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:
Java
// Java program to find Bitwise OR of two// equal halves of an array after performing// K right circular shiftsimport java.util.*;class GFG{ static int MAX = 100005; // Array for storing// the segment treestatic int []seg = new int[4 * MAX];// Function to build the segment treestatic 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 sumstatic 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 + ""); }}// Driver Codepublic 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 |
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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