Calculate possibilities of throwing a coin N times in Python

The challenge

In this challenge, you will be given an integer n, which is the number of times that is thrown a coin. You will have to return an array of strings for all the possibilities (heads[H] and tails[T]). Examples:

`coin(1) should return {"H", "T"}`
`coin(2) should return {"HH", "HT", "TH", "TT"}`
`coin(3) should return {"HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"}`

When finished sort them alphabetically.

In C and C++ just return a `char*` with all elements separated by`,` (without space):
`coin(2) should return "HH,HT,TH,TT"`

INPUT:
`0 < n < 18`

Careful with performance!! You’ll have to pass 3 basic test (n = 1, n = 2, n = 3), many medium tests (3 < n <= 10) and many large tests (10 < n < 18)

The solution in Python code

Option 1:

```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```from itertools import product

def coin(n):
return list(map(''.join, product(*(["HT"]*n))))
```Code language: Python (python)```

Option 2:

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

Option 3:

``````memo = {1 : ["H", "T"]}
for j in range(2, 18):
b = memo[j-1]
total = set()
for i in b:
memo[j] = sorted(total)

def coin(n):
return memo[n]
```Code language: Python (python)```

Test cases to validate our solution

``````test.describe("Basic Tests")
test.assert_equals(coin(1),["H","T"])
test.assert_equals(coin(2),["HH", "HT", "TH", "TT"])
test.assert_equals(coin(3),["HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"])
```Code language: Python (python)```
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