<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 fewest number of coins that you need to make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> coins = [1,2,5], amount = 11
<strong>Output:</strong> 3
<strong>Explanation:</strong> 11 = 5 + 5 + 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> coins = [2], amount = 3
<strong>Output:</strong> -1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> coins = [1], amount = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 12</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>0 &lt;= amount &lt;= 10<sup>4</sup></code></li>
</ul>

"""
problem
+ you have an array of coins representing different demoninations and an amount of money
+ return the smallest number of coins that you need to represent that maount

must use recurssion of some form

insight:
+ different comb -> recurrence relation
+ dp to optimize similar recursive steps

approaches:
+ brute force: go through every possible result and find the smallest number of change
+ optimized (likely sol): go through different results, and use dp to remove redundent calls

optimized recursive (recurrence relation):
parameters -> amount
base case: if amount left == demonimation of some kind, return 1
if amount < all coins, return -1
loop through coins
    recurs on amount minus coin if the coin is less than denmon
        if it returns a valid amount (non negative), return this one up
        else continue on

    if never exited, return -1
"""

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        if amount == 0:
            return 0

        coins.sort(reverse = True)

        coinS = set(coins)
        smallestC = coins[-1]

        @cache
        def getStepsOfAmount_(curr_amount) -> int:
            if curr_amount in coinS:
                return 1
            
            if curr_amount < smallestC:
                return -1
            
            lowest = None

            for coin in coins:
                if coin < curr_amount:
                    steps = getStepsOfAmount_(curr_amount - coin)
                    
                    if steps != -1:
                        lowest = steps + 1 if lowest is None else min(lowest, steps + 1)

            return lowest if lowest is not None else -1
        
        return getStepsOfAmount_(amount)

34011