Introduction

In this blog post, we will explore how to solve the LeetCode problem "1499" 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 array points containing the coordinates of points on a 2D plane, sorted by the x-values, where points[i] = [xi, yi] such that xi < xj for all 1 <= i < j <= points.length. You are also given an integer k. Return the maximum value of the equation yi + yj + |xi - xj| where |xi - xj| <= k and 1 <= i < j <= points.length. It is guaranteed that there exists at least one pair of points that satisfy the constraint |xi - xj| <= k. Example 1: Input: points = [[1,3],[2,0],[5,10],[6,-10]], k = 1 Output: 4 Explanation: The first two points satisfy the condition |xi - xj| <= 1 and if we calculate the equation we get 3 + 0 + |1 - 2| = 4. Third and fourth points also satisfy the condition and give a value of 10 + -10 + |5 - 6| = 1. No other pairs satisfy the condition, so we return the max of 4 and 1. Example 2: Input: points = [[0,0],[3,0],[9,2]], k = 3 Output: 3 Explanation: Only the first two points have an absolute difference of 3 or less in the x-values, and give the value of 0 + 0 + |0 - 3| = 3. Constraints: 2 <= points.length <= 105 points[i].length == 2 -108 <= xi, yi <= 108 0 <= k <= 2 * 108 xi < xj for all 1 <= i < j <= points.length xi form a strictly increasing sequence.

Explanation

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

• Sliding Window with Max Heap: The core idea is to maintain a "sliding window" of points such that for any point j, we only consider points i within a distance k to the left of j (i.e., xj - xi <= k). We use a max heap (priority queue) to efficiently track the maximum value of yi - xi for all points i within the current window.
• Heap Maintenance: As we iterate through the points (representing j), we add yi - xi to the heap. Simultaneously, we remove elements from the heap that are outside the k range, ensuring the heap only contains relevant i values.

• Maximization: For each j, the maximum value of yi + yj + |xi - xj| within the allowed range will be yj + xj + heap[0]. We track the global maximum as we iterate.

• Runtime Complexity: O(N log N) due to heap operations; Storage Complexity: O(N) in the worst case, to store the heap.

Code

    import heapq

def findMaxValueOfEquation(points, k):
    """
    Finds the maximum value of the equation yi + yj + |xi - xj| where |xi - xj| <= k.

    Args:
        points: A list of lists, where each inner list represents a point [x, y].
        k: An integer representing the maximum allowed difference between x values.

    Returns:
        The maximum value of the equation.
    """
    max_val = float('-inf')
    heap = []  # Max heap to store yi - xi

    for xj, yj in points:
        # Remove points outside the k range
        while heap and xj - heap[0][1] > k:
            heapq.heappop(heap)

        # Calculate the potential maximum value
        if heap:
            max_val = max(max_val, yj + xj + (-heap[0][0]))

        # Add the current point to the heap
        heapq.heappush(heap, (-(yj - xj), xj))

    return max_val