algoadvance

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.

Clarifying Questions

  1. Input Range: What is the range of values for the elements in nums?
    • Typically, elements range from -100 to 100, with nums having a length between 1 and 15 for practicality in problems like this.
  2. Duplicates: Are there any specific instructions on handling duplicates within the input array?
    • Duplicates in the array can exist, and they need to be handled such that subsequences considering the same indices aren’t counted multiple times.

Strategy

To solve this problem, we can use a backtracking approach to find all possible non-decreasing subsequences. The key point here is ensuring that subsequences with at least two elements are returned and handling duplicates properly.

  1. Backtracking: Use a recursive function to generate all possible subsequences.
  2. Validation: Ensure each subsequence is non-decreasing.
  3. Avoid Duplicates: Use a set to store subsequences and avoid duplicates.

Code

from typing import List

def findSubsequences(nums: List[int]) -> List[List[int]]:
    def backtrack(start: int, path: List[int]):
        if len(path) > 1:
            subsequences.add(tuple(path))
        for i in range(start, len(nums)):
            if not path or nums[i] >= path[-1]:
                backtrack(i + 1, path + [nums[i]])
    
    subsequences = set()
    backtrack(0, [])
    return [list(seq) for seq in subsequences]

# Example usage:
nums = [4, 6, 7, 7]
print(findSubsequences(nums))

Strategy Explanation

  1. Backtracking Function:
    • The function backtrack takes two parameters: the starting index start and the current path of numbers path.
    • If the length of the path is greater than 1, we add it to the subsequences set to ensure it meets the requirement of being a non-decreasing subsequence with at least two elements.
    • For each element starting from start to the end of nums, we check if it can be appended to the path to maintain the non-decreasing order.
    • We recursively call backtrack with the next starting index and the new path including the current element.
  2. Using a Set:
    • Using a set, subsequences, guarantees that we get unique subsequences since tuples of the same sequence aren’t inserted multiple times into the set.
  3. Result Conversion:
    • Finally, we convert the set of tuples back to a list of lists, which is the required format for the output.

Time Complexity

The time complexity can be analyzed based on the following:

Therefore, the overall time complexity is approximately O(n * 2^n). This is feasible for the given problem constraints where n (length of nums) is between 1 and 15.

Try our interview co-pilot at AlgoAdvance.com