leetcode - 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

Solution

Solved after reading others’ solutions.

This is a classical knapsack problem. We use dp[i][j] to denote the number of combinations that we could get, use first ith coins to form amount j, then the transformation equation would be:
$$dp[i][j]=\{\begin{array}{ll}dp[i-1][j]+dp[i][j-coins[i-1]],& \text{if}j-coins[i-1]\ge 0\\ dp[i-1][j],& \text{if}j-coins[i-1]<0\end{array}$$

Time complexity: $o(n\ast m)$, where n is the amount and m is the length of coins.
Space complexity: $o(n\ast m)$

Code

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