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++ 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;
} |
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!