Teaching Kids Programming – Minimum Number of Arrows to Burst Balloons (Greedy Algorithm) | ninjasquad

**Teaching Kids Programming**: Videos on **Data Structures and Algorithms**

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 <= 10^5

points[i].length == 2

-2^31 <= xstart < xend <= 2^31 – 1

### Minimum Number of Arrows to Burst Balloons (Greedy Algorithm)

The balloons can be treated as intervals, so we are looking for the number of connected intervals. One typical solution to interval problems is to sort them. In this case, we sort the intervals by the end point, and then we check if the current arrow can reach/burst next ballon. We iteratively update the current reach point and the number of needed arrows. This is greedy strategy.

The time complexity is determined by the sorting which takes O(NLogN). The space complexity is O(1) constant.

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class Solution: def findMinArrowShots(self, points: List[List[int]]) -> int: if not points: return 0 points.sort(key=lambda x: x[1]) ans = 1 end = points[0][1] for a, b in points[1:]: if end < a: end = b ans += 1 return ans |

class Solution: def findMinArrowShots(self, points: List[List[int]]) -> int: if not points: return 0 points.sort(key=lambda x: x[1]) ans = 1 end = points[0][1] for a, b in points[1:]: if end < a: end = b ans += 1 return ans

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