# Description

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

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

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

## Explanation

The problem is a typical depth first search coding problem.

We can have a recursion function to add visited subsets to the final results. Remember to make a deep copy when we need to add 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) {

for (int i = startIndex; i < nums.length; 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]]:
result = []

nums = list(sorted(nums))
self.helper(nums, 0, [], result)

return result

def helper(self, nums, start_index, subset, result):

result.append(list(subset))

for i in range(start_index, len(nums)):
subset.append(nums[i])

self.helper(nums, i + 1, subset, result)

subset.pop()

• Time complexity: \mathcal{O}(N \times 2^N)O(N×2N) to generate all subsets and then copy them into output list.
• Space complexity: \mathcal{O}(N \times 2^N)O(N×2N) to keep all the subsets of length NN, since each of NN elements could be present or absent.
• Time complexity: O(N×2^N) to generate all subsets and then copy them into the output list.
• Space complexity: O(N×2^N) to keep all the subsets of length N, since each of N elements could be present or absent.

## 2 Thoughts to “LeetCode 78. Subsets”

1. anushka kasera says:

cannot understand the the remove bit .