<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 inf = 1e4 + 5;
  int res(int x, int amount, vector<int>& coins, vector<vector<int>>& dp){
    // base case
    if(x == 0){
      if(amount%coins[x] == 0){
        return amount/coins[x];
      }
      else{
        return inf;
      }
    }
    // cache check
    if(dp[x][amount] != -1) return dp[x][amount];
    // transition
    int ans = res(x-1, amount, coins, dp);
    if(amount >= coins[x]){
      ans = min({ans, 1 + res(x, amount - coins[x], coins, dp), INT_MAX});
    }
    // save and return
    return dp[x][amount] = ans;
  }
  int coinChange(vector<int>& coins, int amount) {
    int n = coins.size();
    vector<vector<int>> dp(n, vector<int>(amount+1, -1));
    int ans = res(n-1,amount,coins,dp);
    return (ans == inf ? -1 : ans);
  }
};
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