C++ Program to 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 <bits/stdc++.h>
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;
}

Output:

238
230

Time Complexity: O(N + Q*logN)

Auxiliary Space: O(4*MAX)

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