# Leetcode: Bitwise ORs of Subarrays

Bitwise ORs of Subarrays

Similar Problems:

We have an array A of non-negative integers.

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

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

Example 1:

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

Example 2:

```Input: [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:

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

Note:

1. 1 <= A.length <= 50000
2. 0 <= A[i] <= 10^9

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```// Blog link: https://code.dennyzhang.com/bitwise-ors-of-subarrays
// Basic Ideas: dynamic programming
// Complexity: Time O(n*log(w)), Space O(n*log(w))
//             w = the maximum size of elements in A.
func subarrayBitwiseORs(A []int) int {
s_total := map[int]bool{}
s := map[int]bool{}
for _, p := range A {
s2 := map[int]bool{}
s2[p] = true
for q := range s {
r := q|p
s2[r] = true
}
for q:= range s2 { s_total[q] = true }
s = s2
}
return len(s_total)
}
```

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