## Increasing Subsequences

• Time:O(n \cdot 2^n)
• Space:O(n^2)

## C++

class Solution {
public:
vector<vector<int>> findSubsequences(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) {
if (path.size() > 1)
ans.push_back(path);

unordered_set<int> used;

for (int i = s; i < nums.size(); ++i) {
if (used.count(nums[i]))
continue;
if (path.empty() || nums[i] >= path.back()) {
used.insert(nums[i]);
path.push_back(nums[i]);
dfs(nums, i + 1, move(path), ans);
path.pop_back();
}
}
}
};


## JAVA

class Solution {
public List<List<Integer>> findSubsequences(int[] nums) {
return ans;
}

private void dfs(int[] nums, int s, LinkedList<Integer> path, List<List<Integer>> ans) {
if (path.size() > 1)

Set<Integer> used = new HashSet<>();

for (int i = s; i < nums.length; ++i) {
if (used.contains(nums[i]))
continue;
if (path.isEmpty() || nums[i] >= path.getLast()) {
dfs(nums, i + 1, path, ans);
path.removeLast();
}
}
}
}


## Python

class Solution:
def findSubsequences(self, nums: List[int]) -> List[List[int]]:
ans = []

def dfs(s: int, path: List[int]) -> None:
if len(path) > 1:
ans.append(path)

used = set()

for i in range(s, len(nums)):
if nums[i] in used:
continue
if not path or nums[i] >= path[-1]: