Python Programming Practice: LeetCode #6 -- ZigZag Conversion

This episode of Python Programming Practice shows two approaches to LeetCode #6 -- ZigZag Conversion

This is a medium difficulty problem.

Note that the intent of this series to help viewers think through approaches to problems, not to provide copy/paste access to code that will pass LeetCode tests.

[johnbanyan]

man.. ur making sense even to me ...i'm the guy who started coding from last two months...thanks for the content

[neotank]

You are an undiscovered treasure of youtube.

[karamanabdullah]

I know your views are very low but you are very precious for data science.

[fnfal]

Thanks man, your initial assesment made me create a solution before watching the real code in your video.

[devashreesharma]

I loved how you explained it so easily. Thank you so much :)

[sady_]

Man I thought it is 1B but it was 18 views
Wish you the best

[manh]

finally understood this zigzag problem's solution

[tubex]

i loved the way you approach to the problems solving in recursion - can you make a videos that will explain deeply in recursion pls

[vishalsinghsengar]

The content is awesome and deserving of more love

[yashm]

Best Technique !!! thanks for the video :)

[MJ-qrxw]

Was stuck at this problem for 3 hrs, now i feel foolish😅. Thanks for such a good explination

[Lumary]

brilliant solution, thanks for sharing! :)

[sidehat]

I wouldn't have bothered with the intermediate form. Easier just to change one string into the other. Less lines of code.
'''
def zigzag(in_s, row_count, col_count):
in_s_len, out_s, i = len(in_s), "", 0
for col in range(col_count):
for row in range(row_coun)t:
if col % 2 or not row % 2:
out_s += in_s[i]
i += 1
return out_s
'''
God coding on your phone is painful... Zero chance there isn't a bug there but hopefully you get the idea.

I just loop the rows and treat the even and odd ones differently.