<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:
        coins = sorted(coins, reverse=True)
        # O(S*n)  - S amount, n number of coins
        @lru_cache(None)
        def dfs(remaining_amount):
            if remaining_amount < 0:
                return -1
            if remaining_amount == 0:
                return 0
            
            min_cost = float('inf')
            for coin in coins:
                result = dfs(remaining_amount-coin)
                if result != -1:
                    min_cost = min(min_cost, result+1) # +1 for the coin we just used
            
            return min_cost if min_cost != float('inf') else -1
        
        return dfs(amount)














        # cache = [amount+1] * (amount+1)
        # cache[0] = 0

        # for i in range(1, amount+1):
        #     for c in coins:
        #         if i - c >= 0:
        #             # print(i - c)
        #             cache[i] = min(cache[i], cache[i-c] + 1)

        # if cache[amount] <= amount:
        #     return cache[amount]
        # return -1

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