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 - 1The key idea behind solving #452 Minimum Number of Arrows to Burst Balloons is recognizing that overlapping balloon intervals can be burst using a single arrow. Each balloon is represented as an interval [start, end], and an arrow shot at position x will burst all balloons whose intervals contain x. This naturally leads to a greedy strategy.
A common approach is to first sort the intervals based on their ending coordinates. By doing this, you can place an arrow at the end of the first balloon and burst as many overlapping balloons as possible. As you iterate through the sorted intervals, you check whether the current balloon overlaps with the last arrow position. If it does not, a new arrow is required.
This greedy choice works because selecting the earliest possible ending position maximizes the chance of covering future intervals. The sorting step dominates the runtime, resulting in an efficient solution suitable for large inputs.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Greedy with Sorting (sort by end of intervals) | O(n log n) | O(1) or O(log n) depending on sorting implementation |
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This 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.
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.
1import java.util.Arrays;
2
3public class MinimumArrows {
4 public static int findMinArrows(int[][] points)
This Java solution employs the same greedy algorithm as before, sorting the balloons by their end positions and then checking overlaps.
Notice the use of Arrays.sort coupled with a comparator to sort the balloons by their end coordinates efficiently. The iteration over the array allows us to identify when a new arrow is necessary, based on overlaps or lack thereof relative to the 'end' point managed in each loop.
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.
Time Complexity: O(n log n) resulting from sorting operations.
Space Complexity: O(1) when sorting space considerations are minimized.
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Yes, interval-based greedy problems like this are commonly asked in coding interviews at top tech companies. They test a candidate's understanding of sorting, interval overlap, and greedy decision-making.
The problem mainly relies on arrays to store intervals and sorting algorithms to order them. No complex data structures are required beyond simple iteration and comparison of interval boundaries.
The optimal approach uses a greedy strategy combined with sorting. By sorting balloon intervals based on their end coordinates, you can place arrows at strategic points that burst the maximum number of overlapping balloons.
Sorting helps process balloon intervals in a structured order so you can always choose the earliest ending balloon first. This greedy choice ensures that one arrow can cover as many overlapping intervals as possible.
Python’s list-method sort facilitates this solution style by initial ordering on start points establishing an easy visual series.
The greediest overlaps stored by the `last_end` facilitates arrow accounting and minimizes overlapping discernment errors.