<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:
        mem = {}
        n = len(coins)
        
        def dp(i, amount):
            
            if i < 0:
                return 0

            if amount == 0:
                return 1
            
            if (i, amount) in mem:
                return mem[(i, amount)]
            
            if coins[i] > amount:
                mem[(i, amount)] = dp(i-1, amount)
            else:
                mem[(i, amount)] = dp(i-1, amount) + dp(i, amount-coins[i])
                
            return mem[(i, amount)]
        
        #ans = dp(n-1, amount)
        
        mem = [[0 for _ in range(amount + 1)] for _ in range(n + 1)]
        mem = [[0] * (amount + 1) for _ in range(n + 1)]
        
        for i in range(n + 1):
            mem[i][0] = 1
        
        for i in range(n - 1, -1, -1):
            for j in range(1, amount + 1):
                if coins[i] > j:
                    mem[i][j] = mem[i + 1][j]
                else:
                    mem[i][j] = mem[i + 1][j] + mem[i][j - coins[i]]
        
        for row in mem:
            print(row)
        print("ans:", mem[0][amount])
        return mem[0][amount]
        #print(f"mem={mem}")
        #print(f"Final ans={ans}")
        #return ans
    
    
    
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