It's one of the most popular questions on leetcode that seems very easy at first. Coin change is a classic dynamic programming problem. I will proceed with an obvious (albeit wrong) solution and subsequently proceed to an efficient correct solution.
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1. You may assume that you have an infinite number of each kind of coin.
My first instinct is to sort the coins array. Pick the largest coin first and then subtract the largest possible value from the amount. Subsequently, proceed with smaller denominations while keeping track of the count.
However this solution fails the following test case as the minimum sequence is [8, 4]
coins = [1, 2, 4, 8, 9]
amount = 12
Output : 3
Expected: 2
We have to find some subproblem that will help in solving coin-change problem. Given an amount n, we want to generate an exact change using the fewest number of coins of denominations d_{1} < d_{2} < … < d_{m}.
Let Dp(n) represent the minimum number of coins required for a given amount n. Coins d_{j} can be added to amount n - d_{j} only if d_{j} <= n and 0 <= j <= n -1 wher Dp(0) is 0.
\begin{equation} Dp(n) = \min_{j=0}^{d_j<=n} Dp(n - D_j) + 1 \end{equation}
Let us proceed with following test case
coins = [1, 2, 5]
amount = 11
We can vary the amount as i from 0 to amount. The value of Dp[i] for each iteration becomes
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | |
---|---|---|---|---|---|---|---|---|---|---|---|---|
i = 0 | 0 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 1 | 0 | 1 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 2 | 0 | 1 | 1 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 3 | 0 | 1 | 1 | 2 | 12 | 12 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 4 | 0 | 1 | 1 | 2 | 2 | 12 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 5 | 0 | 1 | 1 | 2 | 2 | 1 | 12 | 12 | 12 | 12 | 12 | 12 |
i = 6 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 12 | 12 | 12 | 12 | 12 |
i = 7 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 2 | 12 | 12 | 12 | 12 |
i = 8 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 2 | 3 | 12 | 12 | 12 |
i = 9 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 2 | 3 | 3 | 12 | 12 |
i = 10 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 2 | 3 | 3 | 2 | 12 |
i = 11 | 0 | 1 | 1 | 2 | 2 | 1 | 2 | 2 | 3 | 3 | 2 | 3 |
Time Complexity is \(\theta(n * m)\), Space complexity is \(\theta(n)\)