Sum of Bitwise-OR of all subarrays of a given Array | Set 2

Give an array of positive integers. The task is to find the total sum after performing the bitwise OR operation on all the sub-arrays of the given array.

Examples:

Input : arr[] = {1, 2, 3, 4, 5}
Output : 71

Input : arr[] = {6, 5, 4, 3, 2}
Output : 84

Explanation:



Simple Approach: A simple approach is to find the bitwise OR of each subarray of the given array using two nested loops, and then find the total sum. Time complexity of this approach will be O(N2).

Efficient Approach:

  1. Observe here that if a bit is being set by an element of the array then all subarray having that element will have that bit set. Therefore when we calculate sum of all subarrays having that number, we can directly multiply number of subarrays by the value that bit is making.
  2. Now, to do this an easy way will be to calculate the number of subarrays for which a bit is not set and subtract it from the total number of subarrays.

Let’s see an example:
Let the array A = [1, 2, 3, 4, 5]. Now the 1st bit is not set in the elements 2 and 4 and total number of such subarrays for which the Bitwise-OR will not have the 1st bit set will be 2.

Therefore, total number of subarrays for which the bitwise-OR will have 1st bit as set will be: 15-2 = 13.

Therefore we will add (13 * pow(2, 0)) to sum.

Below is the implementation of the above approach:

C++

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// C++ program to find sum of bitwise OR
// of all subarrays
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sum of bitwise OR
// of all subarrays
int givesum(int A[], int n)
{
    // Find max element of the array
    int max = *max_element(A, A + n);
  
    // Find the max bit position set in
    // the array
    int maxBit = log2(max) + 1;
  
    int totalSubarrays = n * (n + 1) / 2;
  
    int s = 0;
  
    // Traverse from 1st bit to last bit which
    // can be set in any element of the array
    for (int i = 0; i < maxBit; i++) {
        int c1 = 0;
  
        // Vector to store indexes of the array
        // with i-th bit not set
        vector<int> vec;
  
        int sum = 0;
  
        // Traverse the array
        for (int j = 0; j < n; j++) {
  
            // Check if ith bit is not set in A[j]
            int a = A[j] >> i;
            if (!(a & 1)) {
                vec.push_back(j);
            }
        }
  
        // Variable to store count of subarrays
        // whose bitwise OR will have i-th bit
        // not set
        int cntSubarrNotSet = 0;
  
        int cnt = 1;
  
        for (int j = 1; j < vec.size(); j++) {
            if (vec[j] - vec[j - 1] == 1) {
                cnt++;
            }
            else {
                cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
                cnt = 1;
            }
        }
  
        // For last element of vec
        cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
        // If vec is empty then cntSubarrNotSet
        // should be 0 and not 1
        if (vec.size() == 0)
            cntSubarrNotSet = 0;
  
        // Variable to store count of subarrays
        // whose bitwise OR will have i-th bit set
        int cntSubarrIthSet = totalSubarrays - cntSubarrNotSet;
  
        s += cntSubarrIthSet * pow(2, i);
    }
  
    return s;
}
  
// Driver code
int main()
{
    int A[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(A) / sizeof(A[0]);
  
    cout << givesum(A, n);
  
    return 0;
}

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Java

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// Java program to find sum of bitwise OR
// of all subarrays
import java.util.*;
  
class GFG {
  
    // Function to find sum of bitwise OR
    // of all subarrays
    static int givesum(int A[], int n)
    {
  
        // Find max element of the array
        int max = Arrays.stream(A).max().getAsInt();
  
        // Find the max bit position
        // set in the array
        int maxBit = (int)Math.ceil(Math.log(max) + 1);
        int totalSubarrays = n * (n + 1) / 2;
  
        int s = 0;
  
        // Traverse from 1st bit to last bit which
        // can be set in any element of the array
        for (int i = 0; i < maxBit; i++) {
            int c1 = 0;
  
