This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Solution Notebook

Problem: Determine the total number of unique ways to make n cents, given coins of denominations less than n cents.

Constraints

  • Do the coins have to reach exactly n cents?
    • Yes
  • Can we assume we have an infinite number of coins to make n cents?
    • Yes
  • Do we need to report the combination(s) of coins that represent the minimum?
    • No
  • Can we assume the coin denominations are given in sorted order?
    • No
  • Can we assume this fits memory?
    • Yes

Test Cases

  • coins: None or n: None -> Exception
  • coins: [] or n: 0 -> 0
  • coins: [1, 2, 3], n: 5 -> 5

Algorithm

We'll use a bottom-up dynamic programming approach.

The rows (i) represent the coin values.
The columns (j) represent the totals.

  -------------------------
  | 0 | 1 | 2 | 3 | 4 | 5 |
  -------------------------
0 | 1 | 0 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 1 | 2 | 2 | 3 | 3 |
3 | 1 | 1 | 2 | 3 | 4 | 5 |
  -------------------------

Number of ways to get total n with coin[n] equals:
* Number of ways to get total n with coin[n - 1] plus
* Number of ways to get total n - coin[n]

if j == 0:
    T[i][j] = 1
if row == 0:
    T[i][j] = 0
if coins[i] >= j
    T[i][j] = T[i - 1][j] + T[i][j - coins[i]]
else:
    T[i][j] = T[i - 1][j]

The answer will be in the bottom right corner of the matrix.

Complexity:

  • Time: O(i * j)
  • Space: O(i * j)

Code

In [1]:
class CoinChanger(object):

    def make_change(self, coins, total):
        if coins is None or total is None:
            return None
        if not coins or total == 0:
            return 0
        coins = [0] + coins
        num_rows = len(coins)
        num_cols = total + 1
        T = [[None] * num_cols for _ in range(num_rows)]
        for i in range(num_rows):
            for j in range(num_cols):
                if i == 0:
                    T[i][j] = 0
                    continue
                if j == 0:
                    T[i][j] = 1
                    continue
                if coins[i] <= j:
                    T[i][j] = T[i - 1][j] + T[i][j - coins[i]]
                else:
                    T[i][j] = T[i - 1][j]
        return T[num_rows - 1][num_cols - 1]

Unit Test

In [2]:
%%writefile test_coin_change.py
import unittest


class Challenge(unittest.TestCase):

    def test_coin_change(self):
        coin_changer = CoinChanger()
        self.assertEqual(coin_changer.make_change([1, 2], 0), 0)
        self.assertEqual(coin_changer.make_change([1, 2, 3], 5), 5)
        self.assertEqual(coin_changer.make_change([1, 5, 25, 50], 10), 3)
        print('Success: test_coin_change')


def main():
    test = Challenge()
    test.test_coin_change()


if __name__ == '__main__':
    main()
Overwriting test_coin_change.py
In [3]:
%run -i test_coin_change.py
Success: test_coin_change