<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:
    long long int a[(unsigned long)1e4+1][13];
    long long int coinChange(vector<int>& coins, int amount) {
        for(int i=0;i<1e4+1;i++)
        {
            for(int j=0;j<13;j++)
            {
                a[i][j]=-1;
            }
        }
            for(int j=0;j<13;j++)
            {
                a[0][j]=0;
            }
        if(ans(coins, amount,coins.size()-1)>=INT_MAX) return -1;
        return ans(coins, amount,coins.size()-1);
    }
    long long int ans(vector<int>& coins, int amount,int n)
    {
        if(amount<0 || n<0) return INT_MAX;
        if(a[amount][n]!=-1) return a[amount][n];
        a[amount][n]=min(1+ans(coins,amount-coins[n],n),ans(coins,amount,n-1));
        return a[amount][n];
    }
};
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