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Non-decreasing Subsequences⚓︎

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Description⚓︎

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

Solution⚓︎

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class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int> &nums, int startIndex) {
        if (path.size() > 1) res.push_back(path);

        unordered_set<int> used;
        for (int i = startIndex; i < nums.size(); i++) {
            if (used.find(nums[i]) != used.end()) continue;
            if (!path.empty() && nums[i] < path.back()) continue;

            used.insert(nums[i]);
            path.push_back(nums[i]);
            backtracking(nums, i + 1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        backtracking(nums, 0);
        return res;
    }
};
  • Time complexity: \(O(n\times 2^n)\);
  • Space complexity: \(O(n)\).

Optimized solution:

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class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int> &nums, int startIndex) {
        if (path.size() > 1) res.push_back(path);

        bool used[210] = {false};
        for (int i = startIndex; i < nums.size(); i++) {
            if (used[nums[i] + 100]) continue;
            if (!path.empty() && nums[i] < path.back()) continue;

            used[nums[i] + 100] = true;
            path.push_back(nums[i]);
            backtracking(nums, i + 1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        backtracking(nums, 0);
        return res;
    }
};