# Missing Number

## Problem

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

### Cyclic Sort

```
class Solution {
public int missingNumber(int[] nums) {
var i = 0;
var n = nums.length;
while (i < nums.length) {
if (nums[i] == i || nums[i] >= nums.length) {
i++;
} else {
swap(nums, nums[i], i);
}
}
for (i = 0; i < nums.length; i++) {
if (nums[i] != i) {
return i;
}
}
return nums.length;
}
void swap(int[] a, int x, int y) {
a[x] ^= a[y];
a[y] ^= a[x];
a[x] ^= a[y];
}
}
```

### XOR

Another use for that XOR trick!

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