            // Vector to store indexes of the array
            // with i-th bit not set
            Vector<Integer> vec = new Vector<>();
  
            int sum = 0;
  
            // Traverse the array
            for (int j = 0; j < n; j++) {
  
                // Check if ith bit is not set in A[j]
                int a = A[j] >> i;
                if (!(a % 2 == 1)) {
                    vec.add(j);
                }
            }
  
            // Variable to store count of subarrays
            // whose bitwise OR will have i-th bit
            // not set
            int cntSubarrNotSet = 0;
  
            int cnt = 1;
  
            for (int j = 1; j < vec.size(); j++) {
                if (vec.get(j) - vec.get(j - 1) == 1) {
                    cnt++;
                }
                else {
                    cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
                    cnt = 1;
                }
            }
  
            // For last element of vec
            cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
            // If vec is empty then cntSubarrNotSet
            // should be 0 and not 1
            if (vec.size() == 0)
                cntSubarrNotSet = 0;
  
            // Variable to store count of subarrays
            // whose bitwise OR will have i-th bit set
            int cntSubarrIthSet = totalSubarrays - cntSubarrNotSet;
  
            s += cntSubarrIthSet * Math.pow(2, i);
        }
        return s;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 3, 4, 5 };
        int n = A.length;
        System.out.println(givesum(A, n));
    }
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python 3 program to find sum of 
# bitwise OR of all subarrays 
  
# from math lib. import log2 function
from math import log2
  
# Function to find sum of bitwise OR 
# of all subarrays 
def givesum(A, n) :
  
    # Find max element of the array 
    max_element = max(A) 
  
    # Find the max bit position set in 
    # the array 
    maxBit = int(log2(max_element)) + 1
  
    totalSubarrays = n * (n + 1) // 2
  
    s = 0
  
    # Traverse from 1st bit to last bit which 
    # can be set in any element of the array 
    for i in range(maxBit) : 
        c1 = 0
  
        # List to store indexes of the array 
        # with i-th bit not set 
        vec = []
  
        sum = 0
  
        # Traverse the array 
        for j in range(n) : 
  
            # Check if ith bit is not set in A[j] 
            a = A[j] >> i
              
            if (not(a & 1)) :
                vec.append(j)
  
        # Variable to store count of subarrays 
        # whose bitwise OR will have i-th bit 
        # not set 
        cntSubarrNotSet = 0
  
        cnt = 1
  
        for j in range(1, len(vec)) :
              
            if (vec[j] - vec[j - 1] == 1) : 
                cnt += 1
              
            else :
                  
                cntSubarrNotSet += cnt * (cnt + 1) // 2
  
                cnt = 1
              
        # For last element of vec 
        cntSubarrNotSet += cnt * (cnt + 1) // 2
  
        # If vec is empty then cntSubarrNotSet 
        # should be 0 and not 1
        if len(vec) == 0:
            cntSubarrNotSet = 0    
  
        # Variable to store count of subarrays 
        # whose bitwise OR will have i-th bit set 
        cntSubarrIthSet = totalSubarrays - cntSubarrNotSet 
  
        s += cntSubarrIthSet * pow(2, i) 
      
    return s
  
# Driver code 
if __name__ == "__main__" :
  
    A = [ 1, 2, 3, 4, 5 ]
    n = len(A)
  
    print(givesum(A, n))
  
# This code is contributed by Ryuga

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C#

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// C# program to find sum of bitwise OR
// of all subarrays
using System;
using System.Linq;
using System.Collections.Generic;
  
class GFG {
  
    // Function to find sum of bitwise OR
    // of all subarrays
    static int givesum(int[] A, int n)
    {
  
        // Find max element of the array
        int max = A.Max();
  
        // Find the max bit position
        // set in the array
        int maxBit = (int)Math.Ceiling(Math.Log(max) + 1);
        int totalSubarrays = n * (n + 1) / 2;
  
        int s = 0;
  
