LeetCode #90 — MEDIUM

Subsets II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

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

Example 2:

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

Constraints:

1 <= nums.length <= 10
-10 <= nums[i] <= 10

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums that may contain duplicates, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking · Bit Manipulation

Example 1

[1,2,2]

Example 2

[0]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> ans = new ArrayList<>();
        backtrack(0, nums, new ArrayList<>(), ans);
        return ans;
    }

    private void backtrack(int start, int[] nums, List<Integer> cur, List<List<Integer>> ans) {
        ans.add(new ArrayList<>(cur));
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) continue;
            cur.add(nums[i]);
            backtrack(i + 1, nums, cur, ans);
            cur.remove(cur.size() - 1);
        }
    }
}

import "sort"

func subsetsWithDup(nums []int) [][]int {
    sort.Ints(nums)
    ans := [][]int{}
    cur := []int{}

    var backtrack func(start int)
    backtrack = func(start int) {
        ans = append(ans, append([]int{}, cur...))
        for i := start; i < len(nums); i++ {
            if i > start && nums[i] == nums[i-1] {
                continue
            }
            cur = append(cur, nums[i])
            backtrack(i + 1)
            cur = cur[:len(cur)-1]
        }
    }

    backtrack(0)
    return ans
}

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        ans: List[List[int]] = []
        cur: List[int] = []

        def backtrack(start: int) -> None:
            ans.append(cur[:])
            for i in range(start, len(nums)):
                if i > start and nums[i] == nums[i - 1]:
                    continue
                cur.append(nums[i])
                backtrack(i + 1)
                cur.pop()

        backtrack(0)
        return ans

impl Solution {
    pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
        nums.sort();
        let mut ans = Vec::new();
        let mut cur = Vec::new();

        fn backtrack(start: usize, nums: &Vec<i32>, cur: &mut Vec<i32>, ans: &mut Vec<Vec<i32>>) {
            ans.push(cur.clone());
            for i in start..nums.len() {
                if i > start && nums[i] == nums[i - 1] {
                    continue;
                }
                cur.push(nums[i]);
                backtrack(i + 1, nums, cur, ans);
                cur.pop();
            }
        }

        backtrack(0, &nums, &mut cur, &mut ans);
        ans
    }
}

function subsetsWithDup(nums: number[]): number[][] {
  nums.sort((a, b) => a - b);
  const ans: number[][] = [];
  const cur: number[] = [];

  const backtrack = (start: number): void => {
    ans.push([...cur]);
    for (let i = start; i < nums.length; i++) {
      if (i > start && nums[i] === nums[i - 1]) continue;
      cur.push(nums[i]);
      backtrack(i + 1);
      cur.pop();
    }
  };

  backtrack(0);
  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × 2^n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.