LeetCode 14. Longest Common Prefix Solution Explained - Java

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[atefe]

No one explains solutions better than you, thank you so much for these videos, they are helping me a lot in the process of getting ready for technical interviews!

[joanmoralesf]

Its insane how short this video is but you explained it better than all the long fancy videos. Thank you

[jay-wfft]

I think it would be great if you would show 2 solutions. One with a slow run time and then the optimal solution such as this. It really helps to understand it even more. Either way awesome video

[prajwalshenoy]

Nice one. We can have a case inside for loop to break it when prefix length reaches zero to avoid continuing to loop when there's no scope of common prefix further.

[ananyadutta]

bro your explaining method is quite creative, out of 7 videos based on this that I'd WATCHED, this was the best!

[KhayamGondal]

you can add a check for prefix length ==0 at the end of the while loop. Incase nothing matches with the 2nd string then no need to check with rest of the list. And instead of substringing every itteration you can keep track of right index and only substring when returning the result.

[vcfirefox]

Dude, I am probably much older than you, but goddamn I am LOVING your explanations. Getting a chance to prep my interviews very efficiently!

[tanyagoel]

It was so nicely explained.
Thanks a lot!

[satyamgupta]

what a solution man thanks alot nick brother please continue making these videos.

u r just different when it comes to explaining in most optimised ways.

[bhargavvaddepally]

Damn man that explanation was so smooth

[poojaupadhyay]

such a clear approach! everytime i do leetcode and feel like wtf about a question, i come here to watch your tutorial and i understand the solution so clearly! thanks a lot, u have a great help dude!

[simrankak]

Why i didnt find your channel until now. Your explaination is crisp clear and very simple thankyou :)

[saradham]

Best solution for this problem . Great approach and clear explanation Thanks !!!

[kshitijpandey]

You wrapped that up very smoothly. Thanks Nick🙂.

[aadhi_dinesh]

IN PYTHON
s=["flow", "flew", "flower"]
k=s[0]
flag=0
a=""
for x in range(len(k)):
for j in s[1:]:
if k[x]!=j[x]:
flag=1
else:
a=""+k[x]
if flag==1:
break
else:
print(a, end="")

[greatred]

thanks, tried many things for c# but kept falling out of bounds. This works and simple to understand

[Ben-pbct]

Could you explain why the time complexity is O(S), where S is the sum of all characters in all strings?

[brucewayne]

Wow man. I could have never thought of approaching this problem this way. Thanks a ton👍

[ChetanKhairnar]

got through 3 different videos for a good solution but nick is one who always gives the best approach, Appreciate bro

[abhinav]

Kudos to such a clean explanation ! Thank you.