<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:

        # Quick Discard
        if amount == 0: return 1

        # Cache to reduce the repeated searches
        seen      = {}

        # Constants
        MAX_INDEX = len(coins) - 1

        # Usd DFS
        def coinChangeDFS(amount, index, ways):

            # Find whether this permutation has been reached before
            key = tuple([amount, index])
            if key in seen:
                return seen[key]
            #end

            # Determine if we can keep decrementing the amount
            # for the current count
            coin = coins[index]
            if amount > coin :
                ways += coinChangeDFS(amount - coin, index, 0)

            # Determine if we've reached the total... performed this way
            # to prevent an unnecessary recursive call
            elif amount == coin :
                ways += 1
            #end

            # Move on to the next coint
            if index < MAX_INDEX :
                while index < MAX_INDEX :
                    index += 1
                    if amount >= coins[index] :
                        ways += coinChangeDFS(amount, index, 0)
                        break
                    #end
                #end
            #end

            # Add the current permutation to the seen list
            seen[key] = ways

            # Return the Permuation Count
            return ways

        #end

        return coinChangeDFS(amount, 0, 0)

    #end

75890