<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:
        # coins.sort()
        # dp, res = {}, []
        # n = len(coins)

        # def dfs(amount, index):
        #     if amount == 0:
        #         return 1

        #     if (amount, index) in dp:
        #         return dp[(amount, index)]

        #     res = 0
        #     for i in range(index, n):
        #         if amount - coins[i] < 0:
        #             break
        #         res += dfs(amount - coins[i], i)            

        #     dp[(amount, index)] = res
        #     return res    

        # return dfs(amount, 0)    

        dp = {i + 1: {} for i in range(amount)}
        # dp[i] = [ index: [curr, totalSum]]

        for i in range(1, amount + 1):
            sum1 = 0
            for j in range(len(coins) - 1, -1, -1):
                if i - coins[j] < 0:
                    dp[i][j] = [0, sum1]    
                elif i - coins[j] == 0:
                    sum1 += 1
                    dp[i][j] = [1, sum1]    
                else:
                    sum1 += dp[i - coins[j]][j][1]
                    dp[i][j] = [dp[i - coins[j]][j][1], sum1]    
                
            if i == amount:
                return sum1        

        return 1        
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