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57 lines
1.4 KiB
Python
57 lines
1.4 KiB
Python
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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
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eight coins in general circulation:
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1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
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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
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How many different ways can £2 be made using any number of coins?
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Hint:
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> There are 100 pence in a pound (£1 = 100p)
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> There are coins(in pence) are available: 1, 2, 5, 10, 20, 50, 100 and 200.
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> how many different ways you can combine these values to create 200 pence.
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Example:
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to make 6p there are 5 ways
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1,1,1,1,1,1
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1,1,1,1,2
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1,1,2,2
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2,2,2
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1,5
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to make 5p there are 4 ways
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1,1,1,1,1
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1,1,1,2
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1,2,2
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5
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"""
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def solution(pence: int) -> int:
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"""Returns the number of different ways to make X pence using any number of coins.
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The solution is based on dynamic programming paradigm in a bottom-up fashion.
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>>> solution(500)
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6295434
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>>> solution(200)
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73682
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>>> solution(50)
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451
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>>> solution(10)
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11
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"""
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coins = [1, 2, 5, 10, 20, 50, 100, 200]
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number_of_ways = [0] * (pence + 1)
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number_of_ways[0] = 1 # base case: 1 way to make 0 pence
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for coin in coins:
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for i in range(coin, pence + 1, 1):
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number_of_ways[i] += number_of_ways[i - coin]
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return number_of_ways[pence]
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if __name__ == "__main__":
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assert solution(200) == 73682
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