---
comments: true
---
# 14.5 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.
## 14.5.1 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 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.
{ class="animation-figure" }
Figure 14-22 Example data for the unbounded knapsack problem
### 1. Dynamic programming approach
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**.
Under the rules of the unbounded knapsack problem, the state $[i, c]$ can change in two ways.
- **Not putting item $i$ in**: As with the 0-1 knapsack problem, transition to $[i-1, c]$.
- **Putting item $i$ in**: Unlike the 0-1 knapsack problem, transition to $[i, c-wgt[i-1]]$.
The state transition equation thus becomes:
$$
dp[i, c] = \max(dp[i-1, c], dp[i, c - wgt[i-1]] + val[i-1])
$$
### 2. Code implementation
Comparing the code for the two problems, the state transition changes from $i-1$ to $i$, the rest is completely identical:
=== "Python"
```python title="unbounded_knapsack.py"
def unbounded_knapsack_dp(wgt: list[int], val: list[int], cap: int) -> int:
"""Complete knapsack: Dynamic programming"""
n = len(wgt)
# Initialize dp table
dp = [[0] * (cap + 1) for _ in range(n + 1)]
# State transition
for i in range(1, n + 1):
for c in range(1, cap + 1):
if wgt[i - 1] > c:
# If exceeding the knapsack capacity, do not choose item i
dp[i][c] = dp[i - 1][c]
else:
# The greater value between not choosing and choosing item i
dp[i][c] = max(dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1])
return dp[n][cap]
```
=== "C++"
```cpp title="unbounded_knapsack.cpp"
/* Complete knapsack: Dynamic programming */
int unboundedKnapsackDP(vector &wgt, vector &val, int cap) {
int n = wgt.size();
// Initialize dp table
vector> dp(n + 1, vector(cap + 1, 0));
// State transition
for (int i = 1; i <= n; i++) {
for (int c = 1; c <= cap; c++) {
if (wgt[i - 1] > c) {
// If exceeding the knapsack capacity, do not choose item i
dp[i][c] = dp[i - 1][c];
} else {
// The greater value between not choosing and choosing item i
dp[i][c] = max(dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1]);
}
}
}
return dp[n][cap];
}
```
=== "Java"
```java title="unbounded_knapsack.java"
/* Complete knapsack: Dynamic programming */
int unboundedKnapsackDP(int[] wgt, int[] val, int cap) {
int n = wgt.length;
// Initialize dp table
int[][] dp = new int[n + 1][cap + 1];
// State transition
for (int i = 1; i <= n; i++) {
for (int c = 1; c <= cap; c++) {
if (wgt[i - 1] > c) {
// If exceeding the knapsack capacity, do not choose item i
dp[i][c] = dp[i - 1][c];
} else {
// The greater value between not choosing and choosing item i
dp[i][c] = Math.max(dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1]);
}
}
}
return dp[n][cap];
}
```
=== "C#"
```csharp title="unbounded_knapsack.cs"
[class]{unbounded_knapsack}-[func]{UnboundedKnapsackDP}
```
=== "Go"
```go title="unbounded_knapsack.go"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "Swift"
```swift title="unbounded_knapsack.swift"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "JS"
```javascript title="unbounded_knapsack.js"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "TS"
```typescript title="unbounded_knapsack.ts"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "Dart"
```dart title="unbounded_knapsack.dart"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "Rust"
```rust title="unbounded_knapsack.rs"
[class]{}-[func]{unbounded_knapsack_dp}
```
=== "C"
```c title="unbounded_knapsack.c"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "Kotlin"
```kotlin title="unbounded_knapsack.kt"
[class]{}-[func]{unboundedKnapsackDP}
```
=== "Ruby"
```ruby title="unbounded_knapsack.rb"
[class]{}-[func]{unbounded_knapsack_dp}
```
=== "Zig"
```zig title="unbounded_knapsack.zig"
[class]{}-[func]{unboundedKnapsackDP}
```
### 3. Space optimization
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 Figure 14-23 to understand the difference.
