LC518 - Coin Change II

Problem

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

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

Solution

DFS

The naive approach uses a DFS to go through the entire decision tree to evaluate possible paths of coin selections. To prevent duplicates, i.e. 2+2+1 and 2+1+2, paths will only evaluate coins with greater or equal value to the first selection (first node). The time complexity of this approach is $O(m^n)$, where $m$ is the number of coins and $n$ is the total amount.

Cached DFS

Caching (in theory) reduces the time complexity to $O(m\cdot n)$, as repeated evaluations are returned from cache. The cache uses $O(m\cdot n)$ space. However, it seems that it does not perform optimally in this implementation upon evaluation.

def change(self, amount: int, coins: List[int]) -> int:
	@cache
	def dfs(i, target):
		if target == amount:
			return 1
		if target > amount or i == len(coins):
			return 0
		return dfs(i, target + coins[i]) + dfs(i+1, target)
	return dfs(0,0)
		

2D Dynamic Programming

The advantage of using DP is that the space complexity goes down to $O(n)$. For

def change(self, amount: int, coins: List[int]) -> int:
	arr = [0] * (amount + 1)
	arr[0] = 1
	for i in range(len(coins), 0, -1):
		next = [0] * (amount+1)
		next[0] = 1
		for target in range(1, amount+1):
			next[target] = arr[target]
			if target - coins[i-1] >= 0:
				next[target] += next[target-coins[i-1]]
		arr = next
	return arr[amount]