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

Rust Solution

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);
}

Having problems with this solution? Click here to submit an issue on github.