<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 __init__(self):
        self.cache = {}

    def coinChange(self, coins: List[int], amount: int) -> int:
        res = self.find(coins, amount)
        if res == 1000000:
            return -1
        else:
            return res

    def find(self, coins: List[int], amount: int) -> int:
        L = len(coins)
        if (L,amount) not in self.cache:
            if not amount:
                self.cache[(L,amount)] = 0
            elif not coins:
                self.cache[(L,amount)] = 1000000
            else:
                a, b = 1000000, 1000000
                a = self.find(coins[:-1], amount)
                if coins[-1] <=amount:
                    b = self.find(coins, amount-coins[-1])
                self.cache[(L,amount)] = min(a,b+1)
        return self.cache[(L,amount)]
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