Non-overlapping Intervals

Given an array of intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping
Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:
Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

[@mriduljain5801]

in both the cases there is one thing common to be pointed out that both cases that array should be sorted according to ending points
class Solution {
public:
static bool comp(vector a, vector b){
return a[1]

[@tridibeshmisra9426]

Thanks, ma'am for the simple explanation. I was stuck with the implementation for a long time.

[@MitrankShah]

Very very very good explanation! Thank you so much!!!

[@JohnWickea]

ek to itni cute ,uper se coder ...haaye jaan hi logi kya 😇🥰❤❤

[@ajaybabupatel1665]

Thanks for explaining in this very intutive way

[@GauravSingh-ku5xy]

Thanks. Very good explanation.

[@satyamshukla1418]

The leetcode description was not very clear to me so i misunderstood and applied Lis which ran few test cases but this was very helpful .

[@user-le3pr5mr1j]

Amazing Explanation !!

[@manasa1864]

Love (u) the way u teach ❤❤

[@amanshujaiswal4728]

great explanation

[@Raj10185]

very good explanation my code for same :-
sort(intervals.begin(),intervals.end());
int n = intervals.size();
vector temp(2,0);
temp[0]= intervals[0][0];
temp[1]= intervals[0][1];
int count =0;

for(int i = 1 ; i < n ; i++)
{


int currst = intervals[i][0];
int currend = intervals[i][1];


if( currst < temp[1]) //overlapping
{


count++;


//check for full merge overlapping then update the temp
if(currst > temp[0] && currend < temp[1])
{

temp[0]= currst;
temp[1]=currend;

}






}
else //non overlapping just update the temp
{

temp[0]= currst;
temp[1]=currend;



}







}


return count;



}

[@jayanthotwani6200]

tysm nicely explained

[@raushankumar6993]

Thanks a lot

[@ashvinkumhar5819]

Thank you diiiiiiii

[@samyakjain7422]

yrrr tum kitni cute ho...dayum beauty with brains....

[@none2868]

She wants me

[@shoryasinghal5241]

got it

[@ishikaagrawal5648]

in your first example you said intervals are not overlapping but they are overlapping so please modify that example otherwise students will get confused

[@Anonymous____________A721]

See whole video

[@tridevthakur7711]

Feeling Laughs with proverb:" Killing 1 bird with 2 stones "

[@Anonymous____________A721]

Cruel 😖

[@MilindGupta]

The first example which u took [1,5],[3,8],[10,15]. Here 3 is actually overlapping but u told it's not
Anyways u cleared in further video

[@probabilitycodingisfunis1]

Yes Milind, thanks for pointing it out, took the wrong example : )

[@vinamrasangal8436]

abe saale milind jab clear ho gya tha tujhe toh comment karne ki kyu khujli hori

[@shatnuu_]

mam are you alive 🙄

[@naveenkumarkota2090]

are u in a hurry to go somewhere why are u explaining it very fast focus on that

[@Anonymous____________A721]

😮