452. Minimum Number of Arrows to Burst Balloons

1. Description

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.

2. Example

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

3. Constraints

• 0 <= points.length <= $10^4$
• points[i].length == 2
• $-2^{31}$ <= xstart < xend <= $2^{31} - 1$

4. Solutions

My Accepted Solution

n = m_points.size()
Time complexity: O($nlog_2n$)
Space complexity: O(1)

class Solution
{
public:
    // int findMinArrowShots(vector<vector<int>>& points)
    int findMinArrowShots(vector<vector<int>> &m_points)
    {
        if(m_points.size().empty()) return 0;
        
        sort(m_points.begin(), m_points.end(), [](vector<int> left, vector<int> right){return left.back() < right.back();}); // sort as ballons' end

        int result = 1, arrowPosition = m_points.front().back();
        for(int i = 0; i < m_points.size(); i++)
        {
            // we try to use an arrow to burst as many ballons as possible
            // so we try to shot every ballon's end position, which means we are more likely to shoot the next ballon
            // if the next ballon's beginning can't touch our shoot position, we have to use another arrow
            if(m_points[i].front() > arrowPosition) 
            {
                arrowPosition = m_points[i].back();
                result++;
            }
        }

        return result;
    }
};