<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 n;
int coin[12];
int dp[12][10001];
    int rec(int i, int left){
        if(left<0) return 1e9;
        if(i==n){
            if(left==0) return 0;
            else return 1e9;
        }

        if(dp[i][left]!=-1) return dp[i][left];
        int ans=1e9;
        ans=min({ans,rec(i+1,left), rec(i,left-coin[i])+1});

        return dp[i][left]=ans;
    }
    int coinChange(vector<int>& coins, int amount) {
        n=coins.size();
        memset(dp,-1,sizeof(dp));
        for(int i=0; i<n; i++) coin[i]=coins[i];
        if(rec(0,amount)==1e9) return -1;
        return rec(0,amount);
    }
};
15441