# Bitwise ORs Of Subarrays Problem

## Description

LeetCode Problem 898.

We have an array arr of non-negative integers.

For every (contiguous) subarray sub = [arr[i], arr[i + 1], …, arr[j]] (with i <= j), we take the bitwise OR of all the elements in sub, obtaining a result arr[i] | arr[i + 1] | … | arr[j].

Return the number of possible results. Results that occur more than once are only counted once in the final answer.

Example 1:

``````1
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Input: arr = [0]
Output: 1
Explanation: There is only one possible result: 0.
``````

Example 2:

``````1
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Input: arr = [1,1,2]
Output: 3
Explanation: The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.
``````

Example 3:

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Input: arr = [1,2,4]
Output: 6
Explanation: The possible results are 1, 2, 3, 4, 6, and 7.
``````

Constraints:

• 1 <= nums.length <= 5 * 10^4
• 0 <= nums[i]<= 10^9

## Sample C++ Code

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class Solution {
public:
int subarrayBitwiseORs(vector<int>& arr) {
int sz = arr.size();
unordered_set<int> st;
for (int i = 0; i < sz; i++) {
st.insert(arr[i]);
for (int j = i - 1; j >= 0; j--) {
if ((arr[i] | arr[j]) == arr[j])
break;
arr[j] |= arr[i];
st.insert(arr[j]);
}
}
return st.size();
}
};
``````