        // Traverse from 1st bit to last bit which
        // can be set in any element of the array
        for (int i = 0; i < maxBit; i++) {
  
            // Vector to store indexes of the array
            // with i-th bit not set
            List<int> vec = new List<int>();
  
            // Traverse the array
            for (int j = 0; j < n; j++) {
  
                // Check if ith bit is not set in A[j]
                int a = A[j] >> i;
                if (!(a % 2 == 1)) {
                    vec.Add(j);
                }
            }
  
            // Variable to store count of subarrays
            // whose bitwise OR will have i-th bit
            // not set
            int cntSubarrNotSet = 0;
  
            int cnt = 1;
  
            for (int j = 1; j < vec.Count; j++) {
                if (vec[j] - vec[j - 1] == 1) {
                    cnt++;
                }
                else {
                    cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
                    cnt = 1;
                }
            }
  
            // For last element of vec
            cntSubarrNotSet += cnt * (cnt + 1) / 2;
  
            // If vec is empty then cntSubarrNotSet
            // should be 0 and not 1
            if (vec.Count() == 0)
                cntSubarrNotSet = 0;
  
            // Variable to store count of subarrays
            // whose bitwise OR will have i-th bit set
            int cntSubarrIthSet = totalSubarrays - cntSubarrNotSet;
  
            s += (int)(cntSubarrIthSet * Math.Pow(2, i));
        }
        return s;
    }
  
    // Driver code
    public static void Main()
    {
        int[] A = { 1, 2, 3, 4, 5 };
        int n = A.Length;
        Console.WriteLine(givesum(A, n));
    }
}
  
/* This code contributed by PrinciRaj1992 */

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PHP

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<?php
// PHP program to find sum of bitwise OR
// of all subarrays
  
// Function to find sum of bitwise OR
// of all subarrays
function givesum($A, $n)
{
    // Find max element of the array
    $max = max($A);
  
    // Find the max bit position set in
    // the array
    $maxBit = (int)((log($max) / 
                     log10(2)) + 1);
  
    $totalSubarrays = (int)($n * ($n + 1) / 2);
  
    $s = 0;
  
    // Traverse from 1st bit to last bit which
    // can be set in any element of the array
    for ($i = 0; $i < $maxBit; $i++)
    {
        $c1 = 0;
  
        // Vector to store indexes of 
        // the array with i-th bit not set
        $vec = array();
  
        $sum = 0;
  
        // Traverse the array
        for ($j = 0; $j < $n; $j++) 
        {
  
            // Check if ith bit is 
            // not set in A[j]
            $a = $A[$j] >> $i;
            if (!($a & 1))
            {
                array_push($vec, $j);
            }
        }
  
        // Variable to store count of subarrays
        // whose bitwise OR will have i-th bit
        // not set
        $cntSubarrNotSet = 0;
  
        $cnt = 1;
  
        for ($j = 1; $j < count($vec); $j++) 
        {
            if ($vec[$j] - $vec[$j - 1] == 1)
            {
                $cnt++;
            }
            else 
            {
                $cntSubarrNotSet += (int)($cnt
                                         ($cnt + 1) / 2);
  
                $cnt = 1;
            }
        }
  
        // For last element of vec
        $cntSubarrNotSet += (int)($cnt *
                                 ($cnt + 1) / 2);
  
        // If vec is empty then cntSubarrNotSet 
        // should be 0 and not 1
        if (count($vec) == 0)
            $cntSubarrNotSet = 0;
  
        // Variable to store count of subarrays
        // whose bitwise OR will have i-th bit set
        $cntSubarrIthSet = $totalSubarrays
                           $cntSubarrNotSet;
  
        $s += $cntSubarrIthSet * pow(2, $i);
    }
  
    return $s;
}
  
// Driver code
$A = array( 1, 2, 3, 4, 5 );
$n = count($A);
  
echo givesum($A, $n);
  
// This code is contributed by mits
?>

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Output:

71

Time Complexity: O(N*logN)



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