<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> &coins, int amount) {
    /* 1 2 5
     * 11 | 10 9 6 | 5 8 7 4 5 1 | 3 0 (done)
     *
     * amount = x^a + y^b + ...
     */
    if (!amount)
      return 0;

    vector<int> lefts{amount};
    bitset<10000> is_left;

    for (int coin_cnt = 0; !lefts.empty(); ++coin_cnt) {
      vector<int> next_lefts;
      for (auto coin : coins) {
        if (amount / coin > coin_cnt) {
          for (auto left : lefts) { // huge loop
            if (!left)
              return coin_cnt;
            else if (left == coin)
              return coin_cnt + 1;
            else if (left > coin && !is_left[left - coin]) {
              is_left[left - coin] = true;
              next_lefts.emplace_back(left - coin);
            }
          }
        }
      }

      lefts = std::move(next_lefts);
    }

    return -1;
  }
};
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