Given an array of integers arr.
We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).
Let’s define a and b as follows:
a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
Note that ^ denotes the bitwise-xor operation.
Return the number of triplets (i, j and k) Where a == b.
Example 1:
Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)
Example 2:
Input: arr = [1,1,1,1,1]
Output: 10
Example 3:
Input: arr = [2,3]
Output: 0
Example 4:
Input: arr = [1,3,5,7,9]
Output: 3
Example 5:
Input: arr = [7,11,12,9,5,2,7,17,22]
Output: 8
Solution
这个题要求找出满足条件的i,j,k使得arr[i]^arr[i+1]…arr[j-1] 和arr[j]^arr[j+1]…arr[k]的结果一样，换句话说它们的异或结果等于0时是个有效的切片，我们可以通过计算此时j-i 的 差来计算以i开始，以j结尾的i,j,k组合。然后把所有的结果加起来就是最后可能的组合总数。
class Solution:
def countTriplets(self, arr: List[int]) -> int:
ans = 0
for i in range(len(arr)):
xor = arr[i]
for j in range(i+1,len(arr)):
xor = xor ^ arr[j]
if (xor ==0):
ans = ans +(j-i)
return ans
