491. Non-decreasing Subsequences LeetCode Solution

In this guide, you will get 491. Non-decreasing Subsequences LeetCode Solution with the best time and space complexity. The solution to Non-decreasing Subsequences problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Non-decreasing Subsequences solution in C++
Non-decreasing Subsequences solution in Java
Non-decreasing Subsequences solution in Python
Additional Resources

491. Non-decreasing Subsequences LeetCode Solution image

Problem Statement of Non-decreasing Subsequences

Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.

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

Complexity Analysis

Time Complexity: O(n \cdot 2^n)
Space Complexity: O(n^2)

491. Non-decreasing Subsequences LeetCode Solution in 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.contains(nums[i]))
        continue;
      if (path.empty() || nums[i] >= path.back()) {
        used.insert(nums[i]);
        path.push_back(nums[i]);
        dfs(nums, i + 1, std::move(path), ans);
        path.pop_back();
      }
    }
  }
};
/* code provided by PROGIEZ */

491. Non-decreasing Subsequences LeetCode Solution in Java

class Solution {
  public List<List<Integer>> findSubsequences(int[] nums) {
    List<List<Integer>> ans = new LinkedList<>();
    dfs(nums, 0, new LinkedList<>(), ans);
    return ans;
  }

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

    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()) {
        used.add(nums[i]);
        path.addLast(nums[i]);
        dfs(nums, i + 1, path, ans);
        path.removeLast();
      }
    }
  }
}
// code provided by PROGIEZ

491. Non-decreasing Subsequences LeetCode Solution in 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]:
          used.add(nums[i])
          dfs(i + 1, path + [nums[i]])

    dfs(0, [])
    return ans
 # code by PROGIEZ