• 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: Hashtable + Prefix XOR

The problem is asking to find # of subarrays whose element wise xor is 0. We can use a hashtable to store the frequency of each prefix xor value, which reduces this problem to # of Subarray sum equal to k.

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

// Author: Huahua
class Solution {
public:
  long long beautifulSubarrays(vector<int>& nums) {
    long long ans = 0;
    unordered_map<int, int> m{{0, 1}};
    int x = 0;
    for (int i = 0; i < nums.size(); ++i) {
      x ^= nums[i];
      ans += m[x];
      m[x] += 1;
    }
    return ans;
  } 
};

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