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:
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
/* 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;
}
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.
Figure 13-10 Subset search and pruning out of bounds
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
targetis 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. 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.
- When choosing \(3\) in the first round and \(4\) in the second round, all subsets containing these two elements are generated, denoted as \([3, 4, \dots]\).
- 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.
- First two rounds choose \(3\) and \(5\), generating subset \([3, 5, \dots]\).
- First two rounds choose \(4\) and \(5\), generating subset \([4, 5, \dots]\).
- 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.and2..
Figure 13-11 Different choice orders leading to duplicate subsets
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. 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 exceedstargetas subsequent elements are larger and their subset sum will definitely exceedtarget. - Eliminate the element sum variable
total, by performing subtraction ontargetto count the element sum. Whentargetequals \(0\), record the solution.
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
/* 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;
}
Figure 13-12 shows the overall backtracking process after inputting the array \([3, 4, 5]\) and target element \(9\) into the above code.
Figure 13-12 Subset sum I backtracking process
13.3.2 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.
Figure 13-13 Duplicate subsets caused by equal elements
1. 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. Code implementation¶
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
/* 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;
}
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.
Figure 13-14 Subset sum II backtracking process