README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0518.Coin%20Change%20II/README_EN.md	Array Dynamic Programming

518. Coin Change II

中文文档

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: Dynamic Programming (Complete Knapsack)

We define $f[i][j]$ as the number of coin combinations to make up the amount $j$ using the first $i$ types of coins. Initially, $f[0][0] = 1$, and the values of other positions are all $0$.

We can enumerate the quantity $k$ of the last coin used, then we have equation one:

$$ f[i][j] = f[i - 1][j] + f[i - 1][j - x] + f[i - 1][j - 2 \times x] + \cdots + f[i - 1][j - k \times x] $$

where $x$ represents the face value of the $i$-th type of coin.

Let $j = j - x$, then we have equation two:

$$ f[i][j - x] = f[i - 1][j - x] + f[i - 1][j - 2 \times x] + \cdots + f[i - 1][j - k \times x] $$

Substituting equation two into equation one, we can get the following state transition equation:

$$ f[i][j] = f[i - 1][j] + f[i][j - x] $$

The final answer is $f[m][n]$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of types of coins and the total amount, respectively.

Python3

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        m, n = len(coins), amount
        f = [[0] * (n + 1) for _ in range(m + 1)]
        f[0][0] = 1
        for i, x in enumerate(coins, 1):
            for j in range(n + 1):
                f[i][j] = f[i - 1][j]
                if j >= x:
                    f[i][j] += f[i][j - x]
        return f[m][n]

Java

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

C++

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        int m = coins.size(), n = amount;
        unsigned f[m + 1][n + 1];
        memset(f, 0, sizeof(f));
        f[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= coins[i - 1]) {
                    f[i][j] += f[i][j - coins[i - 1]];
                }
            }
        }
        return f[m][n];
    }
};

Go

func change(amount int, coins []int) int {
	m, n := len(coins), amount
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	f[0][0] = 1
	for i := 1; i <= m; i++ {
		for j := 0; j <= n; j++ {
			f[i][j] = f[i-1][j]
			if j >= coins[i-1] {
				f[i][j] += f[i][j-coins[i-1]]
			}
		}
	}
	return f[m][n]
}

TypeScript

function change(amount: number, coins: number[]): number {
    const [m, n] = [coins.length, amount];
    const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i <= m; ++i) {
        for (let j = 0; j <= n; ++j) {
            f[i][j] = f[i - 1][j];
            if (j >= coins[i - 1]) {
                f[i][j] += f[i][j - coins[i - 1]];
            }
        }
    }
    return f[m][n];
}

We notice that $f[i][j]$ is only related to $f[i - 1][j]$ and $f[i][j - x]$. Therefore, we can optimize the two-dimensional array into a one-dimensional array, reducing the space complexity to $O(n)$.

Python3

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        n = amount
        f = [1] + [0] * n
        for x in coins:
            for j in range(x, n + 1):
                f[j] += f[j - x]
        return f[n]

Java

class Solution {
    public int change(int amount, int[] coins) {
        int n = amount;
        int[] f = new int[n + 1];
        f[0] = 1;
        for (int x : coins) {
            for (int j = x; j <= n; ++j) {
                f[j] += f[j - x];
            }
        }
        return f[n];
    }
}

C++

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        int n = amount;
        unsigned f[n + 1];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int x : coins) {
            for (int j = x; j <= n; ++j) {
                f[j] += f[j - x];
            }
        }
        return f[n];
    }
};

Go

func change(amount int, coins []int) int {
	n := amount
	f := make([]int, n+1)
	f[0] = 1
	for _, x := range coins {
		for j := x; j <= n; j++ {
			f[j] += f[j-x]
		}
	}
	return f[n]
}

TypeScript

function change(amount: number, coins: number[]): number {
    const n = amount;
    const f: number[] = Array(n + 1).fill(0);
    f[0] = 1;
    for (const x of coins) {
        for (let j = x; j <= n; ++j) {
            f[j] += f[j - x];
        }
    }
    return f[n];
}