Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Example 1:
Input: nums = [3,0,1] Output: 2 Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1] Output: 2 Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1] Output: 8 Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
n == nums.length1 <= n <= 10^40 <= nums[i] <= n- All the numbers of
numsare unique.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
class Solution:
def missingNumber(self, nums: List[int]) -> int:
xor_sum = 0
# XOR all indices and numbers
for i in range(len(nums)):
xor_sum ^= i ^ nums[i]
# Also XOR the last index (len(nums))
return xor_sum ^ len(nums)Both methods run in O(n) time since they iterate through the array once.
Both methods operate in O(1) space, as they use only a few additional variables regardless of the input size.