Minimum Number of Arrows to Burst Balloons

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Minimum Number of Arrows to Burst Balloons: 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] = [x_start, x_end] denotes a balloon whose horizontal diameter stretches between x_start and x_end. 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 x_start and x_end is burst by an arrow shot at x if x_start <= x <= x_end. 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].

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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.

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Example 2:

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].

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Constraints:

• 1 <= points.length <= 10^5
• points[i].length == 2
• -2^31 <= x_start < x_end <= 2^31 - 1

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Try this Problem on your own or check similar problems:

1. Non-overlapping Intervals

Solution:

Java

public int findMinArrowShots(int[][] points) {
  Arrays.sort(points, (a,b) -> Integer.compare(a[1], b[1]));

  int[] point = points[0];
  int numberOfArrows = 1;

  for(int i = 1; i < points.length; ++i){
    if(points[i][0] > point[1]) {
      point[1] = points[i][1];
      ++numberOfArrows;
    }
  }

  return numberOfArrows;
}

JavaScript

/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function (points) {
  points.sort((a, b) => a[1] - b[1]);

  let point = points[0];
  let numberOfArrows = 1;

  for (let i = 1; i < points.length; i++) {
    if (points[i][0] > point[1]) {
      point[1] = points[i][1];
      numberOfArrows++;
    }
  }

  return numberOfArrows;
};

Python

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort(key=lambda x: x[1])

        point = points[0]
        numberOfArrows = 1

        for i in range(1, len(points)):
            if points[i][0] > point[1]:
                point[1] = points[i][1]
                numberOfArrows += 1

        return numberOfArrows

C++

class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        std::sort(points.begin(), points.end(), [](const std::vector<int>& a, const std::vector<int>& b) {
            return a[1] < b[1];
        });

        std::vector<int> point = points[0];
        int numberOfArrows = 1;

        for (int i = 1; i < points.size(); ++i) {
            if (points[i][0] > point[1]) {
                point[1] = points[i][1];
                ++numberOfArrows;
            }
        }

        return numberOfArrows;
    }
};

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Time/Space Complexity:

• Time Complexity: O(nlogn)
• Space Complexity: O(1)

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Explanation:

We have almost the same implementation as in Non-overlapping Intervals, we sort by the end time so we can accommodate more intervals (intervals finishing earlier will make space for the next intervals) and also since the end point decides intersection points as we move to the right it makes sense to sort the intervals by their end time. Each iteration we check if the intervals overlap, if they don't, we need more arrows and we also select the next interval for the next iteration while we continue checking for overlaps.