minimum-number-of-arrows-to-burst-balloons.py

"""

452. Minimum Number of Arrows to Burst Balloons
Medium

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

"""

# V0

# V1
# IDEA : GREEDY
# https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/editorial/
class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        if not points:
            return 0
        
        # sort by x_end
        points.sort(key = lambda x : x[1])
        
        arrows = 1
        first_end = points[0][1]
        for x_start, x_end in points:
            # if the current balloon starts after the end of another one,
            # one needs one more arrow
            if first_end < x_start:
                arrows += 1
                first_end = x_end
        
        return arrows

# V1 
# https://blog.csdn.net/fuxuemingzhu/article/details/79888836
# https://blog.csdn.net/MebiuW/article/details/53096708
class Solution(object):
    def findMinArrowShots(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        if not points: return 0
        points.sort(key = lambda x : x[1])
        curr_pos = points[0][1]
        ans = 1
        for i in range(len(points)):
            if curr_pos >= points[i][0]:
                continue
            curr_pos = points[i][1]
            ans += 1
        return ans

# V2 
# Time:  O(nlogn)
# Space: O(1)
class Solution(object):
    def findMinArrowShots(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        if not points:
            return 0

        points.sort()

        result = 0
        i = 0
        while i < len(points):
            j = i + 1
            right_bound = points[i][1]
            while j < len(points) and points[j][0] <= right_bound:
                right_bound = min(right_bound, points[j][1])
                j += 1
            result += 1
            i = j
        return result