<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 number of combinations that make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>0</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>The answer is <strong>guaranteed</strong> to fit into a signed <strong>32-bit</strong> integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> amount = 5, coins = [1,2,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> amount = 3, coins = [2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> the amount of 3 cannot be made up just with coins of 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> amount = 10, coins = [10]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 300</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 5000</code></li>
	<li>All the values of <code>coins</code> are <strong>unique</strong>.</li>
	<li><code>0 &lt;= amount &lt;= 5000</code></li>
</ul>

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        self.dp=[{} for i in range(0,len(coins)+1)]
        self.dp[len(coins)]={0:1}
        self.rec(amount,coins,0)
        # print(self.dp) 
        return self.dp[0].get(amount,0)

    def rec(self,amount,coins,n):
        if self.dp[n]!={}:
            return self.dp[n]

        d=self.rec(amount,coins,n+1)
        dN={}
        sum=0
        while(sum<=amount):
            for k,val in d.items():
                if sum+k>amount:
                    continue
                v=dN.get(sum+k,-1)
                if v==-1:
                    dN[sum+k]=val
                else:
                    dN[sum+k]=v+val

            sum=sum+coins[n]
        self.dp[n]=dN
        return dN
            
            



21740