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Solving Leetcode Interviews in Seconds with AI: Find All Good Indices

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "2420" 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 a 0-indexed integer array nums of size n and a positive integer k. We call an index i in the range k <= i < n - k good if the following conditions are satisfied: The k elements that are just before the index i are in non-increasing order. The k elements that are just after the index i are in non-decreasing order. Return an array of all good indices sorted in increasing order. Example 1: Input: nums = [2,1,1,1,3,4,1], k = 2 Output: [2,3] Explanation: There are two good indices in the array: - Index 2. The subarray [2,1] is in non-increasing order, and the subarray [1,3] is in non-decreasing order. - Index 3. The subarray [1,1] is in non-increasing order, and the subarray [3,4] is in non-decreasing order. Note that the index 4 is not good because [4,1] is not non-decreasing. Example 2: Input: nums = [2,1,1,2], k = 2 Output: [] Explanation: There are no good indices in this array. Constraints: n == nums.length 3 <= n <= 105 1 <= nums[i] <= 106 1 <= k <= n / 2

Explanation

Here's the breakdown of the approach and the Python code:

  • High-Level Approach:

    • Precompute two boolean arrays: non_increasing and non_decreasing. non_increasing[i] will be true if the k elements before index i+k are non-increasing, and non_decreasing[i] will be true if the k elements after index i are non-decreasing.
    • Iterate through the possible good indices (from k to n - k - 1) and check if both non_increasing[i-k] and non_decreasing[i] are true.
    • Collect the good indices into a result list and return it.
  • Complexity:

    • Runtime: O(n)
    • Storage: O(n)

Code

    def goodIndices(nums, k):
    n = len(nums)
    non_increasing = [True] * n
    non_decreasing = [True] * n

    # Calculate non_increasing
    for i in range(k, n):
        if nums[i-1] < nums[i]:
            non_increasing[i] = False
        else:
            non_increasing[i] = non_increasing[i-1]

    # Calculate non_decreasing (from right to left)
    for i in range(n - k - 1, -1, -1):
        if nums[i+1] < nums[i]:
            non_decreasing[i] = False
        else:
            non_decreasing[i] = non_decreasing[i+1]

    result = []
    for i in range(k, n - k):
        if non_increasing[i] and non_decreasing[i]:
            result.append(i)

    return result

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