How do you solve Non-decreasing Subsequences on LeetCode?

Non-decreasing Subsequences (LeetCode #491) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(2^n) time complexity and O(n) space complexity. Key insight: Non-decreasing subsequences can be efficiently generated using backtracking.

What is the brute force solution for Non-decreasing Subsequences?

The brute force approach for Non-decreasing Subsequences is Brute Force, with O(2^n) time complexity and O(n) space complexity. The brute-force approach involves generating all possible subsequences of the array and then filtering out the non-decreasing ones. This method is straightforward but inefficient due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(2^n).

What algorithm or data structure is used to solve Non-decreasing Subsequences?

Non-decreasing Subsequences uses the following concepts: Array, Hash Table, Backtracking, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Non-decreasing Subsequences?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as arrays with all identical elements. Practice writing clean and readable code, as clarity is key in interviews.

What are common mistakes when solving Non-decreasing Subsequences?

Not considering subsequences of length less than 2. Failing to handle duplicates properly.

#491

Non-decreasing Subsequences

Medium

ArrayHash TableBacktrackingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(2^n)
Space	O(n)	O(n)

💡

Intuition

Time O(2^n)Space O(n)

The optimal solution uses backtracking with pruning to efficiently generate non-decreasing subsequences. By avoiding unnecessary branches, we can significantly reduce the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a set to store unique subsequences.
2Step 2: Use a backtracking function that builds subsequences while ensuring they are non-decreasing.
3Step 3: For each element, decide to include it in the current subsequence or skip it based on the last added element.

solution.py13 lines

1# Full working Python code
2from typing import List
3
4def findSubsequences(nums: List[int]) -> List[List[int]]:
5    result = set()
6    def backtrack(start, path):
7        if len(path) >= 2:
8            result.add(tuple(path))
9        for i in range(start, len(nums)):
10            if not path or path[-1] <= nums[i]:
11                backtrack(i + 1, path + [nums[i]])
12    backtrack(0, [])
13    return [list(seq) for seq in result]

ℹ

Complexity note: The time complexity remains exponential due to the nature of subsequences, but the pruning significantly reduces the number of valid paths explored.

1Non-decreasing subsequences can be efficiently generated using backtracking.
2Using a set helps in avoiding duplicate subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.