Coin change#

Levels: level-3

Algorithms: dynamic-programming

Practice Link#

LeetCode

Description#

You are given coins of different denominations and a total amount of money amount.
Write a function to compute the fewest number of coins that you need to make up that amount.
If that amount of money cannot be made up by any combination of the coins, return -1.

Note:

You may assume that you have an infinite number of each kind of coin.

Examples#

1Input: coins = [1, 2, 5], amount = 11
2Output: 3
3Explanation: 11 = 5 + 5 + 1

1Input: coins = [2], amount = 3
2Output: -1

Python Solution#

 1class Solution(object):
 2    def coinChange(self, coins, amount):
 3        """
 4        :type coins: List[int]
 5        :type amount: int
 6        :rtype: int
 7        """
 8        # Start with a dp array of size amount + 1. 
 9        # Use amount + 1 as the "max" value. 
10        rs = [amount + 1] * (amount + 1)
11        # No coin needed for amount zero
12        rs[0] = 0
13        # Calculate min counts needed for each target upto amount.
14        for i in range(1, amount + 1):
15            # Check for each available denomination
16            for c in coins:
17                if i >= c:
18                    # Coins needed for target amount i is
19                    # the min of (current calculated min number of coins)
20                    # and (coins needed for i - c target amount + 1) 
21                    rs[i] = min(rs[i], rs[i - c] + 1)
22
23        if rs[amount] == amount + 1:
24            # return -1 if calculated n coins is the "max" value set initially.
25            return -1
26        return rs[amount]
27
28
29tm = [
30    ([1, 2, 5], 11, 3),
31    ([2], 3, -1),
32    ([], 11, -1),
33]
34
35if __name__ == "__main__":
36
37    sol1 = Solution()
38
39    for idx, nin in enumerate(tm):
40        resout = sol1.coinChange(nin[0], nin[1])
41        print(nin, resout, resout == nin[-1])