Subsets I

Problem link: https://leetcode.com/problems/subsets/

Constaint:


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

Idea

Suppose we have a temparary list to store all the elements we processed so far. For each number at index i, we have two posibilities: 1) Add nums[i] to the list; 2) Not add nums[i] to the list. Use DFS to do this recursively for every number. The terminating condition is when the index == nums.length;

DFS solution 1

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        search(nums, 0, new ArrayList<Integer>(), results);
        return results;
    }
    
    private void search(int[] nums, int index, List<Integer> subset, List<List<Integer>> results) {
        if (index == nums.length) {
            results.add(new ArrayList<>(subset));
            return;
        }
        
        search(nums, index + 1, subset, results);
        subset.add(nums[index]);
        search(nums, index + 1, subset, results);
        subset.remove(subset.size() - 1);
    }
}

DFS solution 2

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> results = new ArrayList<>();
        search(nums, 0, new ArrayList<Integer>(), results);
        return results;
    }
    
    private void search(int[] nums, int start, List<Integer> subset, List<List<Integer>> results) {
        results.add(new ArrayList<Integer>(subset));
        for (int i = start; i < nums.length; i++) {
            subset.add(nums[i]);
            search(nums, i + 1, subset, results);
            subset.remove(subset.size() - 1);
        }
    }
}

Time: O(n * 2^n) as there are 2^n possible combinations, and making a copy of each subset takes O(n).
Space: O(n). The recursion stack will be as large as the size of array.