Introduction

In this blog post, we will explore how to solve the LeetCode problem "452" 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

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. You do not know the exact y-coordinates of the balloons. Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path. Given the array points, return the minimum number of arrows that must be shot to burst all balloons. Example 1: Input: points = [[10,16],[2,8],[1,6],[7,12]] Output: 2 Explanation: The balloons can be burst by 2 arrows: - Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6]. - Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12]. Example 2: Input: points = [[1,2],[3,4],[5,6],[7,8]] Output: 4 Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows. Example 3: Input: points = [[1,2],[2,3],[3,4],[4,5]] Output: 2 Explanation: The balloons can be burst by 2 arrows: - Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3]. - Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5]. Constraints: 1 <= points.length <= 105 points[i].length == 2 -231 <= xstart < xend <= 231 - 1

Explanation

Here's the solution:

• Sort by End: Sort the balloons based on their end coordinates. This is crucial for efficiently finding overlapping intervals.

• Greedy Approach: Iterate through the sorted balloons. If a balloon overlaps with the current arrow's range, it's burst by the same arrow. If it doesn't overlap, a new arrow is needed, and its range is set to the new balloon's interval.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) (excluding the input array and space used by sorting).

Code

    def findMinArrowShots(points):
    """
    Finds the minimum number of arrows needed to burst all balloons.

    Args:
        points: A list of balloon intervals.

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

    if not points:
        return 0

    # Sort the balloons by their end coordinates
    points.sort(key=lambda x: x[1])

    arrows = 1
    end = points[0][1]  # The end coordinate of the first balloon

    for i in range(1, len(points)):
        # If the current balloon's start is greater than the end of the previous arrow's range,
        # we need a new arrow.
        if points[i][0] > end:
            arrows += 1
            end = points[i][1]

    return arrows