Размер видео: 1280 X 720853 X 480640 X 360
Показать панель управления
Автовоспроизведение
Автоповтор
Discord Link: 🔗discord.gg/aFBTauq
Sir Can you please give approach for this problem :Given Array of integers (N
explanation is super. plz share those kind of tricks with us. thank you.
Thanks 🙏
Thanks for the explanation man :)
Happy to help!
gr8 explanation can u tell that utkarsh sir video also, thank sir!
That was way back man, not really sure if i can find it now. Will ping the link here if i find it.
@@competitivecoding-newtonsc9601 okay sir thanks a lot!
Sir,can you explain this line cout
draw the graph,you will see that every second node you visit is a special nodefor example for visiting from a to b lets say n nodes are there that means n/2 special nodes will be there as well
please make video editorial for round 845 D
Thank man🔥🔥🔥
No worries ✌🏻
cout
dis[tgt] is the number of nodes on the path, not the edges. So it is always odd. And useful nodes on such a path are (dis[tgt] + 1)/2
Please provide solution code link also thanks
You can find it from my codeforces profile epsilon_573
Discord Link: 🔗
discord.gg/aFBTauq
Sir Can you please give approach for this problem :
Given Array of integers (N
explanation is super. plz share those kind of tricks with us. thank you.
Thanks 🙏
Thanks for the explanation man :)
Happy to help!
gr8 explanation can u tell that utkarsh sir video also, thank sir!
That was way back man, not really sure if i can find it now. Will ping the link here if i find it.
@@competitivecoding-newtonsc9601 okay sir thanks a lot!
Sir,
can you explain this line
cout
draw the graph,you will see that every second node you visit is a special nodefor example for visiting from a to b lets say n nodes are there that means n/2 special nodes will be there as well
please make video editorial for round 845 D
Thank man🔥🔥🔥
No worries ✌🏻
cout
dis[tgt] is the number of nodes on the path, not the edges. So it is always odd. And useful nodes on such a path are (dis[tgt] + 1)/2
Please provide solution code link also thanks
You can find it from my codeforces profile epsilon_573