leetcode_gym/_problems/coinChange2.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Coin Change 2	2020-06-07	dp	https://leetcode.com/problems/coin-change-2	Not Solved

Problem

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.

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

Solution

This video from backtobackswe explains why you subtract the coin.

        0	1	2	3	4	5
[1]     1	1	1	1	^1^	1
[1,2]	1	1	`2`	2	*3*	3
[1,2,5]	1	1	2	2	3	4

*3* = ^1^ + `2`

Think of "ways to make $4 with [1,2]" is equal to:

It's equal to (ways to make $4 with [1] coin) + (ways to make $2 with [1,2] coins)

(1+1+1+1) + ( [1,1] + [2] ) "+2 if we use the $2 coin"

The reason it's "ways to make $2 with [1,2] coins" is because if we USE THE 2 coin, then it's $4-$2 coin, and we solve subproblem of ways to make $2 with [1,2] coins

{% highlight python %}

def change(self, amount: int, coins: List[int]) -> int:

    dp = [ [1]+[0]*amount for _ in range(len(coins)+1)]
    
    for c in range(1, len(coins)+1):
        for a in range(1, amount+1):
            if coins[c-1] <= a:
                dp[c][a] = dp[c-1][a] + dp[c][a-coins[c-1]]
            else:
                dp[c][a] = dp[c-1][a]
    
    print(dp)
    return dp[len(coins)][amount]  # or dp[-1][-1]

{% endhighlight %}

---

This code is simplier, try to analyze this one instead

{% highlight python %}

def change(self, amount: int, coins: List[int]) -> int:
    dp = [0] * (amount + 1)             # DP Array of size (Amount +1) initialized to 0
    dp[0] = 1

    for currCoin in coins:
        for currAmount in range(currCoin, amount + 1):
            dp[currAmount] += dp[currAmount - currCoin]

    return dp[-1]

{% endhighlight %}