Count Unguarded Cells in the Grid - Leetcode 2257 - Python

[@XavierWoods-l9s]

Thanks to your leetcode daily videos and neetcode 150 list I finally got an internship opportunity at Amazon. Thanks again 🙏

[@bhanunani1307]

please give some suggestions for begineers

[@XavierWoods-l9s]

@bhanunani1307 I think he already made a very good video on the same I would suggest you to follow that and keep grinding leetcode.

[@debrajkundu2780]

He teaches DSA and take care of your eyes as well.

[@safirswe]

How about instead of marking a cell guarded, you mark it horizontally or vertically guarded, then when iterating through a guards position horizontally, you stop with all the previous criteria or if it's been horizontally guarded previously, then it'd mean it's guarded from another horizontal direction and we can stop. Same again vertically. I believe it'd speed things up, not enough to change the worst case time complexity, but enough.

[@fortitude242]

Hey @Neetcode, can you group problems by the patterns you have on neetcode, you doing great but just having a playlist by pattern will help us ramp up the prep process as we will only work through patterns which we think we need to work on. What do you think?

[@sdemji]

3:48 it can be done using binary search, the only thing is one have to invert,. "We reverse the question: iterate over all cells and for each one determine if it can be seen by some guard." Mikolaj Murasik. I agree it's annoying to code the mentioned approach but the one Mikolaj used is better tc wise, and is slightly less annoying although less straightforward than the ones from the editorial

[@vignarajd1223]

hey, have a question! In some of your videos, you mentioned that the time complexities shown on leetcode may not always be accurate or relevant, but here you used them to compare two approaches this means, leetcode shows us the correct time complexity?

[@rahulsihara8946]

i thought you will be on break, nice to see you though.

[@bhanunani1307]

i can understand the code ver easily but i am unable to crack the problem and do it myself any help?

[@jwsw]

House Robber V: Escape from jail

[@Akash-rg2oe]

I always wonder how this man solves these questions 2 weeks earlier before they become daily questions.

[@oppie335]

yo !!

[@luizfelipe2330]

I solved this problem differently. I actually used a DFS. Now, after seeing your video, I'm feeling dumb, hahaha.

[@pwn2424]

The first solution is more efficient than the second solution right?

[@floriankubiak7313]

It's sufficient to only mark as 0 or 1, later count the 0s.

[@floriankubiak7313]

14:02 you can save the position of the last wall in variable => only 1 pass

[@floriankubiak7313]

I'll dump my code here, because I prefer the way to iterate over the grids by using direction-arrays. I like the readability (I'm sure it can be better than the way I did it, advice appreciated) class Solution: def countUnguarded(self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]) -> int: grid = [[0] * n for i in range(m)] for w in walls: grid[w[0]][w[1]] = 1 for g in guards: grid[g[0]][g[1]] = 1 directions = [[-1,0],[1,0],[0,1],[0,-1]] for g in guards: for d in directions: y,x = g y += d[0] x += d[1] while 0

[@ilyasramatullaev7416]

thank you so much for making content for us

[@unlucky-777]

I solved it at 650ms in first try. We take those

[@rickastley4_the_second]

nice video

[@devmahad]

thanks :)

[@bhanunani1307]

please solve leetcode 493.Reverse Pairs

[@EdWestfieldJr]

`Time Limit Exceeded`

[@RC23032]

l=[] for i in range(m): li=[] for j in range(n): if [i,j] in guards: li.append('G') continue if [i,j] in walls: li.append('W') continue li.append(0) l.append(li) for i,j in guards: for x,y in [(-1,0),(1,0),(0,-1),(0,1)]: dx,dy=i+x,j+y while 0

[@bthulasikrishna9597]

Same here 😅. We are visiting the cells which are already visited.

[@RC23032]

@@bthulasikrishna9597 yea lol