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.
Constraints:
-
n == nums.length
-
1 <= n <= 104
-
0 <= nums[i] <= n
- All the numbers of
nums
are unique.
Follow up: Could you implement a solution using only O(1)
extra space complexity and O(n)
runtime complexity?
SOLUTION:
class Solution:
def missingNumber(self, nums: List[int]) -> int:
n = len(nums)
currxor = 0
allxor = n
for i in range(n):
currxor ^= nums[i]
allxor ^= i
return currxor ^ allxor
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