In this section, we first solve another common knapsack problem: the unbounded knapsack, and then explore a special case of it: the coin change problem.
Given $n$ items, where the weight of the $i^{th}$ item is $wgt[i-1]$ and its value is $val[i-1]$, and a backpack with a capacity of $cap$. **Each item can be selected multiple times**. What is the maximum value of the items that can be put into the backpack without exceeding its capacity? See the example below.
The unbounded knapsack problem is very similar to the 0-1 knapsack problem, **the only difference being that there is no limit on the number of times an item can be chosen**.
- In the 0-1 knapsack problem, there is only one of each item, so after placing item $i$ into the backpack, you can only choose from the previous $i-1$ items.
- In the unbounded knapsack problem, the quantity of each item is unlimited, so after placing item $i$ in the backpack, **you can still choose from the previous $i$ items**.
Since the current state comes from the state to the left and above, **the space-optimized solution should perform a forward traversal for each row in the $dp$ table**.
This traversal order is the opposite of that for the 0-1 knapsack. Please refer to the following figures to understand the difference.
The knapsack problem is a representative of a large class of dynamic programming problems and has many variants, such as the coin change problem.
!!! question
Given $n$ types of coins, the denomination of the $i^{th}$ type of coin is $coins[i - 1]$, and the target amount is $amt$. **Each type of coin can be selected multiple times**. What is the minimum number of coins needed to make up the target amount? If it is impossible to make up the target amount, return $-1$. See the example below.
- The two problems can be converted into each other: "item" corresponds to "coin", "item weight" corresponds to "coin denomination", and "backpack capacity" corresponds to "target amount".
- The optimization goals are opposite: the unbounded knapsack problem aims to maximize the value of items, while the coin change problem aims to minimize the number of coins.
- The unbounded knapsack problem seeks solutions "not exceeding" the backpack capacity, while the coin change seeks solutions that "exactly" make up the target amount.
**First step: Think through each round's decision-making, define the state, and thus derive the $dp$ table**
The state $[i, a]$ corresponds to the sub-problem: **the minimum number of coins that can make up the amount $a$ using the first $i$ types of coins**, denoted as $dp[i, a]$.
The two-dimensional $dp$ table is of size $(n+1) \times (amt+1)$.
**Second step: Identify the optimal substructure and derive the state transition equation**
**Third step: Define boundary conditions and state transition order**
When the target amount is $0$, the minimum number of coins needed to make it up is $0$, so all $dp[i, 0]$ in the first column are $0$.
When there are no coins, **it is impossible to make up any amount >0**, which is an invalid solution. To allow the $\min()$ function in the state transition equation to recognize and filter out invalid solutions, consider using $+\infty$ to represent them, i.e., set all $dp[0, a]$ in the first row to $+\infty$.
### Code implementation
Most programming languages do not provide a $+\infty$ variable, only the maximum value of an integer `int` can be used as a substitute. This can lead to overflow: the $+1$ operation in the state transition equation may overflow.
For this reason, we use the number $amt + 1$ to represent an invalid solution, because the maximum number of coins needed to make up $amt$ is at most $amt$. Before returning the result, check if $dp[n, amt]$ equals $amt + 1$, and if so, return $-1$, indicating that the target amount cannot be made up. The code is as follows:
Given $n$ types of coins, where the denomination of the $i^{th}$ type of coin is $coins[i - 1]$, and the target amount is $amt$. **Each type of coin can be selected multiple times**, **ask how many combinations of coins can make up the target amount**. See the example below.
### Dynamic programming approach
Compared to the previous problem, the goal of this problem is to determine the number of combinations, so the sub-problem becomes: **the number of combinations that can make up amount $a$ using the first $i$ types of coins**. The $dp$ table remains a two-dimensional matrix of size $(n+1) \times (amt + 1)$.
The number of combinations for the current state is the sum of the combinations from not selecting the current coin and selecting the current coin. The state transition equation is:
$$
dp[i, a] = dp[i-1, a] + dp[i, a - coins[i-1]]
$$
When the target amount is $0$, no coins are needed to make up the target amount, so all $dp[i, 0]$ in the first column should be initialized to $1$. When there are no coins, it is impossible to make up any amount >0, so all $dp[0, a]$ in the first row should be set to $0$.
