<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>

import numpy as np
class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        dp = np.full((len(coins) + 1, amount + 1), -1)
        #print(dp)
        
        def ways_to_make_amt_with_rest_of_coins(i: int, amt: int) -> int:
            if dp[i, amt] != -1:
                return dp[i, amt]
            res = 0
            if i == len(coins):
                res = 1 if amt == 0 else 0
            else:
                take = ways_to_make_amt_with_rest_of_coins(i, amt - coins[i]) if amt >= coins[i] else 0
                skip = ways_to_make_amt_with_rest_of_coins(i + 1, amt)
                res = take + skip
            dp[i, amt] = res
            return res
        ret = ways_to_make_amt_with_rest_of_coins(0, amount)
        #print(dp)
        return ret

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