<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:
        # if not coins:
        #     return 0

        # n  = len(coins)

        # def helper(idx, amount):
        #     if amount == 0:
        #         return 0
            
        #     if idx < n and amount > 0:
        #         min_cost = float('inf')

        #         for x in range(0, amount // coins[idx] + 1):

        #             result = helper(idx + 1, amount - x * coins[idx])
        #             if result != -1:
        #                 min_cost = min (min_cost, result + x)

        #         return -1 if min_cost == float('inf') else min_cost
        #     return -1 

        # return helper(0, amount)
        # dp = [float('inf')] * (amount + 1)
        # dp[0] = 0

        # for coin in coins:
        #     for x in range(coin, amount + 1):
        #         dp[x] = min(dp[x], dp[x - coin] + 1)
        
        # return dp[amount] if dp[amount] != float('inf') else -1
        @lru_cache(None)
        def dfs(rem):
            if rem < 0:
                return - 1
            
            if rem == 0:
                return 0 
            
            min_cost = float('inf')

            for coin in coins:
                res = dfs(rem - coin)
                if res != -1:
                    min_cost = min(min_cost, res + 1)
            
            return min_cost if min_cost != float('inf') else -1 

        return dfs(amount)


        

                    
36828