# Summary:
Given a number, `num`, return the shortest amount of `steps` it would take from 1, to land exactly on that number.

# Description:
A `step` is defined as either:
- Adding 1 to the number: `num += 1`
- Doubling the number: `num *= 2`

You will always start from the number `1` and you will have to return the shortest count of steps it would take to land exactly on that number.

`1 <= num <= 10000`

Examples:

`num == 3` would return `2` steps:

```
1 -- +1 --> 2:        1 step
2 -- +1 --> 3:        2 steps

2 steps
```

`num == 12` would return `4` steps:

```
1 -- +1 --> 2:        1 step
2 -- +1 --> 3:        2 steps
3 -- x2 --> 6:        3 steps
6 -- x2 --> 12:       4 steps

4 steps
```

`num == 16` would return `4` steps:

```
1 -- +1 --> 2:        1 step
2 -- x2 --> 4:        2 steps
4 -- x2 --> 8:        3 steps
8 -- x2 --> 16:       4 steps

4 steps
```

Difficulty: EASY
Starter code: def shortest_steps_to_num(num):
	

Solutions:
def shortest_steps_to_num(num):
	steps = 0
	while num != 1:
		if num % 2:
			num -= 1
		else:
			num //= 2
		steps += 1
	return steps


def shortest_steps_to_num(num):
	return num.bit_length() + bin(num).count('1') - 2


def shortest_steps_to_num(num):
	x = 0
	while num > 1:
		if num % 2:
			num -= 1
		else:
			num /= 2
		x += 1
	return x


def shortest_steps_to_num(n):
	return bin(n).count('1') + n.bit_length() - 2


def shortest_steps_to_num(num):
	return 0 if num == 1 else 1 + shortest_steps_to_num(num - 1 if num % 2 else num / 2)


def shortest_steps_to_num(n):
	return 0 if n == 1 else 1 + shortest_steps_to_num(n - 1) if n % 2 else 1 + shortest_steps_to_num(n // 2)


def shortest_steps_to_num(n):
	step = 0
	while n > 1:
		(n, step) = (n // 2, step + 1 + n % 2)
	return step


def shortest_steps_to_num(num):
	s = bin(num)
	return len(s) + s.count('1') - 4


def shortest_steps_to_num(num):
	steps = 0
	while num > 1:
		if num % 2 == 0:
			num = num // 2
		elif num == 3:
			num -= 1
		else:
			num -= 1
		steps += 1
	return steps


def shortest_steps_to_num(num):
	a = str(bin(num))[3:]
	return len(a) + a.count('1')