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hello-algo/en/docs/chapter_backtracking/subset_sum_problem.md

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---
comments: true
---
# 13.3   Subset sum problem
## 13.3.1   Case without duplicate elements
!!! question
Given an array of positive integers `nums` and a target positive integer `target`, find all possible combinations such that the sum of the elements in the combination equals `target`. The given array has no duplicate elements, and each element can be chosen multiple times. Please return these combinations as a list, which should not contain duplicate combinations.
For example, for the input set $\{3, 4, 5\}$ and target integer $9$, the solutions are $\{3, 3, 3\}, \{4, 5\}$. Note the following two points.
- Elements in the input set can be chosen an unlimited number of times.
- Subsets do not distinguish the order of elements, for example $\{4, 5\}$ and $\{5, 4\}$ are the same subset.
### 1.   Reference permutation solution
Similar to the permutation problem, we can imagine the generation of subsets as a series of choices, updating the "element sum" in real-time during the choice process. When the element sum equals `target`, the subset is recorded in the result list.
Unlike the permutation problem, **elements in this problem can be chosen an unlimited number of times**, thus there is no need to use a `selected` boolean list to record whether an element has been chosen. We can make minor modifications to the permutation code to initially solve the problem:
=== "Python"
```python title="subset_sum_i_naive.py"
def backtrack(
state: list[int],
target: int,
total: int,
choices: list[int],
res: list[list[int]],
):
"""Backtracking algorithm: Subset Sum I"""
# When the subset sum equals target, record the solution
if total == target:
res.append(list(state))
return
# Traverse all choices
for i in range(len(choices)):
# Pruning: if the subset sum exceeds target, skip that choice
if total + choices[i] > target:
continue
# Attempt: make a choice, update elements and total
state.append(choices[i])
# Proceed to the next round of selection
backtrack(state, target, total + choices[i], choices, res)
# Retract: undo the choice, restore to the previous state
state.pop()
def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
"""Solve Subset Sum I (including duplicate subsets)"""
state = [] # State (subset)
total = 0 # Subset sum
res = [] # Result list (subset list)
backtrack(state, target, total, nums, res)
return res
```
=== "C++"
```cpp title="subset_sum_i_naive.cpp"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Java"
```java title="subset_sum_i_naive.java"
/* Backtracking algorithm: Subset Sum I */
void backtrack(List<Integer> state, int target, int total, int[] choices, List<List<Integer>> res) {
// When the subset sum equals target, record the solution
if (total == target) {
res.add(new ArrayList<>(state));
return;
}
// Traverse all choices
for (int i = 0; i < choices.length; i++) {
// Pruning: if the subset sum exceeds target, skip that choice
if (total + choices[i] > target) {
continue;
}
// Attempt: make a choice, update elements and total
state.add(choices[i]);
// Proceed to the next round of selection
backtrack(state, target, total + choices[i], choices, res);
// Retract: undo the choice, restore to the previous state
state.remove(state.size() - 1);
}
}
/* Solve Subset Sum I (including duplicate subsets) */
List<List<Integer>> subsetSumINaive(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // State (subset)
int total = 0; // Subset sum
List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
backtrack(state, target, total, nums, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_i_naive.cs"
[class]{subset_sum_i_naive}-[func]{Backtrack}
[class]{subset_sum_i_naive}-[func]{SubsetSumINaive}
```
=== "Go"
```go title="subset_sum_i_naive.go"
[class]{}-[func]{backtrackSubsetSumINaive}
[class]{}-[func]{subsetSumINaive}
```
=== "Swift"
```swift title="subset_sum_i_naive.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "JS"
```javascript title="subset_sum_i_naive.js"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "TS"
```typescript title="subset_sum_i_naive.ts"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Dart"
```dart title="subset_sum_i_naive.dart"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Rust"
```rust title="subset_sum_i_naive.rs"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i_naive}
```
=== "C"
```c title="subset_sum_i_naive.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Kotlin"
```kotlin title="subset_sum_i_naive.kt"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
=== "Ruby"
```ruby title="subset_sum_i_naive.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i_naive}
```
=== "Zig"
```zig title="subset_sum_i_naive.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
Inputting the array $[3, 4, 5]$ and target element $9$ into the above code yields the results $[3, 3, 3], [4, 5], [5, 4]$. **Although it successfully finds all subsets with a sum of $9$, it includes the duplicate subset $[4, 5]$ and $[5, 4]$**.
This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in Figure 13-10, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
![Subset search and pruning out of bounds](subset_sum_problem.assets/subset_sum_i_naive.png){ class="animation-figure" }
<p align="center"> Figure 13-10 &nbsp; Subset search and pruning out of bounds </p>
To eliminate duplicate subsets, **a straightforward idea is to deduplicate the result list**. However, this method is very inefficient for two reasons.
- When there are many array elements, especially when `target` is large, the search process produces a large number of duplicate subsets.
- Comparing subsets (arrays) for differences is very time-consuming, requiring arrays to be sorted first, then comparing the differences of each element in the arrays.
