<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>

#define ll long long
vector<ll>coins_now;
ll n;
ll dp[100005];

ll func(ll koto)
{
    //cout<<"koto: "<<koto<<endl;
    if(koto==n){return 0;}
    else if(koto>n){return 1e9;}
    if(dp[koto]!=-1)
    {
        return dp[koto];
    }
    ll koita=1e18;
    for(int i=0;i<coins_now.size();i++)
    {
        //cout<<"paisa: "<<coins_now[i]<<endl;
        koita=min(koita, 1+func(koto+coins_now[i]));
    }
    //cout<<"koita: "<<koita<<endl;
    return dp[koto]=koita;
}


class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        for(int i=0;i<coins.size();i++)
        {
            coins_now.push_back(coins[i]);
        }
        n=amount;
       for(int i=0;i<=n+5;i++)
       {
            dp[i]=-1;
       }
        ll now=func(0);
        coins_now.clear();
        if(now<0 || now>=1e9)
        return -1;
        return now;

    }
};
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