mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-07 06:09:28 +00:00
1f8a21d727
* Tighten up psf/black and flake8
* Fix some tests
* Fix some E741
* Fix some E741
* updating DIRECTORY.md
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
57 lines
1.4 KiB
Python
57 lines
1.4 KiB
Python
"""Coin sums
|
||
|
||
In England the currency is made up of pound, £, and pence, p, and there are
|
||
eight coins in general circulation:
|
||
|
||
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
|
||
It is possible to make £2 in the following way:
|
||
|
||
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
|
||
How many different ways can £2 be made using any number of coins?
|
||
|
||
Hint:
|
||
> There are 100 pence in a pound (£1 = 100p)
|
||
> There are coins(in pence) are available: 1, 2, 5, 10, 20, 50, 100 and 200.
|
||
> how many different ways you can combine these values to create 200 pence.
|
||
|
||
Example:
|
||
to make 6p there are 5 ways
|
||
1,1,1,1,1,1
|
||
1,1,1,1,2
|
||
1,1,2,2
|
||
2,2,2
|
||
1,5
|
||
to make 5p there are 4 ways
|
||
1,1,1,1,1
|
||
1,1,1,2
|
||
1,2,2
|
||
5
|
||
"""
|
||
|
||
|
||
def solution(pence: int) -> int:
|
||
"""Returns the number of different ways to make X pence using any number of coins.
|
||
The solution is based on dynamic programming paradigm in a bottom-up fashion.
|
||
|
||
>>> solution(500)
|
||
6295434
|
||
>>> solution(200)
|
||
73682
|
||
>>> solution(50)
|
||
451
|
||
>>> solution(10)
|
||
11
|
||
"""
|
||
coins = [1, 2, 5, 10, 20, 50, 100, 200]
|
||
number_of_ways = [0] * (pence + 1)
|
||
number_of_ways[0] = 1 # base case: 1 way to make 0 pence
|
||
|
||
for coin in coins:
|
||
for i in range(coin, pence + 1, 1):
|
||
number_of_ways[i] += number_of_ways[i - coin]
|
||
return number_of_ways[pence]
|
||
|
||
|
||
if __name__ == "__main__":
|
||
assert solution(200) == 73682
|