### 2. &nbsp; Duplicate subset pruning
**We consider deduplication during the search process through pruning**. Observing Figure 13-11, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
1. When choosing $3$ in the first round and $4$ in the second round, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
2. Later, when $4$ is chosen in the first round, **the second round should skip $3$** because the subset $[4, 3, \dots]$ generated by this choice completely duplicates the subset from step `1.`.
In the search process, each layer's choices are tried one by one from left to right, so the more to the right a branch is, the more it is pruned.
1. First two rounds choose $3$ and $5$, generating subset $[3, 5, \dots]$.
2. First two rounds choose $4$ and $5$, generating subset $[4, 5, \dots]$.
3. If $5$ is chosen in the first round, **then the second round should skip $3$ and $4$** as the subsets $[5, 3, \dots]$ and $[5, 4, \dots]$ completely duplicate the subsets described in steps `1.` and `2.`.
![Different choice orders leading to duplicate subsets](subset_sum_problem.assets/subset_sum_i_pruning.png){ class="animation-figure" }
<p align="center"> Figure 13-11 &nbsp; Different choice orders leading to duplicate subsets </p>
In summary, given the input array $[x_1, x_2, \dots, x_n]$, the choice sequence in the search process should be $[x_{i_1}, x_{i_2}, \dots, x_{i_m}]$, which needs to satisfy $i_1 \leq i_2 \leq \dots \leq i_m$. **Any choice sequence that does not meet this condition will cause duplicates and should be pruned**.
### 3. &nbsp; Code implementation
To implement this pruning, we initialize the variable `start`, which indicates the starting point for traversal. **After making the choice $x_{i}$, set the next round to start from index $i$**. This will ensure the choice sequence satisfies $i_1 \leq i_2 \leq \dots \leq i_m$, thereby ensuring the uniqueness of the subsets.
Besides, we have made the following two optimizations to the code.
- Before starting the search, sort the array `nums`. In the traversal of all choices, **end the loop directly when the subset sum exceeds `target`** as subsequent elements are larger and their subset sum will definitely exceed `target`.
- Eliminate the element sum variable `total`, **by performing subtraction on `target` to count the element sum**. When `target` equals $0$, record the solution.
=== "Python"
```python title="subset_sum_i.py"
def backtrack(
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
):
"""Backtracking algorithm: Subset Sum I"""
# When the subset sum equals target, record the solution
if target == 0:
res.append(list(state))
return
# Traverse all choices
# Pruning two: start traversing from start to avoid generating duplicate subsets
for i in range(start, len(choices)):
# Pruning one: if the subset sum exceeds target, end the loop immediately
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
if target - choices[i] < 0:
break
# Attempt: make a choice, update target, start
state.append(choices[i])
# Proceed to the next round of selection
backtrack(state, target - choices[i], choices, i, res)
# Retract: undo the choice, restore to the previous state
state.pop()
def subset_sum_i(nums: list[int], target: int) -> list[list[int]]:
"""Solve Subset Sum I"""
state = [] # State (subset)
nums.sort() # Sort nums
start = 0 # Start point for traversal
res = [] # Result list (subset list)
backtrack(state, target, nums, start, res)
return res
```
=== "C++"
```cpp title="subset_sum_i.cpp"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Java"
```java title="subset_sum_i.java"
/* Backtracking algorithm: Subset Sum I */
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
// When the subset sum equals target, record the solution
if (target == 0) {
res.add(new ArrayList<>(state));
return;
}
// Traverse all choices
// Pruning two: start traversing from start to avoid generating duplicate subsets
for (int i = start; i < choices.length; i++) {
// Pruning one: if the subset sum exceeds target, end the loop immediately
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
if (target - choices[i] < 0) {
break;
}
// Attempt: make a choice, update target, start
state.add(choices[i]);
// Proceed to the next round of selection
backtrack(state, target - choices[i], choices, i, res);
// Retract: undo the choice, restore to the previous state
state.remove(state.size() - 1);
}
}
/* Solve Subset Sum I */
List<List<Integer>> subsetSumI(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // State (subset)
Arrays.sort(nums); // Sort nums
int start = 0; // Start point for traversal
List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
backtrack(state, target, nums, start, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_i.cs"
[class]{subset_sum_i}-[func]{Backtrack}
[class]{subset_sum_i}-[func]{SubsetSumI}
```
=== "Go"
```go title="subset_sum_i.go"
[class]{}-[func]{backtrackSubsetSumI}
[class]{}-[func]{subsetSumI}
```
=== "Swift"
```swift title="subset_sum_i.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "JS"
```javascript title="subset_sum_i.js"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "TS"
```typescript title="subset_sum_i.ts"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Dart"
```dart title="subset_sum_i.dart"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Rust"
```rust title="subset_sum_i.rs"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i}
```
=== "C"
```c title="subset_sum_i.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Kotlin"
```kotlin title="subset_sum_i.kt"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
=== "Ruby"
```ruby title="subset_sum_i.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i}
```
=== "Zig"
```zig title="subset_sum_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
Figure 13-12 shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
![Subset sum I backtracking process](subset_sum_problem.assets/subset_sum_i.png){ class="animation-figure" }
<p align="center"> Figure 13-12 &nbsp; Subset sum I backtracking process </p>
## 13.3.2 &nbsp; Considering cases with duplicate elements
!!! question
Given an array of positive integers `nums` and a target positive integer `target`, find all possible combinations such that the sum of the elements in the combination equals `target`. **The given array may contain duplicate elements, and each element can only be chosen once**. Please return these combinations as a list, which should not contain duplicate combinations.
