Sum of bitwise AND of all subarrays

Given an array consisting of N positive integers, find the sum of bit-wise and of all possible sub-arrays of the array.

Examples:

Input : arr[] = {1, 5, 8}
Output : 15
Bit-wise AND of {1} = 1
Bit-wise AND of {1, 5} = 1
Bit-wise AND of {1, 5, 8} = 0 
Bit-wise AND of {5} = 5
Bit-wise AND of {5, 8} = 0
Bit-wise AND of {8} = 8

Sum = 1 + 1 + 0 + 5 + 0 + 8 =  15

Input : arr[] =  {7, 1, 1, 5}
Output : 20


Simple Solution: A simple solution will be to generate all the sub-arrays, and sum up the AND values of all the sub-arrays. It will take linear time on an average to find the AND value of a sub-array and thus, the over all time complexity will be O(n3).

Efficient Solution: For the sake of better understanding, let’s assume that any bit of an element is represented by the variable ‘i’ and the variable ‘sum’ is used to store the final sum.

The idea here is, we will try to find the number of AND values(sub-arrays with bit-wise and(&)) with ith bit set. Let us suppose, there are ‘Si‘ number of sub-arrays with ith bit set. For, ith bit, sum can be updated as sum += (2i * S).

We will break the task to multiple steps. At each step, we will try to find the number of AND values with ith bit set. For this, we will simply iterate through the array and find the number of contiguous segments with ith bit set and there lengths. For, each such segment of length ‘l’, value of sum can be updated as sum += (2i * l * (l + 1))/2.

Since, for each bit, we are performing O(N) iterations, the time complexity of this approach will be O(N).

Below is the implementation of the above idea:

C++

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// CPP program to find sum of bitwise AND
// of all subarrays
  
#include <iostream>
#include <vector>
using namespace std;
  
// Function to find the sum of
// bitwise AND of all subarrays
int findAndSum(int arr[], int n)
{
    // variable to store
    // the final sum
    int sum = 0;
  
    // multiplier
    int mul = 1;
  
    for (int i = 0; i < 30; i++) {
        // variable to check if
        // counting is on
        bool count_on = 0;
  
        // variable to store the
        // length of the subarrays
        int l = 0;
  
        // loop to find the contiguous
        // segments
        for (int j = 0; j < n; j++) {
            if ((arr[j] & (1 << i)) > 0)
                if (count_on)
                    l++;
                else {
                    count_on = 1;
                    l++;
                }
  
            else if (count_on) {
                sum += ((mul * l * (l + 1)) / 2);
                count_on = 0;
                l = 0;
            }
        }
  
        if (count_on) {
            sum += ((mul * l * (l + 1)) / 2);
            count_on = 0;
            l = 0;
        }
  
        // updating the multiplier
        mul *= 2;
    }
  
    // returning the sum
    return sum;
}
  
// Driver Code
int main()
{
  
    int arr[] = { 7, 1, 1, 5 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << findAndSum(arr, n);
  
    return 0;
}

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Python3

# Python3 program to find Sum of
# bitwise AND of all subarrays
import math as mt

# Function to find the Sum of
# bitwise AND of all subarrays
def findAndSum(arr, n):

# variable to store the final Sum
Sum = 0

# multiplier
mul = 1

for i in range(30):

# variable to check if counting is on
count_on = 0

# variable to store the length
# of the subarrays
l = 0

# loop to find the contiguous
# segments
for j in range(n):

if ((arr[j] & (1 << i)) > 0):
if (count_on):
l += 1
else:
count_on = 1
l += 1

elif (count_on):
Sum += ((mul * l * (l + 1)) // 2)
count_on = 0
l = 0

if (count_on):
Sum += ((mul * l * (l + 1)) // 2)
count_on = 0
l = 0

# updating the multiplier
mul *= 2

# returning the Sum
return Sum

# Driver Code
arr = [7, 1, 1, 5]

n = len(arr)

print(findAndSum(arr, n))

# This code is contributed by Mohit Kumar

Output:

20

Time Complexity: O(N)



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