LeetCode 735. Asteroid Collision (Solution Explained)

[@pri18551]

your explanations are always so good.

[@sachin_yt]

wow! thank was refreshingly clear and concise! Thank you boss!

[@kaikuregami2859]

Great explanation. thank you nick

[@narayanbhat3279]

Very well explained, it would be better if the while true loop is only for the last case, in all the other cases it just performs an operation and breaks

[@lavanyam3224]

Not really, for this case: [-2, -2, 1, -2] after pushing the first three elements in stack, when you come for the last element, you would have to check all the three conditions again. If you've a solution where the while loop just runs for the last case, share here.

[@z08840]

C++ in place, time O(n^2), space O(1) : 1. simplest, but lots of array moving: vector asteroidCollision(vector& a) { for(int left = 0, right = 1; right < a.size();) { if(a[left] > 0 && a[right] < 0) { int d = a[left] + a[right]; if(d = 0) a.erase(a.begin() + right); if(left < 0) ++left, ++right; } else ++left, ++right; } return a; } 2. efficient: vector asteroidCollision(vector& a) { for(int left = 0, right = 1; right < a.size();) { if(a[left] > 0 && a[right] < 0) { int d = a[left] + a[right]; if(d = 0 && a[left] == 0; --left); } if(d >= 0) { a[right] = 0; ++right; } if(left < 0) left = right++; } else left = right++; } a.resize(stable_partition(a.begin(), a.end(), [](int n){ return n != 0; }) - a.begin()); return a; }

[@ahmedelsayed7762]

+Nick White Please can you solve 1371? Its really hard and it involves Bit masks

[@1murkeybadmayn]

For the first example, input: [5, 10, -5], why is the output not [10] since 5 and -5 are moving in opposite directions and are on the correct sides to meet? Both should explode leaving 10 alone, no? I think they should have a constraint saying the asteroids have to be adjacent.

[@EnterANameReal]

5 and 10 don't collide 10 and -5 collide, which destroys -5 that leaves the first 5, 10 which are going in the same direction so they never collide

[@pinesasyg9894]

Well done.. amazing intuition

[@wandered7]

Thanks dude.. That helps a lot

[@erajjoan4587]

can do it without a while loop. just decrement the for loop's i when the top of the stack is positive and about to be destroyed by a greater negative

[@god-speed03]

You could use while stack is not empty in inner while loop...

[@ugola1]

Thank you, amazing explaination.

[@damodaranm3044]

really well explained . in future videos pls do explain like this and dont pls rush up explain solwly . because begineers like us may get trouble in understanding your fast pace.

[@thewatcherlollol]

If you're a beginner you shouldn't be doing mediums..

[@shivraj8033]

thns for this solution been puzzling me for the whole day

[@chetanthapa4003]

Bro please make video on basic calculator 2 on leetcode

[@rohanreddymelachervu3498]

Very clear explanation :) Thank you!!

[@vaibhavgaware709]

Thanks dude.... nice explnation im going to subscribe your channel.

[@tibettenballs4962]

why not make it a heap?

[@mayankarcturus]

Amazing explanation, this video should go right into the LC discuss section. 👍🏻

[@AndresIpod1]

Is the time complexity O(n^2)?

[@ramazanchasygov3886]

O(n), In the worst case scenario it would be like O(2n) which is O(n). [1, 2, 3, -4] => 2n [1, 2, -3, 4] => 2(n-1) + 1 [1, -2, 3, 4] => 2(n-2) + 2 [-1, 2, 3, 4] => 2(n-3) + 3 [1, 2, 3, 4] => n

[@michaeldang8189]

The question violates physics!! :)