#78

Medium

auto_awesomeBacktracking search with pruning

LeetCode Problem Workspace

Subsets

Generate all subsets of a set of unique integers using backtracking with pruning to avoid duplicates.

Problem Statement

Given an integer array nums consisting of unique elements, you need to return all possible subsets of the array. The solution set must not contain duplicate subsets, and you can return the subsets in any order.

For example, with the input nums = [1,2,3], the output would include all possible subsets such as [], [1], [2], [1,2], [3], [1,3], [2,3], and [1,2,3]. The length of nums is constrained to 1 ≤ nums.length ≤ 10, ensuring feasible time complexity.

Examples

Example 1

Input: nums = [1,2,3]

Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Example details omitted.

Example 2

Input: nums = [0]

Output: [[],[0]]

Example details omitted.

Constraints

1 <= nums.length <= 10
-10 <= nums[i] <= 10
All the numbers of nums are unique.

Solution Approach

Backtracking Search with Pruning

The optimal approach is using backtracking, where you explore each element and decide whether to include it in the current subset. By pruning the recursion tree (not revisiting subsets already explored), we efficiently generate all possible subsets without duplication.

Iterative Subset Construction

An alternative method is to construct subsets iteratively. Start with the empty subset, and for each element, add it to all existing subsets to create new subsets. This approach avoids recursion but still covers all possibilities.

Bit Manipulation Approach

For those comfortable with bit manipulation, this problem can also be tackled by treating each subset as a binary number, where each bit represents whether an element is included. This provides a compact and efficient way to generate all subsets.

Complexity Analysis

Metric	Value
Time	\mathcal{O}(N \times 2^N)
Space	\mathcal{O}(N)

The time complexity is O(N × 2^N), where N is the length of the input array. This is because there are 2^N possible subsets, and we need to generate each subset. The space complexity is O(N) due to the recursive stack and the space used for storing subsets.

What Interviewers Usually Probe

Candidate demonstrates familiarity with backtracking techniques and pruning strategies.
Candidate handles edge cases such as small arrays or empty sets.
Candidate optimizes the solution and considers different approaches like bit manipulation or iterative methods.

Common Pitfalls or Variants

Common pitfalls

Failing to prune unnecessary recursive calls, leading to duplicate subsets.
Incorrectly handling the base case or recursion termination, which can result in incomplete subsets.
Overcomplicating the problem by not leveraging simpler iterative or bit manipulation techniques.

Follow-up variants

Allow subsets to include duplicate elements.
Return the subsets sorted or in a specific order.
Implement a solution using dynamic programming or memoization.

FAQ

What is the key approach for solving the Subsets problem?

The key approach is backtracking with pruning, ensuring that we efficiently explore each subset without duplicating solutions.

How do you avoid duplicate subsets in the Subsets problem?

By pruning unnecessary recursive calls and ensuring that each subset is generated uniquely in the search space.

Can bit manipulation be used for the Subsets problem?

Yes, bit manipulation can represent each subset as a binary number, efficiently generating subsets without recursion.

What is the time complexity of the Subsets problem?

The time complexity is O(N × 2^N), where N is the length of the input array, because there are 2^N subsets.

How can backtracking with pruning optimize the Subsets solution?

Pruning helps avoid redundant calculations by ensuring subsets are generated without revisiting previously explored solutions.

terminal

Solution

Solution 1: DFS (Backtracking)

We design a function $dfs(i)$, which represents starting the search from the $i$th element of the array for all subsets. The execution logic of the function $dfs(i)$ is as follows:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        def dfs(i: int):
            if i == len(nums):
                ans.append(t[:])
                return
            dfs(i + 1)
            t.append(nums[i])
            dfs(i + 1)
            t.pop()

        ans = []
        t = []
        dfs(0)
        return ans

Solution 2: Binary Enumeration

We can also use the method of binary enumeration to get all subsets.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        def dfs(i: int):
            if i == len(nums):
                ans.append(t[:])
                return
            dfs(i + 1)
            t.append(nums[i])
            dfs(i + 1)
            t.pop()

        ans = []
        t = []
        dfs(0)
        return ans

Continue Practicing

#90 subsets ii #473 matchsticks to square #491 non decreasing subsequences