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This approach utilizes the concept of recursive backtracking to explore all potential subsets. We start with an empty list and progressively build subsets by either including or excluding each element.
Time Complexity: O(2^n * n), where n is the length of nums. We have 2^n subsets, and copying each subset takes O(n).
Space Complexity: O(n), where n is the depth of the recursion stack.
1import java.util.ArrayList;
2import java.util.List;
3
4public class Solution {
5 public List<List<Integer>> subsets(int[] nums) {
6 List<List<Integer>> result = new ArrayList<>();
7 backtrack(nums, result, new ArrayList<>(), 0);
8 return result;
9 }
10
11 private void backtrack(int[] nums, List<List<Integer>> result, List<Integer> current, int start) {
12 result.add(new ArrayList<>(current));
13 for (int i = start; i < nums.length; i++) {
14 current.add(nums[i]);
15 backtrack(nums, result, current, i + 1);
16 current.remove(current.size() - 1);
17 }
18 }
19
20 public static void main(String[] args) {
21 Solution sol = new Solution();
22 int[] nums = {1, 2, 3};
23 List<List<Integer>> result = sol.subsets(nums);
24
25 for (List<Integer> subset : result) {
26 System.out.println(subset);
27 }
28 }
29}
30In this Java implementation, we utilize a recursive backtracking approach. The 'backtrack' method iteratively decides whether to include each element, building up the current subset, and then explores further possibilities recursively.
This approach takes advantage of bit manipulation to enumerate subsets. Each subset can correspond to a binary number where each bit decides if an element is present in the subset:
Time Complexity: O(2^n * n), mapping to subset enumeration.
Space Complexity: O(1), as it utilizes constants.
1const subsets
In JavaScript iteration over a complete subset count translated to binary bits selects index proximities for each element in derived subsets. It harnesses shifts to ensure discerning subset computations.