## Problem – Missing Number

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 <= 10`^{4}

`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?

### Missing Number LeetCode Solution in Java

```
public int missingNumber(int[] nums) {
int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}
return xor ^ i;
}
```

### Missing Number LeetCode Solution in C++

```
class Solution {
public:
int missingNumber(vector<int>& nums) {
int result = nums.size();
int i=0;
for(int num:nums){
result ^= num;
result ^= i;
i++;
}
return result;
}
};
```

### Missing Number LeetCode Solution in Python

```
class Solution(object):
def missingNumber(self, nums):
return sum(range(len(nums)+1)) - sum(nums)
```

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