Given an array **A** of non-negative integers, where . The task is to count number of distinct possible results obtained by taking the bitwise OR of all the elements in all possible Subarrays.

**Examples:**

Input: A = [1, 2] Output: 3 Explanation: The possible subarrays are [1], [2], [1, 2]. These Bitwise OR of subarrays are 1, 2, 3. There are 3 distinct values, so the answer is 3. Input: A = [1, 2, 4] Output: 6 Explanation: The possible distinct values are 1, 2, 3, 4, 6, and 7.

**Approach:**

The Naive approach is to generate all possible subarrays and take bitwise OR of all elements in the subarray. Store each result in **set** and return **length of the set**.

**Efficient Approach:**

We can make the above approach better. The Naive approach is to calculate all possible result where, **res(i, j) = A[i] | A[i+1] | … | A[j]**. However we can speed this up by taking note of the fact that **res(i, j+1) = res(i, j) | A[j+1]**. At the **kth** step, say we have all of the **res(i, k)** in some set **pre**. Then we can find the next **pre** set **(for k -> k+1)** by using **res(i, k+1) = res(i, k) | A[k+1]**.

However, the number of unique values in this set **pre** is atmost 32, since the list **res(k, k), res(k-1, k), res(k-2, k), …** is **monotone increasing**, and any subsequent values that are different from previous must have more 1’s in it’s binary representation which can have maximum of **32 ones**.

**Below is the implementation of above approach.**

## Python

`# Python implementation of the above approach ` ` ` `# function to return count of distinct bitwise OR ` `def` `subarrayBitwiseOR(A): ` ` ` ` ` `# res contains distinct values ` ` ` `res ` `=` `set` `() ` ` ` ` ` `pre ` `=` `{` `0` `} ` ` ` ` ` `for` `x ` `in` `A: ` ` ` `pre ` `=` `{x | y ` `for` `y ` `in` `pre} | {x} ` ` ` `res |` `=` `pre ` ` ` ` ` `return` `len` `(res) ` ` ` ` ` `# Driver program ` `A ` `=` `[` `1` `, ` `2` `, ` `4` `] ` ` ` `# print required answer ` `print` `(subarrayBitwiseOR(A)) ` ` ` `# This code is written by ` `# Sanjit_Prasad ` |

*chevron_right*

*filter_none*

Output:

6

**Time Complexity:** O(N*log(K)), where N is the length of A, and K is the maximum size of elements in A.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Total pairs in an array such that the bitwise AND, bitwise OR and bitwise XOR of LSB is 1
- Minimum possible Bitwise OR of all Bitwise AND of pairs generated from two given arrays
- Count ways to generate pairs having Bitwise XOR and Bitwise AND equal to X and Y respectively
- Count pairs with bitwise XOR exceeding bitwise AND from a given array
- Count pairs with equal Bitwise AND and Bitwise OR value
- Sum of bitwise OR of all subarrays
- Sum of Bitwise-OR of all subarrays of a given Array | Set 2
- Sum of bitwise AND of all subarrays
- Count even length subarrays having bitwise XOR equal to 0
- Count subarrays having total distinct elements same as original array
- Count of subarrays having exactly K distinct elements
- Count subarrays having each distinct element occuring at least twice
- Leftover element after performing alternate Bitwise OR and Bitwise XOR operations on adjacent pairs
- Find subsequences with maximum Bitwise AND and Bitwise OR
- Maximize sum of squares of array elements possible by replacing pairs with their Bitwise AND and Bitwise OR
- Number of subarrays have bitwise OR >= K
- Bitwise operations on Subarrays of size K
- Differences between number of increasing subarrays and decreasing subarrays in k sized windows
- Find N distinct numbers whose bitwise Or is equal to K
- Subarrays with distinct elements

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.