# Missing Number LeetCode Solution | Easy Approach Share:

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 <= 104`
• `0 <= nums[i] <= n`
• All the numbers of `nums` are unique.

Time: O(n)
Space: O(1)

### C++

``````class Solution {
public:
int missingNumber(vector<int>& nums) {
int ans = nums.size();

for (int i = 0; i < nums.size(); ++i)
ans ^= i ^ nums[i];

return ans;
}
};
``````

### Java

`````` class Solution {
public int missingNumber(int[] nums) {
int ans = nums.length;

for (int i = 0; i < nums.length; ++i)
ans ^= i ^ nums[i];

return ans;
}
}

``````

### Python

`````` class Solution:
def missingNumber(self, nums: List[int]) -> int:
ans = len(nums)

for i, num in enumerate(nums):
ans ^= i ^ num

return ans

``````

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