=== "<1>"
{ class="animation-figure" }
=== "<2>"
{ class="animation-figure" }
=== "<3>"
{ class="animation-figure" }
=== "<4>"
{ class="animation-figure" }
=== "<5>"
{ class="animation-figure" }
=== "<6>"
{ class="animation-figure" }
Figure 14-23 Dynamic programming process for the unbounded knapsack problem after space optimization
The code implementation is quite simple, just remove the first dimension of the array `dp`:
=== "Python"
```python title="unbounded_knapsack.py"
def unbounded_knapsack_dp_comp(wgt: list[int], val: list[int], cap: int) -> int:
"""Complete knapsack: Space-optimized dynamic programming"""
n = len(wgt)
# Initialize dp table
dp = [0] * (cap + 1)
# State transition
for i in range(1, n + 1):
# Traverse in order
for c in range(1, cap + 1):
if wgt[i - 1] > c:
# If exceeding the knapsack capacity, do not choose item i
dp[c] = dp[c]
else:
# The greater value between not choosing and choosing item i
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])
return dp[cap]
```
=== "C++"
```cpp title="unbounded_knapsack.cpp"
/* Complete knapsack: Space-optimized dynamic programming */
int unboundedKnapsackDPComp(vector &wgt, vector &val, int cap) {
int n = wgt.size();
// Initialize dp table
vector dp(cap + 1, 0);
// State transition
for (int i = 1; i <= n; i++) {
for (int c = 1; c <= cap; c++) {
if (wgt[i - 1] > c) {
// If exceeding the knapsack capacity, do not choose item i
dp[c] = dp[c];
} else {
// The greater value between not choosing and choosing item i
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1]);
}
}
}
return dp[cap];
}
```
=== "Java"
```java title="unbounded_knapsack.java"
/* Complete knapsack: Space-optimized dynamic programming */
int unboundedKnapsackDPComp(int[] wgt, int[] val, int cap) {
int n = wgt.length;
// Initialize dp table
int[] dp = new int[cap + 1];
// State transition
for (int i = 1; i <= n; i++) {
for (int c = 1; c <= cap; c++) {
if (wgt[i - 1] > c) {
// If exceeding the knapsack capacity, do not choose item i
dp[c] = dp[c];
} else {
// The greater value between not choosing and choosing item i
dp[c] = Math.max(dp[c], dp[c - wgt[i - 1]] + val[i - 1]);
}
}
}
return dp[cap];
}
```
=== "C#"
```csharp title="unbounded_knapsack.cs"
[class]{unbounded_knapsack}-[func]{UnboundedKnapsackDPComp}
```
=== "Go"
```go title="unbounded_knapsack.go"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "Swift"
```swift title="unbounded_knapsack.swift"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "JS"
```javascript title="unbounded_knapsack.js"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "TS"
```typescript title="unbounded_knapsack.ts"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "Dart"
```dart title="unbounded_knapsack.dart"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "Rust"
```rust title="unbounded_knapsack.rs"
[class]{}-[func]{unbounded_knapsack_dp_comp}
```
=== "C"
```c title="unbounded_knapsack.c"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "Kotlin"
```kotlin title="unbounded_knapsack.kt"
[class]{}-[func]{unboundedKnapsackDPComp}
```
=== "Ruby"
```ruby title="unbounded_knapsack.rb"
[class]{}-[func]{unbounded_knapsack_dp_comp}
```
=== "Zig"
```zig title="unbounded_knapsack.zig"
[class]{}-[func]{unboundedKnapsackDPComp}
```
## 14.5.2 Coin change problem
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.
{ class="animation-figure" }
Figure 14-24 Example data for the coin change problem
### 1. Dynamic programming approach
**The coin change can be seen as a special case of the unbounded knapsack problem**, sharing the following similarities and differences.
- 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**
This problem differs from the unbounded knapsack problem in two aspects of the state transition equation.
- This problem seeks the minimum, so the operator $\max()$ needs to be changed to $\min()$.
- The optimization is focused on the number of coins, so simply add $+1$ when a coin is chosen.
