Increasing Subsequences – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer array nums, return all the different possible increasing subsequences of the given array with at least two elements. You may return the answer in any order.
The given array may contain duplicates, and two equal integers should also be considered a special case of increasing sequence.
Example 1:
Input: nums = [4,6,7,7] Output: [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
Example 2:
Input: nums = [4,4,3,2,1] Output: [[4,4]]
Constraints:
1 <= nums.length <= 15-100 <= nums[i] <= 100
C++ Increasing Subsequences LeetCode Solution
class Solution {
public:
vector<vector<int>> findSubsequences(vector<int>& nums) {
vector<vector<int>> res;
vector<int> seq;
dfs(res, seq, nums, 0);
return res;
}
void dfs(vector<vector<int>>& res, vector<int>& seq, vector<int>& nums, int pos) {
if(seq.size() > 1) res.push_back(seq);
unordered_set<int> hash;
for(int i = pos; i < nums.size(); ++i) {
if((seq.empty() || nums[i] >= seq.back()) && hash.find(nums[i]) == hash.end()) {
seq.push_back(nums[i]);
dfs(res, seq, nums, i + 1);
seq.pop_back();
hash.insert(nums[i]);
}
}
}
};
Java Increasing Subsequences LeetCode Solution
public class Solution {
public List<List<Integer>> findSubsequences(int[] nums) {
List<List<Integer>> res = new LinkedList<>();
helper(new LinkedList<Integer>(), 0, nums, res);
return res;
}
private void helper(LinkedList<Integer> list, int index, int[] nums, List<List<Integer>> res){
if(list.size()>1) res.add(new LinkedList<Integer>(list));
Set<Integer> used = new HashSet<>();
for(int i = index; i<nums.length; i++){
if(used.contains(nums[i])) continue;
if(list.size()==0 || nums[i]>=list.peekLast()){
used.add(nums[i]);
list.add(nums[i]);
helper(list, i+1, nums, res);
list.remove(list.size()-1);
}
}
}
}
Python 3 Increasing Subsequences LeetCode Solution
def findSubsequences(self, nums):
subs = {()}
for num in nums:
subs |= {sub + (num,)
for sub in subs
if not sub or sub[-1] <= num}
return [sub for sub in subs if len(sub) >= 2]
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