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 <= A.length <= 50000
- 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) }