<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 f(int idx,int amount,vector<int> &coins,vector<vector<int>> &dp){
        if(amount == 0) return 0;
        if(idx == 0){
            if(amount % coins[idx] == 0) return amount/coins[idx];
            return 1e9;
        }
        if(dp[idx][amount] != -1)   return dp[idx][amount];
        int pick1 = 1e9,pick2 = 1e9;
        if(amount >= coins[idx]){
            pick1 = 1+f(idx-1,amount-coins[idx],coins,dp);
            pick2 = 1+f(idx,amount-coins[idx],coins,dp);
        }
        int notPick = f(idx-1,amount,coins,dp);
        return dp[idx][amount] = min({pick1,pick2,notPick});
    }
    int coinChange(vector<int> &coins,int amount){
        int n = coins.size();
        vector<vector<int>> dp(n,vector<int>(amount+1,-1));
        return f(n-1,amount,coins,dp) == 1e9 ? -1 : f(n-1,amount,coins,dp);
    }
};
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