<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:
        n = len(coins)

        mem = {}

        def backtrack(amnt, pos):
            if amnt > amount:
                return 0
            if amnt == amount:
                return 1
            if (amnt, pos) in mem:
                return mem[amnt, pos]
            mem[amnt, pos] = sum([backtrack(amnt + coins[i], i) for i in range(pos, n)])
            return mem[amnt, pos]

        return backtrack(0, 0)

        
60096