<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 coins(self, idx, amt, coins, dp):
        if idx == 0:
            if amt % coins[idx] == 0:
                return amt//coins[idx]
            else:
                return 10**8

        if dp[idx][amt] != -1:
            return dp[idx][amt]
        not_take = self.coins(idx-1, amt, coins, dp)
        take = 10**8
        if coins[idx] <= amt:
            take = 1 + self.coins(idx, amt-coins[idx], coins, dp)
        dp[idx][amt] = min(not_take, take)
        return dp[idx][amt]

    def coinChange(self, coins: List[int], amount: int) -> int:
        n = len(coins)
        dp = [[-1 for _ in range(amount+1)] for _ in range(n)]
        return self.coins(n-1,amount, coins, dp) if self.coins(n-1,amount, coins, dp) != 10**8 else -1
        
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