# 268. Missing Number

## Description

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 $O(1)$.

Solution 2: Mathematics

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 $O(1)$.

• class Solution {
public int missingNumber(int[] nums) {
int n = nums.length;
int ans = n;
for (int i = 0; i < n; ++i) {
ans ^= (i ^ nums[i]);
}
return ans;
}
}

• class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size();
int ans = n;
for (int i = 0; i < n; ++i) {
ans ^= (i ^ nums[i]);
}
return ans;
}
};

• class Solution:
def missingNumber(self, nums: List[int]) -> int:
return reduce(xor, (i ^ v for i, v in enumerate(nums, 1)))


• func missingNumber(nums []int) (ans int) {
n := len(nums)
ans = n
for i, v := range nums {
ans ^= (i ^ v)
}
return
}

• function missingNumber(nums: number[]): number {
const n = nums.length;
let ans = n;
for (let i = 0; i < n; ++i) {
ans ^= i ^ nums[i];
}
return ans;
}


• /**
* @param {number[]} nums
* @return {number}
*/
var missingNumber = function (nums) {
const n = nums.length;
let ans = n;
for (let i = 0; i < n; ++i) {
ans ^= i ^ nums[i];
}
return ans;
};


• class Solution {
/**
* @param Integer[] $nums * @return Integer */ function missingNumber($nums) {
$n = count($nums);
$sumN = (($n + 1) * $n) / 2; for ($i = 0; $i <$n; $i++) {$sumN -= $nums[$i];
}
return \$sumN;
}
}

• impl Solution {
pub fn missing_number(nums: Vec<i32>) -> i32 {
let n = nums.len() as i32;
let mut ans = n;
for (i, v) in nums.iter().enumerate() {
ans ^= (i as i32) ^ v;
}
ans
}
}