<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:
    """
    coins unique, can use infinite times, amount >= 0
    return number of unique combinations, 0 if impossible
    (2+2+1 and 1+2+2 is same) 

    R: dp(i, r): number of unique combinations, coins[i:] sums to r
    dp(i, r) = sum(dp(i+1, r-coins[i]*k) for k in range(r//coins[i]+1))
    base case: if r == 0, return 1, if i==n: return 0 (assert r >= 0)
    return dp(0, amount)
    
    top-down dp, time O(n*amount^2), space O(n*amount)
    """
    def change(self, amount: int, coins: List[int]) -> int:
        @cache
        def dp(i, r):
            if r == 0: return 1
            if i == len(coins): return 0
            return sum(dp(i+1, r-coins[i]*k) for k in range(r//coins[i]+1))
        return dp(0, amount)
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