Solving Leetcode Interviews in Seconds with AI: Find All Good Indices
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_increasingandnon_decreasing.non_increasing[i]will be true if thekelements before indexi+kare non-increasing, andnon_decreasing[i]will be true if thekelements after indexiare non-decreasing. - Iterate through the possible good indices (from
kton - k - 1) and check if bothnon_increasing[i-k]andnon_decreasing[i]are true. - Collect the good indices into a result list and return it.
- Precompute two boolean arrays:
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