<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>

const int N = 15,M = 1e5;
int mem[N][M];
int n;
class Solution {
public:
    int dp(int ind,int amount,vector<int>& coins)
    {
        if(amount==0)
            return 0;
        int &ret = mem[ind][amount];
        if(~ret)return ret;
        ret = 1e9;
        for(int i=0;i<n;i++)
        {
            if(amount>=coins[ind])ret = min(ret,1+dp(ind,amount-coins[ind],coins));
            if(i+1<n)
                ret = min(ret,dp(i+1,amount,coins));
        }
        return ret;

    }
    int coinChange(vector<int>& coins, int amount) {
        n = coins.size();
        memset(mem,-1,sizeof mem);
        int ans = dp(0,amount,coins);
        return (ans>=1e9?-1:ans);
    }
};
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