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Find array sum using Bitwise OR after splitting given array in two halves after K circular shifts
  • Difficulty Level : Hard
  • Last Updated : 02 Aug, 2020

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

238
230


 

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