<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 {
public:
int t[14][400001];
    int dp(vector<int>& coins, int n, int amount) {
        if(amount==0) return 0;
        if(n<0)
        {
          return 1e9;
        }
        if(t[n][amount]!=-1) return t[n][amount];
        int nottake=0+dp(coins,n-1,amount);
        int take=1e9;
        if(coins[n]<=amount) 
        {
            take=1+dp(coins,n,amount-coins[n]);
        }
        return t[n][amount]=min(take,nottake);
    }
    int coinChange(vector<int>& coins, int amount) {
        int n=coins.size();
         memset(t,-1,sizeof(t));
        int ans=dp(coins,n-1,amount);
       
        return ans==1e9 ? -1:ans;
    }
};
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