[leetcode] 898. Bitwise ORs of Subarrays

Description

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

分析

题目的意思是：给定一个数组，求子数组的异或运算获得的值的数目。这里用三个set集合来计算，第一个set集合s用来保存最终的值，prev保存上一次遍历计算获得的异或值，t保存当前计算获得的异或值。而后遍历维护这三个集合就好了。

代码

class Solution:
    def subarrayBitwiseORs(self, A: List[int]) -> int:
        s=set()
        prev=set()
        for x in A:
            t=set()
            for y in prev:
                t.add(x|y)
            t.add(x)
            prev=t
            s|=t
        return len(s)

参考文献

[LeetCode] Python short and clean solution