<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 solve(int i,int n,vector<int> &coins,int a, vector<vector<int>> &dp){


        if(i == n){
            if(a == 0) return  0;
            return 1e9;
        }

        if(dp[i][a] != -1) return dp[i][a];

        int take = 1e9;

        if(coins[i] <= a){
            take = 1 + solve(i,n,coins,a-coins[i],dp);
        }

        int skip = solve(i+1,n,coins,a,dp);

        return dp[i][a] = min(take,skip);
    }
    int coinChange(vector<int>& coins, int amount) {
        
        sort(coins.rbegin(),coins.rend());
        int n = coins.size();

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

        if(amount == 0) return 0;

        int ans = solve(0,n,coins,amount,dp);

        return ans >= 1e9 ? -1 : ans;
    }
};
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