Compared to the previous question, **this question's input array may contain duplicate elements**, introducing new problems. For example, given the array $[4, \hat{4}, 5]$ and target element $9$, the existing code's output results in $[4, 5], [\hat{4}, 5]$, resulting in duplicate subsets.
**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In Figure 13-13, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
![Duplicate subsets caused by equal elements](subset_sum_problem.assets/subset_sum_ii_repeat.png){ class="animation-figure" }
<p align="center"> Figure 13-13 &nbsp; Duplicate subsets caused by equal elements </p>
### 1. &nbsp; Equal element pruning
To solve this issue, **we need to limit equal elements to being chosen only once per round**. The implementation is quite clever: since the array is sorted, equal elements are adjacent. This means that in a certain round of choices, if the current element is equal to its left-hand element, it means it has already been chosen, so skip the current element directly.
At the same time, **this question stipulates that each array element can only be chosen once**. Fortunately, we can also use the variable `start` to meet this constraint: after making the choice $x_{i}$, set the next round to start from index $i + 1$ going forward. This not only eliminates duplicate subsets but also avoids repeated selection of elements.
### 2. &nbsp; Code implementation
=== "Python"
```python title="subset_sum_ii.py"
def backtrack(
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
):
"""Backtracking algorithm: Subset Sum II"""
# When the subset sum equals target, record the solution
if target == 0:
res.append(list(state))
return
# Traverse all choices
# Pruning two: start traversing from start to avoid generating duplicate subsets
# Pruning three: start traversing from start to avoid repeatedly selecting the same element
for i in range(start, len(choices)):
# Pruning one: if the subset sum exceeds target, end the loop immediately
# This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
if target - choices[i] < 0:
break
# Pruning four: if the element equals the left element, it indicates that the search branch is repeated, skip it
if i > start and choices[i] == choices[i - 1]:
continue
# Attempt: make a choice, update target, start
state.append(choices[i])
# Proceed to the next round of selection
backtrack(state, target - choices[i], choices, i + 1, res)
# Retract: undo the choice, restore to the previous state
state.pop()
def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
"""Solve Subset Sum II"""
state = [] # State (subset)
nums.sort() # Sort nums
start = 0 # Start point for traversal
res = [] # Result list (subset list)
backtrack(state, target, nums, start, res)
return res
```
=== "C++"
```cpp title="subset_sum_ii.cpp"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Java"
```java title="subset_sum_ii.java"
/* Backtracking algorithm: Subset Sum II */
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
// When the subset sum equals target, record the solution
if (target == 0) {
res.add(new ArrayList<>(state));
return;
}
// Traverse all choices
// Pruning two: start traversing from start to avoid generating duplicate subsets
// Pruning three: start traversing from start to avoid repeatedly selecting the same element
for (int i = start; i < choices.length; i++) {
// Pruning one: if the subset sum exceeds target, end the loop immediately
// This is because the array is sorted, and later elements are larger, so the subset sum will definitely exceed target
if (target - choices[i] < 0) {
break;
}
// Pruning four: if the element equals the left element, it indicates that the search branch is repeated, skip it
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// Attempt: make a choice, update target, start
state.add(choices[i]);
// Proceed to the next round of selection
backtrack(state, target - choices[i], choices, i + 1, res);
// Retract: undo the choice, restore to the previous state
state.remove(state.size() - 1);
}
}
/* Solve Subset Sum II */
List<List<Integer>> subsetSumII(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // State (subset)
Arrays.sort(nums); // Sort nums
int start = 0; // Start point for traversal
List<List<Integer>> res = new ArrayList<>(); // Result list (subset list)
backtrack(state, target, nums, start, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_ii.cs"
[class]{subset_sum_ii}-[func]{Backtrack}
[class]{subset_sum_ii}-[func]{SubsetSumII}
```
=== "Go"
```go title="subset_sum_ii.go"
[class]{}-[func]{backtrackSubsetSumII}
[class]{}-[func]{subsetSumII}
```
=== "Swift"
```swift title="subset_sum_ii.swift"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "JS"
```javascript title="subset_sum_ii.js"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "TS"
```typescript title="subset_sum_ii.ts"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Dart"
```dart title="subset_sum_ii.dart"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Rust"
```rust title="subset_sum_ii.rs"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_ii}
```
=== "C"
```c title="subset_sum_ii.c"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Kotlin"
```kotlin title="subset_sum_ii.kt"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
=== "Ruby"
```ruby title="subset_sum_ii.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_ii}
```
=== "Zig"
```zig title="subset_sum_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
Figure 13-14 shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
![Subset sum II backtracking process](subset_sum_problem.assets/subset_sum_ii.png){ class="animation-figure" }
<p align="center"> Figure 13-14 &nbsp; Subset sum II backtracking process </p>
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