How do you solve Subsets on LeetCode?

Subsets (LeetCode #78) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n) space complexity. Key insight: Each element can either be included or excluded from a subset.

What is the brute force solution for Subsets?

The brute force approach for Subsets is Brute Force, with O(2^n) time complexity and O(n) space complexity. The brute force approach involves generating all possible combinations of the elements in the array. This can be visualized as making a decision for each element: either to include it in a subset or not. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Subsets?

Subsets uses the following concepts: Array, Backtracking, Bit Manipulation. The recommended patterns to study are: Backtracking, Bit Manipulation.

What are the interview tips for Subsets?

Explain your thought process clearly while coding. Consider edge cases, like an empty array or a single element. Practice generating subsets with smaller arrays before tackling larger ones.

What are common mistakes when solving Subsets?

Not considering the empty subset. Confusing the order of subsets in the output.

#78

Subsets

Medium

ArrayBacktrackingBit ManipulationBacktrackingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(n * 2^n)
Space	O(n)	O(n)

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Intuition

Time O(n * 2^n)Space O(n)

The optimal solution uses bit manipulation to represent subsets as binary numbers. Each bit in a binary number represents whether to include an element from the array or not.

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Algorithm

3 steps

1Step 1: Initialize an empty list to store all subsets.
2Step 2: Iterate from 0 to 2^n (where n is the length of nums).
3Step 3: For each number, convert it to binary and use the bits to decide which elements to include in the current subset.

solution.py10 lines

1def subsets(nums):
2    n = len(nums)
3    result = []
4    for i in range(1 << n):
5        subset = []
6        for j in range(n):
7            if i & (1 << j):
8                subset.append(nums[j])
9        result.append(subset)
10    return result

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Complexity note: The time complexity is O(n * 2^n) because we generate 2^n subsets and each subset can take up to O(n) time to construct. Space complexity is O(n) for the subset storage.

1Each element can either be included or excluded from a subset.
2The number of subsets grows exponentially with the number of elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.