<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 fun(vector<int> &v,int ind,int n,int k,vector<vector<int>> &dp)
    {
        if(ind==n)
        {
            if(k==0)    return 0;
            else        return 1e9;
        }

        if(dp[ind][k]!=-1)  return dp[ind][k];

        // notTake
        int notTake = 0 + fun(v,ind+1,n,k,dp);

        //Take
        int Take = 1e9;

        if(v[ind]<=k)   Take = 1+fun(v,ind,n,k-v[ind],dp);

        return dp[ind][k] = min(Take,notTake);
    }
    int coinChange(vector<int>& v, int k) {
        int n = v.size();
        vector<vector<int>> dp(n+1, vector<int> (k+1, -1));
        int val = fun(v,0,n,k,dp);
        if(val>=1e9)    return -1;
        else            return val;
    }
};
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