<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>& nums, int k) {
        int n = nums.size();
        sort(nums.rbegin(),nums.rend());

        vector<vector<int>> dp(n,vector<int>(k+1,-1));

        auto give = [&] (auto&& give,int i,int sum) -> int {
            if(sum<0) return 1e8;
            if(!sum) return 0;
            if(i==n) return 1e8;
            if(dp[i][sum] != -1) return dp[i][sum];

            return dp[i][sum] = min(give(give,i+1,sum),1+give(give,i,sum-nums[i]));
        };

        int ans = give(give,0,k);
        if(ans == 1e8) return -1;
        return ans;
    }
};
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