hello-algo/en/docs/chapter_dynamic_programming/unbounded_knapsack_problem.md

Unbounded Knapsack Problem

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.

Unbounded Knapsack Problem

!!! question

Given $n$ items, where the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with capacity $cap$. **Each item can be selected multiple times**. What is the maximum value that can be placed in the knapsack within the capacity limit? An example is shown in the figure below.

Dynamic Programming Approach

The unbounded knapsack problem is very similar to the 0-1 knapsack problem, differing only in that there is no limit on the number of times an item can be selected.

• In the 0-1 knapsack problem, there is only one of each type of item, so after placing item i in the knapsack, we can only choose from the first i-1 items.
• In the unbounded knapsack problem, the quantity of each type of item is unlimited, so after placing item i in the knapsack, we can still choose from the first i items.

Under the rules of the unbounded knapsack problem, the changes in state [i, c] are divided into two cases.

• Not putting item $i$: Same as the 0-1 knapsack problem, transfer to [i-1, c].
• Putting item $i$: Different from the 0-1 knapsack problem, transfer to [i, c-wgt[i-1]].

Thus, the state transition equation becomes:

dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])

Code Implementation

Comparing the code for the two problems, there is one change in state transition from i-1 to i, with everything else identical:

[file]{unbounded_knapsack}-[class]{}-[func]{unbounded_knapsack_dp}

Space Optimization

Since the current state is transferred from states on the left and above, after space optimization, each row in the dp table should be traversed in forward order.

This traversal order is exactly opposite to the 0-1 knapsack. Please refer to the figure below to understand the difference between the two.

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The code implementation is relatively simple, just delete the first dimension of the array dp:

[file]{unbounded_knapsack}-[class]{}-[func]{unbounded_knapsack_dp_comp}

Coin Change Problem

The knapsack problem represents a large class of dynamic programming problems and has many variants, such as the coin change problem.

!!! question

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**. 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 $-1$. An example is shown in the figure below.

Dynamic Programming Approach

The coin change problem can be viewed as a special case of the unbounded knapsack problem, with the following connections and differences.

• The two problems can be converted to each other: "item" corresponds to "coin", "item weight" corresponds to "coin denomination", and "knapsack capacity" corresponds to "target amount".
• The optimization goals are opposite: the unbounded knapsack problem aims to maximize item value, while the coin change problem aims to minimize the number of coins.
• The unbounded knapsack problem seeks solutions "not exceeding" the knapsack capacity, while the coin change problem seeks solutions that "exactly" make up the target amount.

Step 1: Think about the decisions in each round, define the state, and thus obtain the dp table

State [i, a] corresponds to the subproblem: the minimum number of coins among the first i types of coins that can make up amount $a$, denoted as dp[i, a].

The two-dimensional dp table has size (n+1) \times (amt+1).

Step 2: Identify the optimal substructure, and then derive the state transition equation

This problem differs from the unbounded knapsack problem in the following two aspects regarding the state transition equation.

• This problem seeks the minimum value, so the operator \max() needs to be changed to \min().
• The optimization target is the number of coins rather than item value, so when a coin is selected, simply execute +1.

dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)

Step 3: Determine 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 equal 0.

When there are no coins, it is impossible to make up any amount $> 0$, which is an invalid solution. To enable the \min() function in the state transition equation to identify and filter out invalid solutions, we 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, and can only use the maximum value of integer type int as a substitute. However, this can lead to large number overflow: the + 1 operation in the state transition equation may cause overflow.

For this reason, we use the number amt + 1 to represent invalid solutions, because the maximum number of coins needed to make up amt is at most amt. Before returning, check whether dp[n, amt] equals amt + 1; if so, return -1, indicating that the target amount cannot be made up. The code is as follows:

[file]{coin_change}-[class]{}-[func]{coin_change_dp}

The figure below shows the dynamic programming process for coin change, which is very similar to the unbounded knapsack problem.

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Space Optimization

The space optimization for the coin change problem is handled in the same way as the unbounded knapsack problem:

[file]{coin_change}-[class]{}-[func]{coin_change_dp_comp}

Coin Change Problem Ii

!!! question

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. **What is the number of coin combinations that can make up the target amount?** An example is shown in the figure below.

Dynamic Programming Approach

Compared to the previous problem, this problem's goal is to find the number of combinations, so the subproblem becomes: the number of combinations among the first i types of coins that can make up amount $a$. The dp table remains a two-dimensional matrix of size (n+1) \times (amt + 1).

The number of combinations for the current state equals 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 need to be selected 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 equal 0.

Code Implementation

[file]{coin_change_ii}-[class]{}-[func]{coin_change_ii_dp}

Space Optimization

The space optimization is handled in the same way, just delete the coin dimension:

[file]{coin_change_ii}-[class]{}-[func]{coin_change_ii_dp_comp}