# Subsets Leetcode Solution

#### ByBrokenprogrammers

Dec 18, 2022

In this post, you will know how to solve the Subsets Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## Problem

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.

### SubsetsLeetcode Solutions in Python

```class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
ans = []
def dfs(s: int, path: List[int]) -> None:
ans.append(path)
for i in range(s, len(nums)):
dfs(i + 1, path + [nums[i]])
dfs(0, [])
return ans
```

### SubsetsLeetcode Solutionsin CPP

```class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> ans;
dfs(nums, 0, {}, ans);
return ans;
}
private:
void dfs(const vector<int>& nums, int s, vector<int>&& path,
vector<vector<int>>& ans) {
ans.push_back(path);
for (int i = s; i < nums.size(); ++i) {
path.push_back(nums[i]);
dfs(nums, i + 1, move(path), ans);
path.pop_back();
}
}
};
```

### SubsetsLeetcode Solutions in Java

```class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
dfs(nums, 0, new ArrayList<>(), ans);
return ans;
}
private void dfs(int[] nums, int s, List<Integer> path, List<List<Integer>> ans) {
for (int i = s; i < nums.length; ++i) {
dfs(nums, i + 1, path, ans);
path.remove(path.size() - 1);
}
}
}
```

Note: This problem Subsets is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.