<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) {
        vector<int> dp(amount + 1, amount + 1);
        dp[0] = 0;
        
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (coin <= i) {
                    dp[i] = min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        
        return dp[amount] > amount ? -1 : dp[amount];
    }
};
// LC rejects this but it has correct O time.
/*using HashPair = decltype([](const pair<int, int>& pair) {
    int first = hash<int>{}(pair.first);
    int second = hash<int>{}(pair.second);
    return first ^ (second >> 1);
});
class Solution {
private:
    unordered_map<pair<int, int>, int, HashPair> memo;
public:
    int dp(vector<int>& coins, int n, int amount) {
        if(amount == 0) {
            return 0;
        } else if((amount < 0) || (n < 0)) {
            return -1;
        } else if(memo.find({n, amount}) != memo.end()) {
            return memo[{n, amount}];
        }

        int poss = amount / coins[n];
        int fewest = -1;
        for(int i = 0; i <= poss; i++) {
            int next = dp(coins, n - 1, amount - i * coins[n]);
            if(next >= 0) {
                if(fewest < 0) {
                    fewest = i + next;
                } else {
                    fewest = min(fewest, i + next);
                }
            }
        }

        memo[{n, amount}] = fewest;
        return fewest;
    }

    int coinChange(vector<int>& coins, int amount) {
        return dp(coins, coins.size() - 1, amount);    
    }
};*/
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