# 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.*

{% hint style="info" %}
For example:

```
Input: [4, 0, 3, 1]
Output: 2
```

```
Input: [8, 3, 5, 2, 4, 6, 0, 1]
Output: 7
```

{% endhint %}

## Solution

```
class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        #1) sort the numbers in the correct position by their index
        #2) find the missing number that is not equal to it's index
        
        i=0
        while i < len(nums):
            j = nums[i] #getting the index that this number should be in
            if nums[i] < len(nums) and nums[j] != nums[i]:
                nums[j],nums[i] = nums[i],nums[j]
            else:
                i+=1
                
        for i in range(len(nums)):
            if i != nums[i]:
                return i
        return i+1
        
        
        
#Since the array has numbers in range [0, n], there
#will possibly exist a num[n] but not an index n so
#we skip this

#Time: O(n)
#Space: O(n)

```


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