README_EN.md

898. Bitwise ORs of Subarrays

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Description

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

Solutions

Python3

class Solution:
    def subarrayBitwiseORs(self, arr: List[int]) -> int:
        s = set()
        prev = 0
        for i, v in enumerate(arr):
            prev |= v
            curr = 0
            for j in range(i, -1, -1):
                curr |= arr[j]
                s.add(curr)
                if curr == prev:
                    break
        return len(s)

Java

class Solution {
    public int subarrayBitwiseORs(int[] arr) {
        Set<Integer> s = new HashSet<>();
        int prev = 0;
        for (int i = 0; i < arr.length; ++i) {
            prev |= arr[i];
            int curr = 0;
            for (int j = i; j >= 0; --j) {
                curr |= arr[j];
                s.add(curr);
                if (curr == prev) {
                    break;
                }
            }
        }
        return s.size();
    }
}

C++

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

Go

func subarrayBitwiseORs(arr []int) int {
	s := map[int]bool{}
	prev := 0
	for i, v := range arr {
		prev |= v
		curr := 0
		for j := i; j >= 0; j-- {
			curr |= arr[j]
			s[curr] = true
			if curr == prev {
				break
			}
		}
	}
	return len(s)
}

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