Optimizing Arrow Shots For Non-Overlapping Balloons
Sort balloons by end point, then iterate to find non-overlapping intervals. Count the number of intervals and return as minimum arrows needed.
Problem
class Solution {
public int findMinArrowShots(int[][] points) {
//remove all the overlapping baloons
//this will lead to only baloons that are not overlapping
//and we will need at min that many arros to burst them
//step1 : sort the give points in ascending order of Xend
Arrays.sort(points,(a,b)-> Integer.compare(a[1],b[1]));
//step2: get the count of non overlapping intervals and that will the no. of arrows we will need
int xstart = points[0][0];
int xend = points[0][1];
int count =1;// atleast one arrow i...