Covert 1D Array Into 2D Array - Leetcode 2022 - Python

0:00 - Read the problem
0:30 - Drawing Explanation
3:38 - Coding Explanation

leetcode 2022

#neetcode #leetcode #python

[ttlbyyi]

Listen man, I need these videos more than oxygen. Please don't leave too much

[mohamed]

bro please dont stop doing the daily challenge videos

[hemanth]

Hey Navi, there aren't many good explanations for LeetCode problems that are as understandable as yours. Please don't stop—you're helping a lot of people out there!

[DJSTEVE]

Thanks for the solution. Another approach would be to create a m * n matrix with default values and then loop through the matrix and fill it with the elements of original, this way you can avoid the calculations of start, end.

res = [ [ 0 for _ in range(n) ] for _ in range(m) ]
i = 0
for r in range(m):
for c in range(n):
res[r][c] = original[i]
i += 1
return res

[prajwals]

welcome back... and thanks for the vid

[anirudhbhat]

Saw this video and got my reason to get out of bed 😂

[pushkarsaini]

Please don't stop making videos for daily problems, espacially for hard ones, there are not much good explainations other than yours

[abhimanyuambastha]

Hey neetcode please don’t skip daily solutions for hard problems, not many good solutions out there

[chinmay]

The day I realised for some problems your solution is better most of the time than leetcode’s own editorials, it has become mandatory to look at your solution after i solve it with my approach.without that i feel uneasy.

My humble request, don’t stop the daily videos!!!

[SUN_HAO_LUN]

Why the time complexity is O(n*m) and space complexity is O(n)? I think that it should be O(m) for time complexity and O(1) for space complexity (if we ignore res. array).

[Provehitum]

It ain't much in terms of time complexity, but it's honest work.
return [original[i:i+n] for i in range(0, len(original), n)] if len(original) == n * m else []

[varun]

My man for few days dodging hard and medium problems

[viveksoni]

Where were you man from last few days?

[MohitSoni-ze]

I DID THIS:

return np.array(original, (-1, n)) if m*n == len(original) else return [ ]

[mohammedsuhail.s]

last two weeks was so frustrated because of daily questions are hard and also u didnt explain the que

[samuelsolomon]

I literally just solved this problem🤣
FYI-This is today's daily problem

[pastori]

it does seem like your avoiding hard problems