Minimum Number of Arrows to Burst Balloons

Minimum Number of Arrows to Burst Balloons

There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.

Given an array points where points[i] = [xstart, xend], 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: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).

Example 2:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2

Example 4:

Input: points = [[1,2]]
Output: 1

Example 5:

Input: points = [[2,3],[2,3]]
Output: 1

Constraints:

• 0 <= points.length <= 104

• points[i].length == 2

• -231 <= xstart < xend <= 231 - 1

Solutions

🧠 Cpp

class Solution
{
    enum {START, END};

public:
    //O(n log n) solution, because of sorting
    //sort intervals by END, (wider first if ENDs are the same)
    //and check for overlaps with current->at(END) < next_iter->at(START)
    int findMinArrowShots(vector<vector<int>>& points)
    {
        //sort by END value, if equal: put wider intervals first
        std::sort(begin(points), end(points),
                  [](auto a, auto b)
                  {
                      return a[END] == b[END] ? 
                      a[START] > b[START] :
                      a[END] < b[END];
                  });

        int arrow_num = 0;
        for(auto current = begin(points); current != end(points);)
        {
            //count all intervals that overlaps with current
            //still O(n) because we moving forward in vector
            for(auto next_iter = next(current); ; ++next_iter)
            {
                if(next_iter == end(points) || 
                   current->at(END) < next_iter->at(START))
                {
                    arrow_num++;
                    current = next_iter;
                    break;
                }
                //else there is overlap with current
            }
        }

        return arrow_num;
    }
};