Coin Change II
Let's solve the Coin Change II problem using Dynamic Programming.
Statement
Suppose you are given a list of coins and a certain amount of money. Each coin in the list is unique and of a different 0
.
Note: You may assume that for each combination you make, you have an infinite number of each coin. In simpler terms, you can use a specific coin as many times as you want.
Let's say you have only two coins,
3 coins of
cents: . 1 coin of
cents and 1 coin of cents:
Constraints:
1 <=
coins.length
<= 3001 <=
coins[i]
<= 5000All the coins have a unique value.
0 <=
amount
<= 5000
Examples
Let's see a few more examples to get a better understanding of the problem statement:
No. | Coins | Amount | Number of ways |
1 | [1, 2, 5] | 7 | 6 |
2 | [1, 5, 10, 25] | 10 | 4 |
3 | [7] | 9 | 0 |
4 | [9,10,11] | 0 | 1 |
Try it yourself
Implement your solution in the following playground.
def count_ways(coins, amount):# replace this placeholder return statement with your codereturn -1
Note: If you clicked the “Submit” button and the code timed out, this means that your solution needs to be optimized in terms of running time.
Hint: Use dynamic programming and see the magic.
Solution
We will first explore the naive recursive solution to this problem and then see how it can be improved using the Unbounded Knapsack dynamic programming pattern.
Naive approach
A naive approach to solve this problem would be to make all the possible combinations of coins or generate all subsets of coin denominations that sum up to the required amount.
While making the combinations, a point to keep in mind is that we should try to avoid repeatedly counting the same combinations. For example,