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Given a collection of balloons defined by their horizontal diameter start and end coordinates, the goal is to find the minimum number of arrows required to burst all the balloons. Each balloon is represented by a pair of integers, where points[i] = [start, end] denotes the horizontal start and end position of the ith balloon.

Example:

Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: The balloons can be burst with 2 arrows, one at position 6 (covering [1,6] and [2,8]) and another at position 12 (covering [7,12] and [10,16]).

Clarifying Questions:

  1. Q: Can the balloons have overlapping intervals or could they be fully nested within each other? A: Yes, the balloons can overlap and some balloons can be nested within others.
  2. Q: What are the constraints on the coordinates? A: The coordinates are integers, and the number of balloons can be quite large.

Strategy:

The problem can be approached using a greedy algorithm. To minimize the number of arrows, we should try to burst as many balloons as possible with one arrow. Here is the plan:

  1. Sort the Balloons: Sort the intervals based on their end coordinates. This will help in ensuring that once an arrow bursts a balloon, it is efficiently positioned to burst as many subsequent balloons as possible.

  2. Iterate through the Balloons: Begin by shooting an arrow at the end of the first balloon in the sorted list. Then continue to the next balloon and check if the current arrow can also burst it (i.e., if the balloon starts before or when the previous arrow was shot).

  3. Count Arrows: If the current arrow cannot burst the next balloon, shoot another arrow at the end of this new balloon, updating the number of arrows used.

Code:

def findMinArrowShots(points):
    if not points:
        return 0

    # Sort the balloons based on their end coordinates
    points.sort(key=lambda x: x[1])
    
    arrows = 1
    current_end = points[0][1]
    
    for i in range(1, len(points)):
        if points[i][0] > current_end:
            arrows += 1
            current_end = points[i][1]
    
    return arrows

# Example usage
points = [[10,16],[2,8],[1,6],[7,12]]
print(findMinArrowShots(points))  # Output: 2

Time Complexity:

Hence, the overall time complexity is (O(N \log N)).

This approach ensures that the minimum number of arrows required to burst all the balloons is computed efficiently.

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