README_EN.md

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true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0322.Coin%20Change/README_EN.md	Breadth-First Search Array Dynamic Programming

322. Coin Change

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 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 the coins, return -1.

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

 

Example 1:

Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

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

Example 3:

Input: coins = [1], amount = 0
Output: 0

 

Constraints:

• 1 <= coins.length <= 12
• 1 <= coins[i] <= 231 - 1
• 0 <= amount <= 104

Solutions

Solution 1: Dynamic Programming (Complete Knapsack)

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

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

$$ f[i][j] = \min(f[i - 1][j], f[i - 1][j - x] + 1, \cdots, f[i - 1][j - k \times x] + k) $$

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

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

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

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

$$ f[i][j] = \min(f[i - 1][j], f[i][j - x] + 1) $$

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 coinChange(self, coins: List[int], amount: int) -> int:
        m, n = len(coins), amount
        f = [[inf] * (n + 1) for _ in range(m + 1)]
        f[0][0] = 0
        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] = min(f[i][j], f[i][j - x] + 1)
        return -1 if f[m][n] >= inf else f[m][n]

Java

class Solution {
    public int coinChange(int[] coins, int amount) {
        final int inf = 1 << 30;
        int m = coins.length;
        int n = amount;
        int[][] f = new int[m + 1][n + 1];
        for (var g : f) {
            Arrays.fill(g, inf);
        }
        f[0][0] = 0;
        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] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
                }
            }
        }
        return f[m][n] >= inf ? -1 : f[m][n];
    }
}

C++

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int m = coins.size(), n = amount;
        int f[m + 1][n + 1];
        memset(f, 0x3f, sizeof(f));
        f[0][0] = 0;
        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] = min(f[i][j], f[i][j - coins[i - 1]] + 1);
                }
            }
        }
        return f[m][n] > n ? -1 : f[m][n];
    }
};

Go

func coinChange(coins []int, amount int) int {
	m, n := len(coins), amount
	f := make([][]int, m+1)
	const inf = 1 << 30
	for i := range f {
		f[i] = make([]int, n+1)
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0
	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] = min(f[i][j], f[i][j-coins[i-1]]+1)
			}
		}
	}
	if f[m][n] > n {
		return -1
	}
	return f[m][n]
}

TypeScript

function coinChange(coins: number[], amount: number): number {
    const m = coins.length;
    const n = amount;
    const f: number[][] = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(1 << 30));
    f[0][0] = 0;
    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] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
            }
        }
    }
    return f[m][n] > n ? -1 : f[m][n];
}

Rust

impl Solution {
    pub fn coin_change(coins: Vec<i32>, amount: i32) -> i32 {
        let n = amount as usize;
        let mut f = vec![n + 1; n + 1];
        f[0] = 0;
        for &x in &coins {
            for j in x as usize..=n {
                f[j] = f[j].min(f[j - (x as usize)] + 1);
            }
        }
        if f[n] > n {
            -1
        } else {
            f[n] as i32
        }
    }
}

JavaScript

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function (coins, amount) {
    const m = coins.length;
    const n = amount;
    const f = Array(m + 1)
        .fill(0)
        .map(() => Array(n + 1).fill(1 << 30));
    f[0][0] = 0;
    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] = Math.min(f[i][j], f[i][j - coins[i - 1]] + 1);
            }
        }
    }
    return f[m][n] > n ? -1 : 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 $O(n)$.

Similar problems:

• 279. Perfect Squares

Python3

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

Java

class Solution {
    public int coinChange(int[] coins, int amount) {
        final int inf = 1 << 30;
        int n = amount;
        int[] f = new int[n + 1];
        Arrays.fill(f, inf);
        f[0] = 0;
        for (int x : coins) {
            for (int j = x; j <= n; ++j) {
                f[j] = Math.min(f[j], f[j - x] + 1);
            }
        }
        return f[n] >= inf ? -1 : f[n];
    }
}

C++

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = amount;
        int f[n + 1];
        memset(f, 0x3f, sizeof(f));
        f[0] = 0;
        for (int x : coins) {
            for (int j = x; j <= n; ++j) {
                f[j] = min(f[j], f[j - x] + 1);
            }
        }
        return f[n] > n ? -1 : f[n];
    }
};

Go

func coinChange(coins []int, amount int) int {
	n := amount
	f := make([]int, n+1)
	for i := range f {
		f[i] = 1 << 30
	}
	f[0] = 0
	for _, x := range coins {
		for j := x; j <= n; j++ {
			f[j] = min(f[j], f[j-x]+1)
		}
	}
	if f[n] > n {
		return -1
	}
	return f[n]
}

TypeScript

function coinChange(coins: number[], amount: number): number {
    const n = amount;
    const f: number[] = Array(n + 1).fill(1 << 30);
    f[0] = 0;
    for (const x of coins) {
        for (let j = x; j <= n; ++j) {
            f[j] = Math.min(f[j], f[j - x] + 1);
        }
    }
    return f[n] > n ? -1 : f[n];
}

JavaScript

/**
 * @param {number[]} coins
 * @param {number} amount
 * @return {number}
 */
var coinChange = function (coins, amount) {
    const n = amount;
    const f = Array(n + 1).fill(1 << 30);
    f[0] = 0;
    for (const x of coins) {
        for (let j = x; j <= n; ++j) {
            f[j] = Math.min(f[j], f[j - x] + 1);
        }
    }
    return f[n] > n ? -1 : f[n];
};