<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:
    vector<int> arr;
    int n ;
    int target;
    int t[13][10001];
    int fun(int idx,long currSum){
        if (idx == n) return INT_MAX-1;
        if (currSum > target) return INT_MAX-1;
        if (currSum == target) return 0;
        if (t[idx][currSum] != -1) return t[idx][currSum];
        long take = 1 + fun(idx,currSum+arr[idx]);
        long take2 = 1 + fun(idx+1,currSum+arr[idx]);
        long skip = 0 + fun(idx+1,currSum);
        return t[idx][currSum]= min({take,skip,take2});
    }
    int coinChange(vector<int>& coins, int amount) {
        arr = coins;
        n = coins.size();
        target =amount;
        memset(t,-1,sizeof(t));
        long ans  = fun(0,0);
        if (ans != INT_MAX-1) return ans;
        return -1;
    }
};
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