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?
Solutions
Solution 1: Bitwise Operation
The XOR operation has the following properties:
Any number XOR 0 is still the original number, i.e., $x \oplus 0 = x$;
Any number XOR itself is 0, i.e., $x \oplus x = 0$;
Therefore, we can traverse the array, perform XOR operation between each element and the numbers $[0,..n]$, and the final result will be the missing number.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
We can also solve this problem using mathematics. By calculating the sum of $[0,..n]$, subtracting the sum of all numbers in the array, we can obtain the missing number.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.