Count the Number of Beautiful Subarrays

Problem

You are given a 0-indexed integer array nums. In one operation, you can:

Choose two different indices i and j such that 0 <= i, j < nums.length.
Choose a non-negative integer k such that the kth bit (0-indexed) in the binary representation of nums[i] and nums[j] is 1.
Subtract 2k from nums[i] and nums[j].

A subarray is beautiful if it is possible to make all of its elements equal to 0 after applying the above operation any number of times.

Return **the number of *beautiful subarrays* in the array** nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [4,3,1,2,4]
Output: 2
Explanation: There are 2 beautiful subarrays in nums: [4,3,1,2,4] and [4,3,1,2,4].
- We can make all elements in the subarray [3,1,2] equal to 0 in the following way:
  - Choose [3, 1, 2] and k = 1. Subtract 21 from both numbers. The subarray becomes [1, 1, 0].
  - Choose [1, 1, 0] and k = 0. Subtract 20 from both numbers. The subarray becomes [0, 0, 0].
- We can make all elements in the subarray [4,3,1,2,4] equal to 0 in the following way:
  - Choose [4, 3, 1, 2, 4] and k = 2. Subtract 22 from both numbers. The subarray becomes [0, 3, 1, 2, 0].
  - Choose [0, 3, 1, 2, 0] and k = 0. Subtract 20 from both numbers. The subarray becomes [0, 2, 0, 2, 0].
  - Choose [0, 2, 0, 2, 0] and k = 1. Subtract 21 from both numbers. The subarray becomes [0, 0, 0, 0, 0].

Example 2:

Input: nums = [1,10,4]
Output: 0
Explanation: There are no beautiful subarrays in nums.

Constraints:

1 <= nums.length <= 105
0 <= nums[i] <= 106

Solution (Java)

class Solution {
    public long beautifulSubarrays(int[] nums) {
        long count = 0;
        int xor = 0;
        Map<Integer,Integer> map = new HashMap<>();
        map.put(0,1);

        for(int num : nums){
            count += map.merge(xor ^= num, 1, Integer::sum)-1;
        }
        return count;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).