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:2Explanation: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:2Explanation: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:8Explanation: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**.

` ````
```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

` ````
```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

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