<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 helper(self,coins,idx,amount,dp):
        if idx == 0:
            if amount % coins[0] == 0:
                return amount // coins[0]
            return 1e8
        if amount == 0:
            return 0
        if (idx,amount) in dp:
            return dp[(idx,amount)]
        take = 1e8
        if coins[idx] <= amount:
            take = 1 + self.helper(coins,idx,amount-coins[idx],dp)
        notTake = self.helper(coins,idx-1,amount,dp)
        dp[(idx,amount)] = min(take,notTake)
        return min(take,notTake)


    def coinChange(self, coins: List[int], amount: int) -> int:
        n = len(coins)
        dp = {}
        ans =  self.helper(coins,n-1,amount,dp)
        if ans == 1e8:
            return -1
        return ans

    # def coinChange(self, coins: List[int], amount: int) -> int:
    #     n = len(coins)

    
        
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