Introduction

In this blog post, we will explore how to solve the LeetCode problem "2334" 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

You are given an integer array nums and an integer threshold. Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k. Return the size of any such subarray. If there is no such subarray, return -1. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,3,4,3,1], threshold = 6 Output: 3 Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2. Note that this is the only valid subarray. Example 2: Input: nums = [6,5,6,5,8], threshold = 7 Output: 1 Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned. Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions. Therefore, 2, 3, 4, or 5 may also be returned. Constraints: 1 <= nums.length <= 105 1 <= nums[i], threshold <= 109

Explanation

• Iterate and Check: Iterate through the array using a sliding window approach. For each possible subarray length k, check if every element in the subarray is greater than threshold / k.
  • Early Exit: If a valid subarray is found, immediately return its length. No need to search further.
  • Optimize Iteration: Iterate k from n (length of array) down to 1. The problem statement only requires to return ANY such subarray, so by starting from the largest, we can return a result faster.

• Runtime Complexity: O(n^2), Storage Complexity: O(1)

Code

    def validSubarraySize(nums, threshold):
    n = len(nums)
    for k in range(n, 0, -1):
        req = threshold / k
        for i in range(n - k + 1):
            valid = True
            for j in range(i, i + k):
                if nums[j] <= req:
                    valid = False
                    break
            if valid:
                return k
    return -1