Introduction

In this blog post, we will explore how to solve the LeetCode problem "3111" 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 2D integer array points, where points[i] = [xi, yi]. You are also given an integer w. Your task is to cover all the given points with rectangles. Each rectangle has its lower end at some point (x1, 0) and its upper end at some point (x2, y2), where x1 <= x2, y2 >= 0, and the condition x2 - x1 <= w must be satisfied for each rectangle. A point is considered covered by a rectangle if it lies within or on the boundary of the rectangle. Return an integer denoting the minimum number of rectangles needed so that each point is covered by at least one rectangle. Note: A point may be covered by more than one rectangle. Example 1: Input: points = [[2,1],[1,0],[1,4],[1,8],[3,5],[4,6]], w = 1 Output: 2 Explanation: The image above shows one possible placement of rectangles to cover the points: A rectangle with a lower end at (1, 0) and its upper end at (2, 8) A rectangle with a lower end at (3, 0) and its upper end at (4, 8) Example 2: Input: points = [[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6]], w = 2 Output: 3 Explanation: The image above shows one possible placement of rectangles to cover the points: A rectangle with a lower end at (0, 0) and its upper end at (2, 2) A rectangle with a lower end at (3, 0) and its upper end at (5, 5) A rectangle with a lower end at (6, 0) and its upper end at (6, 6) Example 3: Input: points = [[2,3],[1,2]], w = 0 Output: 2 Explanation: The image above shows one possible placement of rectangles to cover the points: A rectangle with a lower end at (1, 0) and its upper end at (1, 2) A rectangle with a lower end at (2, 0) and its upper end at (2, 3) Constraints: 1 <= points.length <= 105 points[i].length == 2 0 <= xi == points[i][0] <= 109 0 <= yi == points[i][1] <= 109 0 <= w <= 109 All pairs (xi, yi) are distinct.

Explanation

• Greedy Approach: Sort the points based on their x-coordinates. Iterate through the sorted points. For each uncovered point, create a rectangle that covers it. Extend the rectangle's right edge as far as possible (within the width constraint w) to cover as many subsequent points as possible.

• Optimization: Because the points are pre-sorted and the width w is fixed, we can efficiently determine which points can be covered by a single rectangle without redundant checks.

• Complexity: Time: O(n log n), Space: O(n), where n is the number of points. The O(n log n) is dominated by sorting. The space complexity is for sorting and storing the sorted array of points.

Code

    def min_rectangles(points, w):
    """
    Calculates the minimum number of rectangles needed to cover all points.

    Args:
        points: A 2D integer array where points[i] = [xi, yi].
        w: The maximum width of a rectangle (x2 - x1 <= w).

    Returns:
        The minimum number of rectangles needed.
    """

    points.sort()  # Sort points by x-coordinate
    rectangles = 0
    covered = [False] * len(points)

    for i in range(len(points)):
        if not covered[i]:
            rectangles += 1
            x1 = points[i][0]
            x2 = x1 + w
            for j in range(i, len(points)):
                if points[j][0] <= x2:
                    covered[j] = True
                else:
                    break  # No need to check further points

    return rectangles