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"""

Problem 31: https://projecteuler.net/problem=31

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

"""

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)

"""

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

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