$$
dp[i, a] = \min(dp[i-1, a], dp[i, a - coins[i-1]] + 1)
$$
**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 $+\infty$.
### 2. 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:
=== "Python"
```python title="coin_change.py"
def coin_change_dp(coins: list[int], amt: int) -> int:
"""Coin change: Dynamic programming"""
n = len(coins)
MAX = amt + 1
# Initialize dp table
dp = [[0] * (amt + 1) for _ in range(n + 1)]
# State transition: first row and first column
for a in range(1, amt + 1):
dp[0][a] = MAX
# State transition: the rest of the rows and columns
for i in range(1, n + 1):
for a in range(1, amt + 1):
if coins[i - 1] > a:
# If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a]
else:
# The smaller value between not choosing and choosing coin i
dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1)
return dp[n][amt] if dp[n][amt] != MAX else -1
```
=== "C++"
```cpp title="coin_change.cpp"
/* Coin change: Dynamic programming */
int coinChangeDP(vector &coins, int amt) {
int n = coins.size();
int MAX = amt + 1;
// Initialize dp table
vector> dp(n + 1, vector(amt + 1, 0));
// State transition: first row and first column
for (int a = 1; a <= amt; a++) {
dp[0][a] = MAX;
}
// State transition: the rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a];
} else {
// The smaller value between not choosing and choosing coin i
dp[i][a] = min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1);
}
}
}
return dp[n][amt] != MAX ? dp[n][amt] : -1;
}
```
=== "Java"
```java title="coin_change.java"
/* Coin change: Dynamic programming */
int coinChangeDP(int[] coins, int amt) {
int n = coins.length;
int MAX = amt + 1;
// Initialize dp table
int[][] dp = new int[n + 1][amt + 1];
// State transition: first row and first column
for (int a = 1; a <= amt; a++) {
dp[0][a] = MAX;
}
// State transition: the rest of the rows and columns
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a];
} else {
// The smaller value between not choosing and choosing coin i
dp[i][a] = Math.min(dp[i - 1][a], dp[i][a - coins[i - 1]] + 1);
}
}
}
return dp[n][amt] != MAX ? dp[n][amt] : -1;
}
```
=== "C#"
```csharp title="coin_change.cs"
[class]{coin_change}-[func]{CoinChangeDP}
```
=== "Go"
```go title="coin_change.go"
[class]{}-[func]{coinChangeDP}
```
=== "Swift"
```swift title="coin_change.swift"
[class]{}-[func]{coinChangeDP}
```
=== "JS"
```javascript title="coin_change.js"
[class]{}-[func]{coinChangeDP}
```
=== "TS"
```typescript title="coin_change.ts"
[class]{}-[func]{coinChangeDP}
```
=== "Dart"
```dart title="coin_change.dart"
[class]{}-[func]{coinChangeDP}
```
=== "Rust"
```rust title="coin_change.rs"
[class]{}-[func]{coin_change_dp}
```
=== "C"
```c title="coin_change.c"
[class]{}-[func]{coinChangeDP}
```
=== "Kotlin"
```kotlin title="coin_change.kt"
[class]{}-[func]{coinChangeDP}
```
=== "Ruby"
```ruby title="coin_change.rb"
[class]{}-[func]{coin_change_dp}
```
=== "Zig"
```zig title="coin_change.zig"
[class]{}-[func]{coinChangeDP}
```
Figure 14-25 show the dynamic programming process for the coin change problem, which is very similar to the unbounded knapsack problem.
=== "<1>"
{ class="animation-figure" }
=== "<2>"
{ class="animation-figure" }
=== "<3>"
{ class="animation-figure" }
=== "<4>"
{ class="animation-figure" }
=== "<5>"
{ class="animation-figure" }
=== "<6>"
{ class="animation-figure" }
=== "<7>"
{ class="animation-figure" }
=== "<8>"
{ class="animation-figure" }
=== "<9>"
{ class="animation-figure" }
=== "<10>"
{ class="animation-figure" }
=== "<11>"
{ class="animation-figure" }
=== "<12>"
{ class="animation-figure" }
=== "<13>"
{ class="animation-figure" }
=== "<14>"
{ class="animation-figure" }
=== "<15>"
{ class="animation-figure" }
Figure 14-25 Dynamic programming process for the coin change problem
### 3. Space optimization
The space optimization for the coin change problem is handled in the same way as for the unbounded knapsack problem:
=== "Python"
```python title="coin_change.py"
