Given an array (a list in Python) of integers and an integer `n`, find all occurrences of `n` in the given array and return another array containing all the index positions of `n` in the given array.

If `n` is not in the given array, return an empty array `[]`.

Assume that `n` and all values in the given array will always be integers.

Example:
```python
find_all([6, 9, 3, 4, 3, 82, 11], 3)
> [2, 4]
```

Difficulty: EASY
Starter code: def find_all(array, n):
	

Solutions:
def find_all(array, n):
	return [index for (index, item) in enumerate(array) if item == n]


def find_all(array, n):
	res = []
	for i in range(len(array)):
		if array[i] == n:
			res.append(i)
	return res


def find_all(array, n):
	return [i for (i, v) in enumerate(array) if v == n]


def find_all(array, n):
	return [i for (i, x) in enumerate(array) if x == n]


def find_all(array, n):
	return [i for i in range(len(array)) if array[i] == n]


def find_all(array, n):
	return [i[0] for i in enumerate(array) if i[1] == n]


def find_all(array, n):
	l = []
	for (i, v) in enumerate(array):
		if v == n:
			l.append(i)
	return l


import numpy as np

def find_all(array, n):
	return list(np.where(np.array(array) == n)[0])


def find_all(array, n):
	return [pos for (pos, number) in enumerate(array) if number == n]


def find_all(array, n):
	return [i for (i, number) in enumerate(array) if number == n]