Coin Change 2

Similar Problems:

- Coin Change
- CheatSheet: Leetcode For Code Interview
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- Tag: #knapsack, #coin

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

- 0 <= amount <= 5000
- 1 <= coin <= 5000
- the number of coins is less than 500
- the answer is guaranteed to fit into 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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/coin-change-2 class Solution(object): ## Basic Ideas: Dynamic programming ## dp(i) = {coin1: count1, coin2: count2, coin3: count3, ...} ## count_j is the combination with the maximum value as coin_j ## ## Complexity: Time O(k*n), Space O(n) def change(self, amount, coins): """ :type amount: int :type coins: List[int] :rtype: int """ if amount == 0: return 1 if len(coins) == 0: return 0 dp = [None]*(1+amount) for i in range(0, amount+1): dp[i] = collections.defaultdict(lambda: 0) for coin in coins: dp[0][coin] = 1 coins.sort() for i in range(1, amount+1): for coin in coins: if i<coin: break v = i-coin dp[i][coin] = dp[v][coin] dp[i][0] = 1 for j in range(1, len(coins)): dp[i][coins[j]] += dp[i][coins[j-1]] return dp[amount][coins[-1]]