Subsets
Detailed guide and Python implementation for the 'Subsets' problem.
1. Concept Overview
The 'Subsets' problem is a key challenge in the Backtracking section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Subsets.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Subsets carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given an integer array nums of unique elements, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Implement a function subsets(nums: list) -> list.
- •1 <= nums.length <= 10
- •-10 <= nums[i] <= 10
- •All the numbers of nums are unique
Examples
[1,2,3]
[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
All 2^3 = 8 subsets of [1,2,3] are generated, including the empty set and the full set.
[0]
[[],[0]]
The two subsets of [0] are the empty set [] and [0].
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def subsets_opt(nums: list[int]) -> list[list[int]]:
res = [[]]
for n in nums:
res += [subset + [n] for subset in res]
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def subsets_brute(nums: list[int]) -> list[list[int]]:
res = []
def dfs(i, subset):
if i == len(nums):
res.append(subset[:])
return
subset.append(nums[i])
dfs(i + 1, subset)
subset.pop()
dfs(i + 1, subset)
dfs(0, [])
return resAlgorithm Pattern Checklist
When dealing with Backtracking data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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