<p>You are given an integer array <code>coins</code> representing coins of different denominations and an integer <code>amount</code> representing a total amount of money.</p>

<p>Return <em>the fewest number of coins that you need to make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> coins = [1,2,5], amount = 11
<strong>Output:</strong> 3
<strong>Explanation:</strong> 11 = 5 + 5 + 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> coins = [2], amount = 3
<strong>Output:</strong> -1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> coins = [1], amount = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 12</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>0 &lt;= amount &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        coinSet = set(coins)
        # the minimum number of coins required to get to amount.
        # dp[amount] = min number of coins to get here
        # if you cannot form that value, -1.
        dp = defaultdict(lambda:float("inf"))

        # base case. Only use 1 coin for that coin value.
        for coinValue in coinSet:
            dp[coinValue] = 1

        dp[0] = 0

        for i in range(1, amount + 1):
            bestCandidate = float('inf')

            # find what is the best way to get to each possible previous value
            for coinValue in coinSet:
                candidate = dp[i - coinValue] # could be inf.
                bestCandidate = min(bestCandidate, candidate)

            # we've now tried every coin. your value is just previous value plus the one coin you tried.
            dp[i] = 1 + bestCandidate
        if dp[amount] == float('inf'):
            return -1
        return dp[amount]


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