518. Coin Change II

Description

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

 

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10]
Output: 1

 

Constraints:

  • 1 <= coins.length <= 300
  • 1 <= coins[i] <= 5000
  • All the values of coins are unique.
  • 0 <= amount <= 5000

Solutions

Solution 1

Python Code
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class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        for coin in coins:
            for j in range(coin, amount + 1):
                dp[j] += dp[j - coin]
        return dp[-1]

Java Code
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class Solution {
    public int change(int amount, int[] coins) {
        int m = coins.length;
        int[][] dp = new int[m + 1][amount + 1];
        dp[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= amount; ++j) {
                for (int k = 0; k * coins[i - 1] <= j; ++k) {
                    dp[i][j] += dp[i - 1][j - coins[i - 1] * k];
                }
            }
        }
        return dp[m][amount];
    }
}

C++ Code
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class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1);
        dp[0] = 1;
        for (auto coin : coins) {
            for (int j = coin; j <= amount; ++j) {
                dp[j] += dp[j - coin];
            }
        }
        return dp[amount];
    }
};

Go Code
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func change(amount int, coins []int) int {
	dp := make([]int, amount+1)
	dp[0] = 1
	for _, coin := range coins {
		for j := coin; j <= amount; j++ {
			dp[j] += dp[j-coin]
		}
	}
	return dp[amount]
}

TypeScript Code
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function change(amount: number, coins: number[]): number {
    let dp = new Array(amount + 1).fill(0);
    dp[0] = 1;
    for (let coin of coins) {
        for (let i = coin; i <= amount; ++i) {
            dp[i] += dp[i - coin];
        }
    }
    return dp.pop();
}

Solution 2

Java Code
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class Solution {
    public int change(int amount, int[] coins) {
        int m = coins.length;
        int[][] dp = new int[m + 1][amount + 1];
        dp[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            int v = coins[i - 1];
            for (int j = 0; j <= amount; ++j) {
                dp[i][j] = dp[i - 1][j];
                if (j >= v) {
                    dp[i][j] += dp[i][j - v];
                }
            }
        }
        return dp[m][amount];
    }
}

Solution 3

Java Code
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class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for (int coin : coins) {
            // 顺序遍历，0-1背包问题是倒序遍历
            for (int j = coin; j <= amount; j++) {
                dp[j] += dp[j - coin];
            }
        }
        return dp[amount];
    }
}