452. Minimum Number of Arrows to Burst Balloons | Delete overlapping intervals

[@fraserdab]

Sort from start point, update the end point based on minimum end point each iteration,8:08 if the 3rd point is within the range of 1st point but is lesser than 1st and 2nd point's minimum end point then it will be considered seperately for another arrow

[@UECAshutoshKumar]

Thank you 😊

[@abhaykumar9612]

bro it can also be solved by sorting on the basis of starting point

[@coderunner743]

thanks bro. today i solved my first interval based question by myself successfully because of your yesterday POTD explanation video.
here is my code based on sorting on the starting point
class Solution {
public int findMinArrowShots(int[][] points) {
int n = points.length;

Arrays.sort(points, (a, b) -> Integer.compare(a[0], b[0]));
int arrows = 0;
int i = 0;
while (i < n) {
int end = points[i][1];
int j = i + 1;
while (j < n && points[j][0]

[@atharvasuryawanshi7675]

Bhai faltu bakwas jyada karta he

[@nerdInsaan]

Just after seeing the number lines I got the logic , below is my code
class Solution {
public int findMinArrowShots(int[][] points) {
Arrays.sort(points, Comparator.comparingInt((int[] arr) -> arr[0]).thenComparingInt(arr -> arr[1]));
int itr = 0;
int len = points.length;
int [] prev = points[itr++];
int count = 1;
while(itr < len){
if(prev[1] >= points[itr][0]){
prev[0] = Math.min(prev[0],points[itr][0]);
prev[1] = Math.min(prev[1],points[itr][1]);
itr++;
} else {
prev = points[itr++];
count++;
}
}
return count;
}
}
P.S. we can also do the same by sorting by starting point with minimal change of taking min of ending of both the points ( If someone didn't get then use notebook and dry run then you got the point)

[@sivalokesh3997]

me being too lazy stored all these in ans[][] & returned it's size. 🙃