Introduction

In this blog post, we will explore how to solve the LeetCode problem "1673" 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 and a positive integer k, return the most competitive subsequence of nums of size k. An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array. We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 is less than 5. Example 1: Input: nums = [3,5,2,6], k = 2 Output: [2,6] Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive. Example 2: Input: nums = [2,4,3,3,5,4,9,6], k = 4 Output: [2,3,3,4] Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 109 1 <= k <= nums.length

Explanation

Here's an efficient solution to find the most competitive subsequence:

• Greedy Approach with Stack: The core idea is to maintain a stack (or deque) that represents the potential most competitive subsequence. We iterate through the input array and for each element, we compare it with the top of the stack. If the current element is smaller and there are enough remaining elements to form a subsequence of size k, we pop elements from the stack until either the stack is empty or the current element is no longer smaller than the top of the stack. Then, if the stack size is less than k, we push the current element onto the stack.

• Ensuring Subsequence Length: The check len(stack) + (len(nums) - i) > k is crucial. It ensures that even after potentially removing elements from the stack, we still have enough remaining elements in the input array to form a subsequence of size k.

• Monotonic Stack Property: By strategically popping elements from the stack, we maintain a monotonically increasing stack (from bottom to top). This ensures that the subsequence is as small as possible in the earlier positions, which is what defines a "more competitive" subsequence.

• Runtime & Storage Complexity: O(N) time complexity, where N is the length of nums. O(K) storage complexity, where K is the size of the competitive subsequence, due to the stack.

Code

    from collections import deque

def mostCompetitive(nums, k):
    stack = deque()
    for i, num in enumerate(nums):
        while stack and num < stack[-1] and len(stack) + (len(nums) - i) > k:
            stack.pop()
        if len(stack) < k:
            stack.append(num)
    return list(stack)