Question

I'm trying to implement a solution to the following problem:

Given a number of dollars, n, and a list of dollar values for m distinct coins, find and print the number of different ways you can make change for n dollars if each coin is available in an infinite quantity.

The complete problem statement is here

Before presenting my solution, I should say that I know dynamic programming is probably the optimal way to solve this problem. The solution I've come up with does not use dynamic programming, and I know it is far from the most efficient solution out there. However, as far as I can tell it is correct.

The problem is that my solution underestimates the number of ways to make change. For example given n = 75 and {25 10 11 29 49 31 33 39 12 36 40 22 21 16 37 8 18 4 27 17 26 32 6 38 2 30 34} as coin values, the solution returns 182 when it should return 16694. It seems to work for smaller test cases, though.

def make_change(coins, n):    
    solns = 0
    num_coins = len(coins)
    for i in range(num_coins):
        solns = solns + make_change_rec(coins, n-coins[0], coins[0])

        # We've found all solutions involving coin. Remove it
        # from consideration
        coins.pop(0)

    return solns

def make_change_rec(coins, n, parent):
    # If n == 0 we've found a solution
    if n == 0:
        return 1

    solns = 0
    # For each coin
    for coin in coins:
        # If coin > n, can't make change using coin
        if coin > n or coin < parent: # coin < parent rule ensures we don't count same solution twice
            continue

        # Use the coin to make change 
        solns = solns + make_change_rec(coins, n-coin, coin)

    return solns

Can someone help me understand what I'm doing wrong? I'm using Python 3.

Answer 1

我认为您需要做的唯一更改是对coins数组进行排序。现在你的递归肯定会耗尽时间。所以我建议你选择经典的动态编程。

def make_change(coins, n):
    dp = [1] + [0] * n
    for coin in coins:
        for i in range(coin, n + 1):
            dp[i] += dp[i - coin]
    return dp[n]

Answer 2

您需要在输入硬币之前对其进行排序。如果你的递归对于较大的值（它将会）运行缓慢，你可以像这样添加memoization：

from functools import lru_cache

#leave your function as it is, but add this:
@lru_cache(maxsize=None)
def make_change_rec(coins, n, parent):

Make change for n dollars with m coins

2 个答案: