def rec_coin_dynam(target,coins,known_results):
'''
INPUT: This funciton takes in a target amount and a list of possible coins to use.
It also takes a third parameter, known_results, indicating previously calculated results.
The known_results parameter shoud be started with [0] * (target+1)
OUTPUT: Minimum number of coins needed to make the target.
'''
# Default output to target
min_coins = target
# Base Case
if target in coins:
known_results[target] = 1
return 1
# Return a known result if it happens to be greater than 1
elif known_results[target] > 0:
return known_results[target]
else:
# for every coin value that is <= than target
for i in [c for c in coins if c <= target]:
# Recursive call, note how we include the known results!
num_coins = 1 + rec_coin_dynam(target-i,coins,known_results)
# Reset Minimum if we have a new minimum
if num_coins < min_coins:
min_coins = num_coins
# Reset the known result
known_results[target] = min_coins
return min_coins
This runs perfectly fine but I have few questions about it.
We give it the following input to run:
target = 74
coins = [1,5,10,25]
known_results = [0]*(target+1)
rec_coin_dynam(target,coins,known_results)
why are we initalising the know result with zeros of length target+1? why can't we just write
know_results = []
Notice that the code contains lines such as:
known_results[target] = 1
return known_results[target]
known_results[target] = min_coins
Now, let me demonstrate the difference between [] and [0]*something in the python interactive shell:
>>> a = []
>>> b = [0]*10
>>> a
[]
>>> b
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
>>>
>>> a[3] = 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: list assignment index out of range
>>>
>>> b[3] = 1
>>>
>>> a
[]
>>> b
[0, 0, 0, 1, 0, 0, 0, 0, 0, 0]
The exception IndexError: list assignment index out of range was raised because we tried to access cell 3 of list a, but a has size 0; there is no cell 3. We could put a value in a using a.append(1), but then the 1 would be at position 0, not at position 3.
There was no exception when we accessed cell 3 of list b, because b has size 10, so any index between 0 and 9 is valid.
Conclusion: if you know in advance the size that your array will have, and this size never changes during the execution of the algorithm, then you might as well begin with an array of the appropriate size, rather than with an empty array.
What is the size of known_results? The algorithm needs results for values ranging from 0 to target. How many results is that? Exactly target+1. For instance, if target = 2, then the algorithm will deal with results for 0, 1 and 2; that's 3 different results. Thus known_results must have size target+1. Note that in python, just like in almost every other programming language, a list of size n has n elements, indexed 0 to n-1. In general, in an integer interval [a, b], there are b-a+1 integers. For instance, there are three integers in interval [8, 10] (those are 8, 9 and 10).