Missing number

#Method 1
'''sum of 1 to n using (n*(n+1))//2 and then subtract the sum of the array, even if all number from 1 to n is there,still answer would be 0 which is right answer'''
class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        n = len(nums)
        return n * (n + 1) // 2 - sum(nums)



"""
Intuition:
The array contains n distinct numbers from the range [0, n], so exactly one number is missing.
If we XOR all values from 0 to n and also XOR all elements in nums, every number that appears
will cancel out (because a ^ a = 0). The only value left after all cancellations is the missing
number. This gives an O(n) time solution using O(1) extra space.

Trick: initialise the answer with len of the array
"""


class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        ans=len(nums) #Here ans starts with len(nums) ie. n because if all elements are in the range from o to n-1,then n will be the ans.
        for i,x in enumerate(nums):
            ans=ans^i^x
        return ans