Coin Change

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Given an integer array coins representing coins of different denominations and integer amount representing a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of coins, return $-1$.

Example:

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Input: coins = [1, 2, 5], amount = 11;
Output: 3;
Input: coins = [2], amount = 3;
Output: -1;

Constraints:

  • $1 \le l_{coins} \le 12$
  • $1 \le coins_i \le 2^{31}-1$
  • $0 \le amount \le 10^4$

Solution

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class CoinChange{
public:
    int coinChange(vector<int>& coins, int amount) {
        int *dp = new int[amount + 1];
    	fill_n(dp, amount + 1, amount + 1);
	dp[0] = 0;
	for (int i = 1; i <= amount; i++) {
	    for (int j = 0; j < coins.size(); j++) {
	        if (coins[j] <= i)
		    dp[i] = dp[i - coins[j]] + 1 < dp[i] ? dp[i - coins[j]] + 1 : dp[i];
	    }
	}
	return dp[amount] > amount ? -1 : dp[amount];
    }
}

Explanation:

$dp_i$ represents the minimum number of coins that makes up the amount $i$. Assume that the values from $dp_0$ to $dp_{i-1}$ are already known, thus the transition equation of $dp_i$ can be deduced as
$$
dp_i = min_{j=0}^{l_{coins}-1}dp_{i-coins_j}+1
$$
which means the value of $dp_i$ is determined by $dp_{i-coins_j}$. If $dp_i$ can transfer from $dp_{i-coins_j}$ via $coins_j$, the number of coins that used should add $1$.