def coin_change_dp_comp(coins: list[int], amt: int) -> int:
"""Coin change: Space-optimized dynamic programming"""
n = len(coins)
MAX = amt + 1
# Initialize dp table
dp = [MAX] * (amt + 1)
dp[0] = 0
# State transition
for i in range(1, n + 1):
# Traverse in order
for a in range(1, amt + 1):
if coins[i - 1] > a:
# If exceeding the target amount, do not choose coin i
dp[a] = dp[a]
else:
# The smaller value between not choosing and choosing coin i
dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1)
return dp[amt] if dp[amt] != MAX else -1
```
=== "C++"
```cpp title="coin_change.cpp"
/* Coin change: Space-optimized dynamic programming */
int coinChangeDPComp(vector &coins, int amt) {
int n = coins.size();
int MAX = amt + 1;
// Initialize dp table
vector dp(amt + 1, MAX);
dp[0] = 0;
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[a] = dp[a];
} else {
// The smaller value between not choosing and choosing coin i
dp[a] = min(dp[a], dp[a - coins[i - 1]] + 1);
}
}
}
return dp[amt] != MAX ? dp[amt] : -1;
}
```
=== "Java"
```java title="coin_change.java"
/* Coin change: Space-optimized dynamic programming */
int coinChangeDPComp(int[] coins, int amt) {
int n = coins.length;
int MAX = amt + 1;
// Initialize dp table
int[] dp = new int[amt + 1];
Arrays.fill(dp, MAX);
dp[0] = 0;
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[a] = dp[a];
} else {
// The smaller value between not choosing and choosing coin i
dp[a] = Math.min(dp[a], dp[a - coins[i - 1]] + 1);
}
}
}
return dp[amt] != MAX ? dp[amt] : -1;
}
```
=== "C#"
```csharp title="coin_change.cs"
[class]{coin_change}-[func]{CoinChangeDPComp}
```
=== "Go"
```go title="coin_change.go"
[class]{}-[func]{coinChangeDPComp}
```
=== "Swift"
```swift title="coin_change.swift"
[class]{}-[func]{coinChangeDPComp}
```
=== "JS"
```javascript title="coin_change.js"
[class]{}-[func]{coinChangeDPComp}
```
=== "TS"
```typescript title="coin_change.ts"
[class]{}-[func]{coinChangeDPComp}
```
=== "Dart"
```dart title="coin_change.dart"
[class]{}-[func]{coinChangeDPComp}
```
=== "Rust"
```rust title="coin_change.rs"
[class]{}-[func]{coin_change_dp_comp}
```
=== "C"
```c title="coin_change.c"
[class]{}-[func]{coinChangeDPComp}
```
=== "Kotlin"
```kotlin title="coin_change.kt"
[class]{}-[func]{coinChangeDPComp}
```
=== "Ruby"
```ruby title="coin_change.rb"
[class]{}-[func]{coin_change_dp_comp}
```
=== "Zig"
```zig title="coin_change.zig"
[class]{}-[func]{coinChangeDPComp}
```
## 14.5.3 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, **ask how many combinations of coins can make up the target amount**. See the example below.
{ class="animation-figure" }
Figure 14-26 Example data for Coin Change Problem II
### 1. 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 $0$.
### 2. Code implementation
=== "Python"
```python title="coin_change_ii.py"
def coin_change_ii_dp(coins: list[int], amt: int) -> int:
"""Coin change II: Dynamic programming"""
n = len(coins)
# Initialize dp table
dp = [[0] * (amt + 1) for _ in range(n + 1)]
# Initialize first column
for i in range(n + 1):
dp[i][0] = 1
# State transition
for i in range(1, n + 1):
for a in range(1, amt + 1):
if coins[i - 1] > a:
# If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a]
else:
# The sum of the two options of not choosing and choosing coin i
dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]]
return dp[n][amt]
```
=== "C++"
```cpp title="coin_change_ii.cpp"
/* Coin change II: Dynamic programming */
int coinChangeIIDP(vector &coins, int amt) {
int n = coins.size();
// Initialize dp table
vector> dp(n + 1, vector(amt + 1, 0));
// Initialize first column
for (int i = 0; i <= n; i++) {
dp[i][0] = 1;
}
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a];
} else {
// The sum of the two options of not choosing and choosing coin i
dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]];
}
}
}
return dp[n][amt];
}
```
=== "Java"
```java title="coin_change_ii.java"
/* Coin change II: Dynamic programming */
int coinChangeIIDP(int[] coins, int amt) {
