# 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:
[
,
,
,
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]```

## 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(nums, results, [], 0)

return results

def helper(self, nums, results, subset, start):
results.append(list(subset))

for i in range(start, len(nums)):
subset.append(nums[i])
self.helper(nums, results, 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.

## 2 Thoughts to “LeetCode 78. Subsets”

1. anushka kasera says:

cannot understand the the remove bit .