` `    ``// Function to build the segment Tree``    ``build(``1``, ``0``, n - ``1``, a);` `    ``// Loop to handle q queries``    ``for``(``int` `j = ``0``; j < q; j++)``    ``{``        ` `        ``// Effective number of``        ``// right circular shifts``        ``int` `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]``        ``int` `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]``        ``int` `first = (query(``1``, ``0``, n - ``1``, ``0``,``                           ``n / ``2` `- ``1` `- i, a) |``                     ``query(``1``, ``0``, n - ``1``,``                           ``n - i, n - ``1``, a));` `        ``int` `temp = sec + first;` `        ``// Print final answer to the query``        ``System.out.print(temp + ``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{` `    ``int` `a[] = { ``7``, ``44``, ``19``, ``86``, ``65``, ``39``, ``75``, ``101` `};``    ``int` `n = a.length;``    ``int` `q = ``2``;` `    ``int` `k[] = { ``4``, ``2` `};` `    ``orsum(a, n, q, k);``}``}` `// This code is contributed by 29AjayKumar`

## 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 `

## C#

 `// C# program to find Bitwise OR of two ``// equal halves of an array after performing ``// K right circular shifts ``using` `System;``class` `GFG{ ``    ` `static` `int` `MAX = 100005; ` `// Array for storing ``// the segment tree ``static` `int` `[]seg = ``new` `int``[4 * MAX]; ` `// Function to build the segment tree ``static` `void` `build(``int` `node, ``int` `l, ``                  ``int` `r, ``int` `[]a) ``{ ``    ``if` `(l == r) ``        ``seg[node] = a[l]; ` `    ``else``    ``{ ``        ``int` `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] ``static` `int` `query(``int` `node, ``int` `l, ``int` `r, ``                 ``int` `start, ``int` `end, ``int` `[]a) ``{ ``    ` `    ``// Check for out of bound condition ``    ``if` `(l > end || r < start) ``        ``return` `0; ` `    ``if` `(start <= l && r <= end) ``        ``return` `seg[node]; ` `    ``// Find middle of the range ``    ``int` `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 ``static` `void` `orsum(``int` `[]a, ``int` `n, ``                  ``int` `q, ``int` `[]k) ``{ ``    ` `    ``// Function to build the segment Tree ``    ``build(1, 0, n - 1, a); ` `    ``// Loop to handle q queries ``    ``for``(``int` `j = 0; j < q; j++) ``    ``{ ``        ` `        ``// Effective number of ``        ``// right circular shifts ``        ``int` `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] ``        ``int` `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] ``        ``int` `first = (query(1, 0, n - 1, 0, ``                         ``n / 2 - 1 - i, a) | ``                    ``query(1, 0, n - 1, ``                          ``n - i, n - 1, a)); ` `        ``int` `temp = sec + first; ` `        ``// Print readonly answer to the query ``        ``Console.Write(temp + ``"\n"``); ``    ``} ``} ` `// Driver Code ``public` `static` `void` `Main(String[] args) ``{ ``    ``int` `[]a = { 7, 44, 19, 86, 65, 39, 75, 101 }; ``    ``int` `n = a.Length; ``    ``int` `q = 2; ` `    ``int` `[]k = { 4, 2 }; ` `    ``orsum(a, n, q, k); ``} ``} ` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:
```238
230```

Time Complexity: O(N + Q*log(N))

Auxiliary Space: O(4*MAX)

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