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95 lines
7.3 KiB
Markdown
95 lines
7.3 KiB
Markdown
# Subset sum problem
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## Case without duplicate elements
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!!! question
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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.
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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.
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- Elements in the input set can be chosen an unlimited number of times.
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- Subsets do not distinguish the order of elements, for example $\{4, 5\}$ and $\{5, 4\}$ are the same subset.
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### Reference permutation solution
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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.
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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:
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```src
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[file]{subset_sum_i_naive}-[class]{}-[func]{subset_sum_i_naive}
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```
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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 $[5, 4]$**.
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This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in the figure below, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
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![Subset search and pruning out of bounds](subset_sum_problem.assets/subset_sum_i_naive.png)
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To eliminate duplicate subsets, **a straightforward idea is to deduplicate the result list**. However, this method is very inefficient for two reasons.
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- When there are many array elements, especially when `target` is large, the search process produces a large number of duplicate subsets.
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- 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.
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### Duplicate subset pruning
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**We consider deduplication during the search process through pruning**. Observing the figure below, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
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1. When choosing $3$ in the first round and $4$ in the second round, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
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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.`.
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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.
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1. First two rounds choose $3$ and $5$, generating subset $[3, 5, \dots]$.
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2. First two rounds choose $4$ and $5$, generating subset $[4, 5, \dots]$.
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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.`.
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![Different choice orders leading to duplicate subsets](subset_sum_problem.assets/subset_sum_i_pruning.png)
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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**.
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### Code implementation
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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.
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Besides, we have made the following two optimizations to the code.
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- 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`.
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- Eliminate the element sum variable `total`, **by performing subtraction on `target` to count the element sum**. When `target` equals $0$, record the solution.
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```src
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[file]{subset_sum_i}-[class]{}-[func]{subset_sum_i}
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```
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The figure below shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
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![Subset sum I backtracking process](subset_sum_problem.assets/subset_sum_i.png)
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## Considering cases with duplicate elements
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!!! question
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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.
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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.
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**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In the figure below, 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.
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![Duplicate subsets caused by equal elements](subset_sum_problem.assets/subset_sum_ii_repeat.png)
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### Equal element pruning
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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.
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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.
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### Code implementation
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```src
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[file]{subset_sum_ii}-[class]{}-[func]{subset_sum_ii}
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```
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The figure below 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.
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![Subset sum II backtracking process](subset_sum_problem.assets/subset_sum_ii.png)
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