LeetCode-in-Java
898. Bitwise ORs of Subarrays
Medium
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 * 1040 <= nums[i] <= 109
Solution
import java.util.HashSet;
import java.util.Set;
public class Solution {
public int subarrayBitwiseORs(int[] arr) {
Set<Integer> set = new HashSet<>();
for (int i = 0; i < arr.length; i++) {
set.add(arr[i]);
for (int j = i - 1; j >= 0; j--) {
if ((arr[i] | arr[j]) == arr[j]) {
break;
}
arr[j] |= arr[i];
set.add(arr[j]);
}
}
return set.size();
}
}