Why this solution isn't working for the coin change algorithm?

Question

Write a program that, given the amount N to make change for and the number of types of m infinitely available coins, and a list of m coins, prints out how many different ways you can make change from the coins to STDOUT.

My intuition was, for each coin, to try that coin, and recurse on n - c where c is the value of the coin, returning 1 if I get to zero and 0 if I get below zero. I passed in the previously used coin and only recursed on coins less than or equal to the previous coin to prevent duplicates. I am confused why this approach is incorrect, and how I can correct it.

Here is my code:

def calc_change(n, coins):
    cache = {}
    c = max(coins)
    nums = calc_ways(n, coins, cache, c)
    return nums

def calc_ways(n, coins, cache, current_coin):
    if n < 0:
        return 0
    if n == 0:
        return 1
    if n not in cache:
        cache[n] = sum(calc_ways(n - c, coins, cache, c) for c in coins if c <= current_coin)
    return cache[n]

answer = calc_change(n, coins) 
print answer

Thank you for any help.

I am running it on hackerrank and getting that the result is incorrect — Jared, May 24 '16 at 03:15
I think your problem is that `cache` stores how many possible combinations there are for `n`, but it doesn't consider the `current_coin`. — Pablo, May 24 '16 at 03:25

score 3 · Accepted Answer · answered May 24 '16 at 03:29

You are indexing cache according to the amount you want to add up to n. The problem is that the amount of combinations for the same n can change depending on the set of coins that you want to consider. (e.g. n=10 and coins=[10,5] has two possible combinations, but n=10 and coins=[5] has only one combination. You need your cache to consider the current_coin variable.

def calc_change(n, coins):
    cache = {}
    c = max(coins)
    nums = calc_ways(n, coins, cache, c)
    return nums

def calc_ways(n, coins, cache, current_coin):
    if n < 0:
        return 0
    if n == 0:
        return 1
    if (n,current_coin) not in cache:
        cache[(n,current_coin)] = sum(calc_ways(n - c, coins, cache, c) for c in coins if c <= current_coin)
    return cache[(n,current_coin)]

answer = calc_change(n, coins) 
print answer

Why this solution isn't working for the coin change algorithm?

1 Answers1