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Create sol2.py
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"""Coin sums | ||
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In England the currency is made up of pound, £, and pence, p, and there are | ||
eight coins in general circulation: | ||
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1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). | ||
It is possible to make £2 in the following way: | ||
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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? | ||
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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. | ||
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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 | ||
""" | ||
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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. | ||
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>>> 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 | ||
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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] | ||
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if __name__ == "__main__": | ||
assert solution(200) == 73682 |
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