<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>

import sys
class Solution:
    def helper(self,n,coins,w,res,idx):
        if n == idx:
            if w == 0:
                return res
            return sys.maxsize
        if w>=coins[idx]:
            return min(self.helper(n,coins,w-coins[idx],res+1,idx),self.helper(n,coins,w-coins[idx],res+1,idx+1),self.helper(n,coins,w,res,idx+1))
        else:
            return self.helper(n,coins,w,res,idx+1)
    def coinChange(self, coins: List[int], amount: int) -> int:
        # res=0
        
        # ans =  self.helper(n,coins,amount,res,0)
        # if ans == sys.maxsize:
        #     return -1
        # return ans
        n=len(coins)
        dp=[[float('inf')]*(amount+1) for _ in range(n+1)]
        for i in range(n+1):
            dp[i][0] = 0
        for i in range(1,n+1):
            for j in range(amount+1):
                if j>=coins[i-1]:
                    dp[i][j] = min(dp[i][j-coins[i-1]]+1,dp[i-1][j])
                else:
                    dp[i][j] = dp[i-1][j]
        if dp[n][amount] == float('inf'):
            return -1
        return dp[n][amount]
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