Leetcode: Bitwise ORs of Subarrays

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. 1 <= A.length <= 50000
  2. 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)
}
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