322. Coin Change

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.

Python3JavaC++GoTypeScriptRustJavaScript

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]

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];
    }
}

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];
    }
};

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]
}

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];
}

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
        }
    }
}

/**
 * @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

Python3JavaC++GoTypeScriptJavaScript

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]

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];
    }
}

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];
    }
};

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]
}

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];
}

/**
 * @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];
};

322. Coin Change

Description

Solutions

Solution 1: Dynamic Programming (Complete Knapsack)

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