Introduction

In this blog post, we will explore how to solve the LeetCode problem "491" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Depth-First Search (DFS) with Backtracking: Explore all possible subsequences using a recursive DFS approach.
  • Non-decreasing Constraint: Maintain the non-decreasing order while building subsequences. Discard branches that violate this constraint.
  • Set for Uniqueness: Use a set to store the subsequences to avoid duplicates. Tuples are used as set elements since lists aren't hashable.

• Runtime Complexity: O(2ⁿ), where n is the length of the input array. Although we use a set to prevent duplicates, in the worst case (an array full of the same number), we will still have to explore all possible subsequences.
• Storage Complexity: O(2ⁿ) in the worst case to store the subsequences in the set.

Code

    def findSubsequences(nums):
    result = set()
    current_sequence = []

    def backtrack(index):
        if len(current_sequence) >= 2:
            result.add(tuple(current_sequence))

        if index == len(nums):
            return

        for i in range(index, len(nums)):
            if not current_sequence or nums[i] >= current_sequence[-1]:
                current_sequence.append(nums[i])
                backtrack(i + 1)
                current_sequence.pop()

    backtrack(0)
    return [list(seq) for seq in result]