Kijiye Meeting Meeting, karte rahiye meeting meeting, halua meeting meeting

@ajaynaik2480

5 күн бұрын

🤣

Goat is back

Sir please start making videos on strings and stacks

Waiting for strings playlist

tysm sir

Yayyy!

Understood

Why is

understood

@takeUforward I have a question regarding this. Why minimum platform doesnt work here? I mean if I try to calculate minimum meeting rooms and then assign 1 meetings to each of one rooms and rest to just one room

std :: cout

How can u say that greedy will always work

Please also include code explanation in c++ which you used to do, that would be really helpful!

As you said sort the array based on meeting timing, and process the problem of sorting values, for example if the shortest meeting starts at 20 and other meeting timings are lesser than 20, how would this approach work?

@siddhantshukla2154

11 күн бұрын

Sorting on basis of ending times, not on the actual length.

Striver after eating mummy ke haath ka khaana abhi kitne video banane hai bana lenge sab hojayega ,welcome back

Simple Java Code For Only Meeting Count: class Meeting{ int start; int end; public Meeting(int s, int e){ this.start=s; this.end=e; } } class Solution { public static int maxMeetings(int start[], int end[], int n) { List meetings = new ArrayList(); for(int i=0; i m.end)); int lastEnd= -1; int count=0; for(Meeting meet: meetings){ if(meet.start > lastEnd){ lastEnd=meet.end; count++; } } return count; } }

Stacks

int maximumMeetings(vector &start, vector &end) { vector v; int n = start.size(); for(int i=0;i time) { ans++; time = item.second; } } return ans; } int this code , i dont need to store the position anywhere , just store both of those in a vector of pairs, i think this code is better than the given video

@after_dark_777

Ай бұрын

It's not better, as the space complexity is now O(n^2) rather than O(n) in the solution

@ghufran_khan_

6 күн бұрын

@@after_dark_777 bro space complexity is O(2n) not O(n^2)

a better code static bool cmp(pair a,pair b){ return a.second

Easy java solution: public static int maxMeetings(int start[], int end[], int n) { List list = new ArrayList(); for(int i=0;i lst.get(2))); int freeTime = 0; int count = 0; for(int i=0;i

static class meeting{ int start; int end; public meeting(int s,int e){ this.start=s; this.end=e; } } public static int maxMeetings(int start[], int end[], int n) { meeting m[]=new meeting[n]; for(int i=0;ia.end-b.end); int count=0; int prev=0; int s=0; int e=0; for(int i=0;im[prev].end){ // count++; // prev=i; // } if(m[i].start>e){ count++; e=m[i].end; } } return count; }

Understood