Problem

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.

Examples

Example 1:

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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:

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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:

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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

Solution

Method 1 – Greedy by Sorting End Points

Intuition

The key idea is to always shoot the next arrow at the earliest possible position that bursts the most balloons. By sorting balloons by their end coordinate, we can always shoot an arrow at the end of the current balloon, ensuring we burst as many overlapping balloons as possible with each arrow.

Approach

  1. Sort the balloons by their end coordinate in ascending order.
  2. Initialize a counter ans for arrows and a variable last to track the position of the last arrow shot (initially set to negative infinity).
  3. Iterate through each balloon:
    • If the start of the current balloon is after last, it means this balloon is not burst by the previous arrow.
    • Shoot a new arrow at the end of the current balloon, increment ans, and update last.
  4. Return the total number of arrows used.

This approach ensures each arrow bursts the maximum number of overlapping balloons.

Code

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class Solution {
public:
     int findMinArrowShots(vector<vector<int>>& points) {
          sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
                return a[1] < b[1];
          });
          int ans = 0;
          long long last = LLONG_MIN;
          for (const auto& p : points) {
                int a = p[0], b = p[1];
                if (a > last) {
                     ans++;
                     last = b;
                }
          }
          return ans;
     }
};
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public class Solution {
    public int FindMinArrowShots(int[][] points) {
        Array.Sort(points, (a, b) => a[1] < b[1] ? -1 : a[1] > b[1] ? 1 : 0);
        int ans = 0;
        long last = long.MinValue;
        foreach (var point in points) {
            if (point[0] > last) {
                ++ans;
                last = point[1];
            }
        }
        return ans;
    }
}
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func findMinArrowShots(points [][]int) (ans int) {
	sort.Slice(points, func(i, j int) bool { return points[i][1] < points[j][1] })
	last := -(1 << 60)
	for _, p := range points {
		a, b := p[0], p[1]
		if a > last {
			ans++
			last = b
		}
	}
	return
}
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class Solution {
     public int findMinArrowShots(int[][] points) {
          Arrays.sort(points, Comparator.comparingInt(a -> a[1]));
          int ans = 0;
          long last = Long.MIN_VALUE;
          for (int[] p : points) {
                int a = p[0], b = p[1];
                if (a > last) {
                     ans++;
                     last = b;
                }
          }
          return ans;
     }
}
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class Solution:
     def findMinArrowShots(self, points: list[list[int]]) -> int:
          points.sort(key=lambda x: x[1])
          ans = 0
          last = float('-inf')
          for a, b in points:
                if a > last:
                     ans += 1
                     last = b
          return ans
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function findMinArrowShots(points: number[][]): number {
    points.sort((a, b) => a[1] - b[1]);
    let ans = 0;
    let last = -Infinity;
    for (const [a, b] of points) {
        if (last < a) {
            ans++;
            last = b;
        }
    }
    return ans;
}

Complexity

  • Time: O(N log N) (for sorting, where N is the number of balloons)
  • Space: O(1) (ignoring sort space, or O(N) if sort is not in-place)