Python/project_euler/problem_31/sol2.py

"""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