898. Bitwise ORs of Subarrays

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:

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

Example 2:

```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:

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

Constraints:

• `1 <= nums.length <= 5 * 104`
• `0 <= nums[i] <= 109`

898. Bitwise ORs of Subarrays
``````struct Solution;
use std::collections::HashSet;

impl Solution {
fn subarray_bitwise_o_rs(a: Vec<i32>) -> i32 {
let mut res: HashSet<i32> = HashSet::new();
let mut prev: HashSet<i32> = HashSet::new();
for x in a {
let mut cur: HashSet<i32> = HashSet::new();
cur.insert(x);
for y in prev {
cur.insert(y | x);
}
for &x in &cur {
res.insert(x);
}
prev = cur;
}
res.len() as i32
}
}

#[test]
fn test() {
let a = vec![0];
let res = 1;
assert_eq!(Solution::subarray_bitwise_o_rs(a), res);
let a = vec![1, 1, 2];
let res = 3;
assert_eq!(Solution::subarray_bitwise_o_rs(a), res);
let a = vec![1, 2, 4];
let res = 6;
assert_eq!(Solution::subarray_bitwise_o_rs(a), res);
}
``````