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 <= 105points[i].length == 2-231 <= xstart < xend <= 231 - 1This approach involves sorting the balloons by their end points. Once sorted, we shoot an arrow at the end point of the first balloon, and then continue moving through the list, checking if subsequent balloons overlap with the burst of the arrow. If a balloon does not overlap, we need an additional arrow for it.
This C solution uses standard libraries to sort a list of balloon intervals by their end points, ensuring we minimize the number of arrows used. Sorting ensures that at every step, we maximize the number of balloons burst by a single arrow.
The function compare is used with qsort to sort the intervals by their end points. Then, we maintain an `end` variable to store the position of the last arrow shot. We iterate through the intervals, and for each interval that does not overlap with the current `end`, we shoot a new arrow and update `end`.
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Time Complexity: O(n log n) due to the sorting step, where n is the number of balloons.
Space Complexity: O(1) if we ignore the space used for sorting.
This approach is adapted from the interval scheduling maximization pattern, where we attempt to find the maximum number of non-overlapping intervals. By sorting the intervals, we can focus on selecting the maximum number of compatible balloons.
The C implementation sorts balloons by their starting x coordinate to maximize the intervals which can overlap and thus require fewer arrows.
Overlapping intervals are managed by checking the end position of current overlaps. If overlaps occur, the continuation of overlapping checks determines the next 'strongly ending' balloon that indicates a need for a new arrow, thereby ensuring no excess arrows.
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Java
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C#
JavaScript
Time Complexity: O(n log n) resulting from sorting operations.
Space Complexity: O(1) when sorting space considerations are minimized.
| Approach | Complexity |
|---|---|
| Greedy Approach Using Sorting by End Points | Time Complexity: O(n log n) due to the sorting step, where n is the number of balloons. |
| Interval Scheduling Maximization | Time Complexity: O(n log n) resulting from sorting operations. |
Minimum Number of Arrows to Burst Balloons - Leetcode 452 - Python • NeetCodeIO • 23,295 views views
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