# 2588. Count the Number of Beautiful Subarrays

## Description

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

## Solutions

Solution 1: Prefix XOR + Hash Table

We observe that a subarray can become an array of all $0$s if and only if the number of $1$s on each binary bit of all elements in the subarray is even.

If there exist indices $i$ and $j$ such that $i \lt j$ and the subarrays $nums[0,..,i]$ and $nums[0,..,j]$ have the same parity of the number of $1$s on each binary bit, then we can turn the subarray $nums[i + 1,..,j]$ into an array of all $0$s.

Therefore, we can use the prefix XOR method and a hash table $cnt$ to count the occurrences of each prefix XOR value. We traverse the array, for each element $x$, we calculate its prefix XOR value $mask$, then add the number of occurrences of $mask$ to the answer. Then, we increase the number of occurrences of $mask$ by $1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public long beautifulSubarrays(int[] nums) {
Map<Integer, Integer> cnt = new HashMap<>();
cnt.put(0, 1);
long ans = 0;
for (int x : nums) {
}
return ans;
}
}

• class Solution {
public:
long long beautifulSubarrays(vector<int>& nums) {
unordered_map<int, int> cnt{ {0, 1} };
long long ans = 0;
for (int x : nums) {
}
return ans;
}
};

• class Solution:
def beautifulSubarrays(self, nums: List[int]) -> int:
cnt = Counter({0: 1})
for x in nums:
return ans


• func beautifulSubarrays(nums []int) (ans int64) {
cnt := map[int]int{0: 1}
for _, x := range nums {
}
return
}

• function beautifulSubarrays(nums: number[]): number {
const cnt = new Map();
cnt.set(0, 1);
let ans = 0;
for (const x of nums) {
}
return ans;
}