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dp O(price * len(denominations)) solution in Speedy category for Making Change by lezeroq
def checkio(price, denominations):
m = [0] + [100500] * price
for i in range(1, price + 1):
for note in denominations:
if note <= i and m[i-note] + 1 < m[i]:
m[i] = m[i-note] + 1
return None if m[price] == 100500 else m[price]
March 22, 2016