0452-minimum-number-of-arrows-to-burst-balloons

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Description

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

Solution(Python)

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        #  points = [[10,16],[2,8],[1,6],[7,12]]
        #   1           6
        #       2          8
        #                     10   16
        #                7       12
        # [[10,16],[2,8],[1,6],[7,12]]
        # sort by start
        #  1, 6 2,8, 7,12, 10, 16
        #  count = 1
        #  current_end track current winterval end
        # 
        #  check overlap by checking start of interval < next_interval end
        # if overlap   update current end 
        # else increase arrow and set current end as interval end
        if not points:
            return 0
        
        # Sort by x_start
        points.sort(key=lambda x: x[0])
        
        arrows = 1
        # Track the end of the current overlapping interval group
        current_end = points[0][1]
        
        for i in range(1, len(points)):
            x_start, x_end = points[i]
            
            # If the next balloon starts after the current overlap ends,
            # we must use a new arrow.
            if x_start > current_end:
                arrows += 1
                current_end = x_end
            else:
                # If they overlap, shrink the allowed end boundary 
                # to the minimum of both ends.
                current_end = min(current_end, x_end)
                
        return arrows

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