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60. Subsets

mediumAsked at Reddit

Generate all 2^n subsets of a set of distinct integers. Reddit uses this to test backtracking and bitmask enumeration — the same enumeration used when computing all flag-combination buckets for feature-rollout cohorts.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit on-site favorite for backend roles.
  • Blind (2025-12)— Reported in Reddit infra-team rounds.

Problem

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.

Constraints

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

Examples

Example 1

Input
nums = [1,2,3]
Output
[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Example 2

Input
nums = [0]
Output
[[],[0]]

Approaches

1. Iterative power-doubling

Start with [[]]. For each number, append it to copies of existing subsets.

Time
O(n * 2^n)
Space
O(n * 2^n)
function subsets(nums) {
  let result = [[]];
  for (const n of nums) {
    const next = result.map(s => [...s, n]);
    result = [...result, ...next];
  }
  return result;
}

Tradeoff: Clean. Doubles output per element.

2. Backtracking with start-index (optimal)

Recurse picking any i >= start. Push current at every entry.

Time
O(n * 2^n)
Space
O(n) recursion
function subsets(nums) {
  const out = [];
  function backtrack(start, current) {
    out.push([...current]);
    for (let i = start; i < nums.length; i++) {
      current.push(nums[i]);
      backtrack(i + 1, current);
      current.pop();
    }
  }
  backtrack(0, []);
  return out;
}

Tradeoff: Memory-efficient (current is shared).

Reddit-specific tips

Reddit interviewers will accept either. Bonus signal: mention the bitmask version (iterate i from 0 to 2^n; for each bit b in i, include nums[b]) and discuss when it's cleaner.

Common mistakes

  • Forgetting to push the empty subset.
  • Pushing current directly without copy.
  • Using start=current.length+1 (off by one).

Follow-up questions

An interviewer at Reddit may pivot to one of these next:

  • Subsets II (LC 90) — with duplicates.
  • Permutations (LC 46).
  • Combinations (LC 77).

Solve it now

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Output

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FAQ

Bitmask version?

for (let mask=0; mask<(1<<n); mask++) collect nums[b] for each set bit b. Useful when you want subsets in deterministic order.

Why push current at the start of backtrack?

Each subset is the snapshot of current at some recursion entry — including the empty one at depth 0.

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