<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:
        
        # let the vlue of dp[i] be the number of ways to make up that coin
        # if we have an arbitrary dp[i-2]
        # and we have coin 2
        # then dp[i] can be +1 since dp[i-2] + 2 is another way of making up the coin
        
        dp = [0] * (amount + 1)
        dp[0] = 1
        # if 2 in coins, then dp[2] = 2 since [(2), (1,1)]
        # else, then dp[2] = 1 since [(1,1)]
        
        
        for coin in coins: # coin = 2
            for i in range(coin, amount + 1): # i = 2
                dp[i] += dp[i - coin]
                # dp[1] += 1 --> now its 1
                # dp[2] += 1 -- now its 1
                ...
                # basically saying that for all amounts its possible to get 
                # that amount by using the 1 coins. 
                
                # dp[2] += 1 --> now 2. 
                # dp[3] += 2 --> now 3 cause (1,1,1), (1,2)
            print(dp)
            
        return dp[amount]
        
19484