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## The problem

Write a perform that takes a shuffled listing of distinctive numbers from `1`

to `n`

with one ingredient lacking (which might be any quantity together with `n`

). Return this lacking quantity.

**Observe**: big lists can be examined.

**Examples:**

```
[1, 3, 4] => 2
[1, 2, 3] => 4
[4, 2, 3] => 1
```

## The answer in Python code

Choice 1:

```
def find_missing_number(a):
n = len(a) + 1
return n * (n + 1) // 2 - sum(a)
```

Choice 2:

```
def find_missing_number(nums):
return sum(vary(1,len(nums)+2))-sum(nums)
```

Choice 3:

```
def find_missing_number(numbers):
if numbers == []:return 1
diff = listing(set(vary(1, len(numbers)+1))- set(numbers))
if diff == []:return max(numbers)+1
return diff[0]
```

## Check circumstances to validate our resolution

```
take a look at.assert_equals(find_missing_number([2, 3, 4]), 1)
take a look at.assert_equals(find_missing_number([1, 3, 4]), 2)
take a look at.assert_equals(find_missing_number([1, 2, 4]), 3)
take a look at.assert_equals(find_missing_number([1, 2, 3]), 4)
```

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