int n = coins.length;
// Initialize dp table
int[][] dp = new int[n + 1][amt + 1];
// Initialize first column
for (int i = 0; i <= n; i++) {
dp[i][0] = 1;
}
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[i][a] = dp[i - 1][a];
} else {
// The sum of the two options of not choosing and choosing coin i
dp[i][a] = dp[i - 1][a] + dp[i][a - coins[i - 1]];
}
}
}
return dp[n][amt];
}
```
=== "C#"
```csharp title="coin_change_ii.cs"
[class]{coin_change_ii}-[func]{CoinChangeIIDP}
```
=== "Go"
```go title="coin_change_ii.go"
[class]{}-[func]{coinChangeIIDP}
```
=== "Swift"
```swift title="coin_change_ii.swift"
[class]{}-[func]{coinChangeIIDP}
```
=== "JS"
```javascript title="coin_change_ii.js"
[class]{}-[func]{coinChangeIIDP}
```
=== "TS"
```typescript title="coin_change_ii.ts"
[class]{}-[func]{coinChangeIIDP}
```
=== "Dart"
```dart title="coin_change_ii.dart"
[class]{}-[func]{coinChangeIIDP}
```
=== "Rust"
```rust title="coin_change_ii.rs"
[class]{}-[func]{coin_change_ii_dp}
```
=== "C"
```c title="coin_change_ii.c"
[class]{}-[func]{coinChangeIIDP}
```
=== "Kotlin"
```kotlin title="coin_change_ii.kt"
[class]{}-[func]{coinChangeIIDP}
```
=== "Ruby"
```ruby title="coin_change_ii.rb"
[class]{}-[func]{coin_change_ii_dp}
```
=== "Zig"
```zig title="coin_change_ii.zig"
[class]{}-[func]{coinChangeIIDP}
```
### 3. Space optimization
The space optimization approach is the same, just remove the coin dimension:
=== "Python"
```python title="coin_change_ii.py"
def coin_change_ii_dp_comp(coins: list[int], amt: int) -> int:
"""Coin change II: Space-optimized dynamic programming"""
n = len(coins)
# Initialize dp table
dp = [0] * (amt + 1)
dp[0] = 1
# State transition
for i in range(1, n + 1):
# Traverse in order
for a in range(1, amt + 1):
if coins[i - 1] > a:
# If exceeding the target amount, do not choose coin i
dp[a] = dp[a]
else:
# The sum of the two options of not choosing and choosing coin i
dp[a] = dp[a] + dp[a - coins[i - 1]]
return dp[amt]
```
=== "C++"
```cpp title="coin_change_ii.cpp"
/* Coin change II: Space-optimized dynamic programming */
int coinChangeIIDPComp(vector &coins, int amt) {
int n = coins.size();
// Initialize dp table
vector dp(amt + 1, 0);
dp[0] = 1;
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[a] = dp[a];
} else {
// The sum of the two options of not choosing and choosing coin i
dp[a] = dp[a] + dp[a - coins[i - 1]];
}
}
}
return dp[amt];
}
```
=== "Java"
```java title="coin_change_ii.java"
/* Coin change II: Space-optimized dynamic programming */
int coinChangeIIDPComp(int[] coins, int amt) {
int n = coins.length;
// Initialize dp table
int[] dp = new int[amt + 1];
dp[0] = 1;
// State transition
for (int i = 1; i <= n; i++) {
for (int a = 1; a <= amt; a++) {
if (coins[i - 1] > a) {
// If exceeding the target amount, do not choose coin i
dp[a] = dp[a];
} else {
// The sum of the two options of not choosing and choosing coin i
dp[a] = dp[a] + dp[a - coins[i - 1]];
}
}
}
return dp[amt];
}
```
=== "C#"
```csharp title="coin_change_ii.cs"
[class]{coin_change_ii}-[func]{CoinChangeIIDPComp}
```
=== "Go"
```go title="coin_change_ii.go"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "Swift"
```swift title="coin_change_ii.swift"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "JS"
```javascript title="coin_change_ii.js"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "TS"
```typescript title="coin_change_ii.ts"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "Dart"
```dart title="coin_change_ii.dart"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "Rust"
```rust title="coin_change_ii.rs"
[class]{}-[func]{coin_change_ii_dp_comp}
```
=== "C"
```c title="coin_change_ii.c"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "Kotlin"
```kotlin title="coin_change_ii.kt"
[class]{}-[func]{coinChangeIIDPComp}
```
=== "Ruby"
```ruby title="coin_change_ii.rb"
[class]{}-[func]{coin_change_ii_dp_comp}
```
=== "Zig"
```zig title="coin_change_ii.zig"
[class]{}-[func]{coinChangeIIDPComp}
```