<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>

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        
        # dp[i][j] = number of coins from first i coins needed to make amount j
        
        n = len(coins)
        dp = [[float('inf') for j in range(amount+1)] for i in range(n)]
        
        for i in range(n): 
            dp[i][0] = 0 # no coins needed to make amount 0
            if coins[i] <= amount:
                dp[i][coins[i]] = 1 # 1 coin needed to make coins[i] amongst first i coins
                        
        for j in range(1, amount+1):
            for i in range(n):
                
                if coins[i] <= j:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-coins[i]]+1)  # dp[i-1][j] -> not using ith coin, or using it 
                else:
                    dp[i][j] = dp[i-1][j] # not using coins[i], just using first i-1 coins
                    
        if dp[n-1][amount] == float('inf'):
            return -1
        return dp[n-1][amount]
21131