### LeetCode Challenge #268. 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.
##### Video Solution
###### Java Solution
```				```
class Solution {
public int missingNumber(int[] nums) {
int range = nums.length;
int actualSum =  (range * (range + 1))/2;
int currSum = 0 ;
for(int i=0;i<nums.length;i++){
currSum = currSum + nums[i];
}
int ans = actualSum - currSum ;
return ans ;
}
}
```
```
###### Pseudo Code of Solution
```				```
Algorithm missingNumber(nums: Array of Integer) -> Integer
Declare range as Integer
Declare actualSum as Integer
Declare currSum as Integer
Declare ans as Integer

// Calculate the range of the array
range <- Length of nums

// Calculate the sum of all numbers from 0 to range
actualSum <- (range * (range + 1)) / 2

// Initialize the sum of the elements in the array to 0
currSum <- 0

// Loop through the array to sum up its elements
For i from 0 to Length of nums - 1 Do
currSum <- currSum + nums[i]
End For

// Calculate the missing number
ans <- actualSum - currSum

// Return the missing number
Return ans
End Algorithm

```
```