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June 5, 2019 15:18
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# given an amount and coin denominations, | |
# compute the number of ways to make amount with coins of the given denoms | |
# 4, [1,2,3] -> 4 | |
def ways_to_make(amount, denoms): | |
# ways to make 4 | |
# ~ ways(4 - 3) + ways(4 - 2) + ways(4 - 1) ~ | |
# ways(1, largest=1) + ways(2, largest=1) | |
# dim 1: amount | |
# dim 2: largest denomination used | |
# 1: 1 | |
# 2: 11, 2 | |
# 3: 111, 12, 3 | |
# 4: | |
# amount | |
# | |
# 0 1 2 3 4 | |
# 1 1 1 1 1 1 | |
# largest 2 1 1 2 2 3 | |
# 3 1 1 2 3 4 | |
# 4 1 1 2 3 5 | |
# ways(4, 3) = | |
# # use 3 | |
# ways(4 - 3, 3) = 1 | |
# # use 2 | |
# ways(4 - 2, 2) = 2 | |
# # use 1 | |
# ways(4 - 1, 1) = 1 | |
denoms.sort() | |
# there's one way to make 0, no matter what denominations we use | |
ways = [[1] * len(denoms)] | |
# determine how many ways there are to make current_amount (using only | |
# denominations or size `row` or smaller) | |
for current_amount in range(1, amount + 1): | |
column = [] | |
for row in denoms: | |
count = 0 | |
# for each denomination less than or equal to the current row, add | |
# the number of ways to make `current_amount - denom`. | |
for denom_ix, denom in enumerate(denoms): | |
# this and all larger denominations are too big | |
if denom > row or denom > current_amount: | |
break | |
else: | |
count += ways[current_amount - denom][denom_ix] | |
column.append(count) | |
ways.append(column) | |
return ways[amount][-1] | |
if __name__ == '__main__': | |
print(ways_to_make(0, [1])) | |
print(ways_to_make(0, [1,2,3])) | |
print(ways_to_make(4, [1,2,3])) | |
print(ways_to_make(4, [1,2,3,7,8,9,1000])) | |
print(ways_to_make(4, [1,2,3,4])) | |
print(ways_to_make(4, list(range(1,100000)))) |
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