mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-21 02:30:15 +00:00
4700297b3e
* Enable ruff RUF002 rule * Fix --------- Co-authored-by: Christian Clauss <cclauss@me.com>
60 lines
1.5 KiB
Python
60 lines
1.5 KiB
Python
"""
|
|
Problem 31: https://projecteuler.net/problem=31
|
|
|
|
Coin sums
|
|
|
|
In England the currency is made up of pound, f, and pence, p, and there are
|
|
eight coins in general circulation:
|
|
|
|
1p, 2p, 5p, 10p, 20p, 50p, f1 (100p) and f2 (200p).
|
|
It is possible to make f2 in the following way:
|
|
|
|
1xf1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
|
|
How many different ways can f2 be made using any number of coins?
|
|
|
|
Hint:
|
|
> There are 100 pence in a pound (f1 = 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 = 200) -> 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
|