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## The problem

On this problem, you’ll be given an integer n, which is the variety of occasions that’s thrown a coin. You’ll have to return an array of strings for all the probabilities (heads[H] and tails[T]). Examples:

`coin(1) ought to return {"H", "T"}`

`coin(2) ought to return {"HH", "HT", "TH", "TT"}`

`coin(3) ought to return {"HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"}`

When completed type them alphabetically.

In C and C++ simply return a `char*`

with all parts separated by`,`

(with out house):`coin(2) ought to return "HH,HT,TH,TT"`

INPUT:`0 < n < 18`

Cautious with efficiency!! You’ll must move 3 fundamental take a look at (n = 1, n = 2, n = 3), many medium exams (3 < n <= 10) and plenty of massive exams (10 < n < 18)

## The answer in Python code

Possibility 1:

```
from itertools import product
def coin(n):
return listing(map(''.be a part of, product(*(["HT"]*n))))
```

Possibility 2:

```
def coin(n): return [x + v for x in coin(n-1) for v in 'HT'] if n - 1 else ['H','T']
```

Possibility 3:

```
memo = {1 : ["H", "T"]}
for j in vary(2, 18):
b = memo[j-1]
complete = set()
for i in b:
complete.add("T"+i)
complete.add(i+"T")
complete.add("H"+i)
complete.add(i+"H")
memo[j] = sorted(complete)
def coin(n):
return memo[n]
```

## Check circumstances to validate our answer

```
take a look at.describe("Primary Checks")
take a look at.assert_equals(coin(1),["H","T"])
take a look at.assert_equals(coin(2),["HH", "HT", "TH", "TT"])
take a look at.assert_equals(coin(3),["HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"])
```

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