<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>

# Top down
class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        @cache
        def dp(i, curr):
            if i >= len(coins) or curr > amount:
                return float("inf")
            elif curr == amount:
                return 0
            return min(
                1 + dp(i, curr+coins[i]),
                dp(i+1, curr)
            )

        min_num_coins = dp(0,0) 
        return min_num_coins if min_num_coins != float("inf") else -1

# Bottom Up
class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:        
        """
        Note
        - i maps to curr in top down soln
            - i-coins[j] maps to curr+coins[i] in dp(i, curr+coins[i]) 
        - j maps to i (index of coins) in top down soln
            - j-1 maps to i+1 in dp(i+1, curr)

        dp = [ [float("inf") for j in range(len(coins)+1)] for i in range(amount+1) ]
        - for i in range(amount+1) : amount can be 0 
        - for j in range(len(coins)+1) : min nos of coins to take can be 0
        """

        dp = [ [float("inf") for j in range(len(coins)+1)] for i in range(amount+1) ]
        
        for j in range(len(coins)+1):
            dp[0][j] = 0

        for i in range(amount+1):
            for j in range(len(coins)+1):
                if j < len(coins) and i - coins[j] >= 0:
                    dp[i][j] = min(dp[i][j], 1 + dp[i - coins[j]][j])
                if j - 1 >= 0:
                    dp[i][j] = min(dp[i][j], dp[i][j-1])

        return dp[-1][-1] if dp[-1][-1] != float("inf") else -1

       
# Bottom Up (space optimised)
# class Solution:
#     def coinChange(self, coins: List[int], amount: int) -> int:
#         """
#         Overall Approach
#         - dp[i] 
#             - nos of ways to form i amt w/o taking new coins
#         - 1 + dp[i - c]
#             - nos of ways to form i amt by taking coin c (1) + nos of ways to form i-c amt
#         """
        
#         dp = [float("inf")] * (amount+1)
#         dp[0] = 0
#         for i in range(len(dp)):
#             for c in coins:
#                 if i - c in range(len(dp)):
#                     dp[i] = min(dp[i], 1 + dp[i - c])
              
#         return dp[-1] if dp[-1] != float("inf") else -1

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