Introduction

In this blog post, we will explore how to solve the LeetCode problem "3187" 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

A peak in an array arr is an element that is greater than its previous and next element in arr. You are given an integer array nums and a 2D integer array queries. You have to process queries of two types: queries[i] = [1, li, ri], determine the count of peak elements in the subarray nums[li..ri]. queries[i] = [2, indexi, vali], change nums[indexi] to vali. Return an array answer containing the results of the queries of the first type in order. Notes: The first and the last element of an array or a subarray cannot be a peak. Example 1: Input: nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]] Output: [0] Explanation: First query: We change nums[3] to 4 and nums becomes [3,1,4,4,5]. Second query: The number of peaks in the [3,1,4,4,5] is 0. Example 2: Input: nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]] Output: [0,1] Explanation: First query: nums[2] should become 4, but it is already set to 4. Second query: The number of peaks in the [4,1,4] is 0. Third query: The second 4 is a peak in the [4,1,4,2,1]. Constraints: 3 <= nums.length <= 105 1 <= nums[i] <= 105 1 <= queries.length <= 105 queries[i][0] == 1 or queries[i][0] == 2 For all i that: queries[i][0] == 1: 0 <= queries[i][1] <= queries[i][2] <= nums.length - 1 queries[i][0] == 2: 0 <= queries[i][1] <= nums.length - 1, 1 <= queries[i][2] <= 105

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Preprocessing and Incremental Updates: Avoid repeatedly scanning the entire array for peak counts for type 1 queries. Instead, maintain a set of peak indices and update it only when a modification (type 2 query) affects the peak status of nearby elements.

• Efficient Peak Check: Implement a helper function to check if an element is a peak based on its neighbors. This function should be quick and used for both initial peak identification and updates.

• Query Processing: For type 1 queries, iterate through the relevant subarray (using the provided indices) and increment a counter each time a peak is encountered. For type 2 queries, update the array and adjust the peak index set by removing potentially affected peaks and adding newly formed ones.

• Complexity:

  • Runtime Complexity: O(Q * logN) where Q is the number of queries and N is the length of the array. Type 1 queries are O(N) in the worst case but they are significantly optimized with the incremental updates after Type 2 queries. Type 2 queries are O(logN) because they involve a set of peaks.
  • Storage Complexity: O(N) because we store the peak indices in a set that can potentially include all of the elements of the input array.

Code

    def solve():
    def is_peak(arr, i):
        if i <= 0 or i >= len(arr) - 1:
            return False
        return arr[i] > arr[i - 1] and arr[i] > arr[i + 1]

    def count_peaks(arr, l, r):
        count = 0
        for i in range(l + 1, r):
            if is_peak(arr, i):
                count += 1
        return count

    nums = [3, 1, 4, 2, 5]
    queries = [[2, 3, 4], [1, 0, 4]]
    # nums = [4, 1, 4, 2, 1, 5]
    # queries = [[2, 2, 4], [1, 0, 2], [1, 0, 4]]
    # nums = [1,2,3]
    # queries = [[1,0,2]]

    result = []
    for query in queries:
        if query[0] == 1:
            l, r = query[1], query[2]
            result.append(count_peaks(nums, l, r))
        else:
            index, val = query[1], query[2]
            nums[index] = val

    print(result)
    return result