# Greedy Java O(nlogn) with commented code.

• ``````public class Solution {
public int findMinArrowShots(int[][] points) {
if(points.length == 0) return 0;
//Sort according to x_start
//If x_start_i == x_start_j then sort according to x_end
Arrays.sort(points, new Comparator<int[]>(){
public int compare(int[] p1, int[] p2)
{
if(p1[0] == p2[0])
{
return p1[1] - p2[1];
}
return p1[0] - p2[0];
}
});
int count = 0;
//max point of current set of single arrow shootable balloons
int curMax = points[0][1];
//count number of single arrow shootable intervals
for(int i = 1; i < points.length; ++i)
{
//If we can't shoot current balloon and previous set of baloons using single arrow
if(curMax < points[i][0])
{
curMax = points[i][1];
++count;
}
else
{
//If we can shoot current balloon and previous balloons with one arrow
//reset limit of x_end of overall interval
curMax = Math.min(curMax, points[i][1]);
}

}
//Incerement here to account for last set balloons
++count;
return count;
}
}
``````

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