<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 coinChange(vector<int>& a, int k) 
    {
        if(k==0)
            return 0;
        int n=a.size();
        sort(a.begin(),a.end());
        vector<long long int> dp(10005,INT_MAX);
        dp[0]=0;
        for(int i=0;i<10005;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i<a[j])
                    break;
                if(i==a[j])
                {
                    dp[i]=1;
                    break;
                }
                dp[i]=min(dp[i],(long long)1+dp[i-a[j]]);
            }
        } 
        if(dp[k]==INT_MAX)
            return -1; 
        return (int)dp[k];
    }
};
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