<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:

        # memoization - cache

        R = len(coins)
        C = amount

        # + 1 to account for 0
        dp = [[0] * (C + 1) for _ in range(R)]
        
        # amt 0 basecase | always one way to make 0, choosing 0 coins
        for i in range(R):
            dp[i][0] = 1

        # find the first row with first coin
        # from 1 to Amount (need the plus 1 bcs the 0)
        for i in range(1, C + 1):
            if i % coins[0] == 0: 
                dp[0][i] = 1
            else: 
                dp[0][i] = 0

        # fill out the rest of the table
        for i in range(1, R):
            for j in range(1, C + 1): 
                # current coin value <= current amount

                if coins[i] <= j:
                    # if less
                    # Number of ways without including the current coin + 
                    # number of ways while including coin
                    dp[i][j] = dp[i - 1][j] + dp[i][j - coins[i]]
                else: 
                    # keep num of ways without including
                    dp[i][j] = dp[i - 1][j]
        
        return dp[R - 1][C]
33021