Grokking Dynamic Programming Interview in Python/

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Coin Change II

Coin Change II

Let's solve the Coin Change II problem using Dynamic Programming.

We'll cover the following...

Statement

Suppose you are given a list of coins and a certain amount of money. Each coin in the list is unique and of a different denominationA denomination is a unit of classification for the stated or face value of financial instruments such as currency notes or coins.. You are required to count the number of ways the provided coins can sum up to represent the given amount. If there is no possible way to represent that amount, then return 0.

Note: You may assume that for each combination you make, you have an infinite number of each coin. In simpler terms, you can use a specific coin as many times as you want.

Let's say you have only two coins, 1010 and 2020 cents, and you want to represent the total amount of 3030 cents using these coins. There are only two ways to do this, you can use either of the following combinations:

  • 3 coins of 1010 cents: 10+10+10=3010+10+10=30.

  • 1 coin of 1010 cents and 1 coin of 2020 cents: 10+20=30.10+20=30.

Constraints:

  • 1 <= coins.length <= 300

  • 1 <= coins[i] <= 5000

  • All the coins have a unique value.

  • 0 <= amount <= 5000

Examples

Let's see a few more examples to get a better understanding of the problem statement:

No.

Coins

Amount

Number of ways

1

[1, 2, 5]

7

6

2

[1, 5, 10, 25]

10

4

3

[7]

9

0

4

[9,10,11]

0

1

Try it yourself

Implement your solution in the following playground.

Press + to interact
Python
usercode > main.py
def count_ways(coins, amount):

  # replace this placeholder return statement with your code

  return -1
Coin Change II

Note: If you clicked the “Submit” button and the code timed out, this means that your solution needs to be optimized in terms of running time.

Hint: Use dynamic programming and see the magic.

Solution

We will first explore the naive recursive solution to this problem and then see how it can be improved using the Unbounded Knapsack dynamic programming pattern.

Naive approach

A naive approach to solve this problem would be to make all the possible combinations of coins or generate all subsets of coin denominations that sum up to the required amount.

While making the combinations, a point to keep in mind is that we should try to avoid repeatedly counting the same combinations. For example, 10+2010+20 cents add up to 3030 cents, and so do 20+1020+10. In the context of combinations, both these sequences are the same, and we will count them as one combination. We can achieve this by using a subset of coins every time we consider them for combinations. The first time, we can use all nn coins, the second time, we can use n−1n−1 coins, that is, by excluding the largest one, and so on. This way, it is not possible to consider the same denomination more than once.

As a base case, we return 11 when the target amount is zero because there is only one way to represent zero irrespective of the number and kinds of coins available. Similarly, for the second base case, if at any point during our search for combinations, the remaining value needed to reach the total amount becomes less than zero, we return 00.

The slides below will help you visualize the process.

Now, let's look at the code in the code widget below for the solution we just discussed.

def count_ways_rec(coins, amount, maximum):
  if amount == 0:     # base case 1
    return 1
  if amount < 0:      # base case 2
    return 0
  ways = 0
  # iterate over coins
  for coin in coins:    
    # to avoid repetition of similar sequences, use coins smaller than maximum
    if coin <= maximum and amount - coin >= 0:  
      
      # notice how maximum is set to the current value of coin in recursive call    
      ways += count_ways_rec(coins, amount-coin, coin)  
  return ways
def count_ways(coins, amount):
  return count_ways_rec(coins, amount, max(coins))
def main():
    coins_lists = [[7], [1, 2, 5], [10], [49, 233, 96, 132, 188], 
                   [310, 278, 99, 326, 5, 575, 569, 15, 141, 54], 
                   [2823, 4551, 1750, 49, 3256, 405, 380, 4785, 3893, 874],
                   [17, 1422, 30, 1153, 1275], [1460], [9, 10, 11]]
    amounts = [9, 5, 10, 225, 350, 3200, 700, 2000, 0]
    # You can uncomment the lines below and check how this recursive solution causes a time-out
    # Length of coins_list used here is 45
    # coins_lists.append([4021, 2158, 3577, 2157, 3928, 1229, 3888, 
    #                     1464, 56, 4945, 138, 3148, 699, 3001, 83, 
    #                     170, 4673, 941, 1539, 4904, 4118, 1953, 17, 
    #                     2963, 4406, 2359, 2363, 2288, 4147, 1070, 
    #                     4006, 3824, 467, 1785, 3365, 264, 3598, 
    #                     1132, 3093, 3634, 3122, 3618, 4545, 3256, 836])
    # amounts.append(5000)
    for i in range(len(amounts)):
        print(i+1, ".", "\tCoins: ", coins_lists[i], sep="")
        print("\tAmount: ", amounts[i], sep="")
        num_ways = count_ways(coins_lists[i], amounts[i])
        print("\tNumber of Ways: ", num_ways, sep="")
        print("-" * 100)
        print()
if __name__ == '__main__':
    main()
Coin Change II using recursion

Note: Please observe that if you include the test case commented out in the driver code, the solution is likely to time out. Try to solve the larger problem using the dynamic programming solutions provided below and see the difference.

Time complexity

If the length of the list coins is ...