<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:
        n=len(coins)
        dp=[[-1 for i in range(amount+1)]for j in range(n+1)]
        ans=self.helper(n-1,amount,coins,n,dp)
        if(ans==sys.maxsize):
            return -1
        return ans

    def helper(self,ind,amount,coins,n,dp):
        if(ind==0):
            if(amount%coins[ind]==0):
                return amount//coins[ind]
            else:
                return sys.maxsize
        if(dp[ind][amount]!=-1):
            return dp[ind][amount]
        #two options either to pick or notpick
        notpick=self.helper(ind-1,amount,coins,n,dp)
        pick=sys.maxsize
        if(coins[ind]<=amount):
            #here we have infinite supply of coins
            pick=1+self.helper(ind,amount-coins[ind],coins,n,dp)
        dp[ind][amount]=min(pick,notpick)
        return dp[ind][amount]
        








    





    


        
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