452. 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] = [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 <= 105
points[i].length == 2
-231 <= xstart < xend <= 231 - 1

Approach

First, let’s consider the following example:
Input: points = [[10,12],[2,8],[1,6],[7,12]].
It is clear that the balloons can be burst by 2 arrows. The first arrow can be shot within the range of 2 to 6.
For instance, shooting an arrow at x = 6 will burst the balloons [2,8] and [1,6].
The last arrow can be shot within the range of 10 to 12.
For example, shooting an arrow at x = 11 will burst the balloons [10,16] and [7,12].

However, it can be challenging to write the given array in a specific order.
To burst balloons, sort the points by their start index (the first index of the coordinates).

For example, if we want to burst a balloon with coordinates [1,6], the arrow can be shot within a range of 1 to 6.
If we want to burst two balloons with coordinates [1,6] and [2,8], which are overlapping, they can be burst by one arrow. The overlapping range is the shoot’s range, which is [2,6].

code

bool compare(vector<int> &a, vector<int> &b);
class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        sort(points.begin() , points.end() ,compare );
        int arrowEnd = points[0][1];
        int minNum =1;

        for(int i=0;i<points.size();i++)
        {
            if(arrowEnd>=points[i][0] )
            {
                arrowEnd = min(points[i][1] , arrowEnd);
            }
            else
            {
                minNum++;
                arrowEnd=points[i][1];
            }
        }
        return minNum;
    }
};

bool compare(vector<int> &a, vector<int> &b)
{
    if(a[0] != b[0])
    {
        return a[0]<b[0];
    }
    return a[1]<b[1];
}

leetcode 452. Minimum Number of Arrows to Burst Balloons

https://www.youtube.com/watch?v=_WIFehFkkig