Magic Squares In Grid - Leetcode 840 - Python

[@yang5843]

You didn't think the Leetcode problems could get worse, yet here we are.

[@gmh14]

this week is just the worst problems week, there's no category, and leetcode knows it

[@NishaRathod-eu3rw]

Didn't feel like doing this problem, but streak :(

[@saishivamgupta4584]

same

[@jeffhappens1]

That second solution is awesome. Idk, I get excited seeing math solutions with these logic tricks. It's like finding a trap door. If this wasn't for leetcode, i'd be feeling more positively about this

[@MP-ny3ep]

Thank you so much for the daily leetcode.

[@gowtham_codes]

Me to me after going through 15:52: that’s cool, let’s settle for the FIRST OPTIMAL SOLUTION.

[@rambabupatidar3092]

the way you explained second approach seems a piece of cake now. Thank you very much🙌

[@parthdeshwal4419]

really?.... i'm so dumb then

[@satyamjha68]

Solved it!!

[@gary1614]

Thanks for the explanation! But I don't think the question mentioned that the sum of each row/col/diag is 15.

[@NeetCodeIO]

Aw I forgot to explain that part, since sum(1..9) = 45, and each row has to have the same sum, it must be 15. Same for columns and diagonals

[@ajitpalsingh606]

@@NeetCodeIO nice observation

[@hithambasheir3283]

Got the intuition and did a slightly different solution, did a few optimizations (with less readability) in the isMagic method
def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
if len(grid) < 3 or len(grid[0]) < 3:
return 0
def isMagic(r, c):
cols = [0 for _ in range(3)]
rows = [0 for _ in range(3)]
diags = [0 for _ in range(2)]
seen = set()
for i in range(r, r+3):
for j in range(c, c+3):
if grid[i][j] in seen or not (1

[@arnavchauhan9637]

Solution 1: 😀
Solution 2: 💀

[@arnavchauhan9637]

implementing the brute force approach gives
0ms Beats 100.00% in java wow
😂😂

[@Graveness4920]

simpler optimization is calc prefix sum of rows and cols then check sub matrix is 45 if yes run check for magic square

[@elliottkau7089]

In the second solution, won't it fail on a grid that follows the rotational pattern but puts the odds in the corners and evens in the edges? Should there also be a second condition on pattern checking that makes sure [r-1][c] is odd?

[@tunno4586]

Got a similar brute force Approach but nice explaination!

[@rajdippal51]

Second solution is just crazy 🤯🤯

[@chien-yuyeh9386]

🎉🎉 Nice one

[@jad_c]

this problem sucks

[@JamesBond-mq7pd]

Why? It's easiest problem.

[@galkk3]

I liked the math solution, hard to think about it but it's pretty nice

[@FahmiJemal-t7l]

we were challenged by this problem when we were a kid

[@sundew_ii]

two 20+ min solution vids in one week. god help us

[@NeetCodeIO]

Yeah aren't the problems supposed to be easy at the beginning of the month

[@Pegasus02Kr]

I think the problem itself is easy enough if you stick to the first solution.
A bit tedious but definitely much better than that Integer to English one...

[@InquisitiveLittleSteps]

Is the possible equal sum always 15? How can we arrive at that conclusion? Please Help...

[@michaelroditis1952]

today I got my first "beats 100%" with a runtime of 25 ms without watching this video!!

[@roderickli6505]

why do the rows cols and diag sum up to 15?

[@THK070]

That math solution is pretty nice, but hard to come up with.
Can anyone tell which tool neet uses for drawing explanation?

[@varunallagh7462]

can u tell whee is it written that sum should be 15

[@pastori2672]

i lost it when you started rotating and flipping the matrix xd (i am losing it)

[@YahyaNaeem-m7r]

Why you took 1 first instead of 2,3,4,5,6,7,8,9 and then made patterns on 14 ?

[@NeetCodeIO]

Just to go through them in order, it's arbitrary but easier to keep track of starting at 1.
We want them to sum to 15, hence 15 - 1 = 14

[@vim2741]

do u have any coupon code for lifetime acess pls

[@parthdeshwal4419]

i'm gonna settle with the first solution

[@gowtham_codes]

Raise your hands if your brain got rotated at 19:00 😂🙋‍♂️

[@zero180795]

first time i'm the first viewer PogChamp