LeetCode 78. Subsets

Description

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

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.

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]]

Constraints:

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

Explanation

The problem is a typical backtracking coding problem. We can have a recursion helper function to add visited subsets to the final results. Remember to make a deep copy when we are adding a subset to the results.

Java Solution

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

Python Solution

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        results = []
        
        self.helper(results, nums, [], 0)
        
        return results
    
    def helper(self, results, nums, subset, start):
        results.append(list(subset))
        
        
        for i in range(start, len(nums)):
            num = nums[i]
            subset.append(num)
            self.helper(results, nums, subset, i + 1)
            subset.pop()
        
  • Time complexity: O(N * 2^N) to generate all subsets and then copy them into the output list. The number for ecursive calls T(n) satisfies the recurrence T(n) = T(n-1) + T(n-2) + T(1) + T(0), which solves to T(n) = O(2^n) .Since we spend O(n) time within a call , the time complexity is O(N * 2^N).
  • Space complexity: O(N * 2^N) to keep all the subsets of length N, since each of N elements could be present or absent.

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