Longest Common Prefix (LeetCode 14) | Full solution with animations and examples | Study Algorithms

[@dhritiarora4147]

All the 15 mins were worth it. Great job!

[@nikoo28]

Thanks for the support and motivation

[@SuryaPandey-e2s]

You always provide the best, easiest, and simplest approaches. I watch your explanations and then try to write the code independently, and I succeed most of the time. Keep making such great content. It really helps us a lot.

[@nikoo28]

that is so nice to hear. I hope to keep bringing quality content for you guys.

[@PawanKamat-q3i]

I never thought of sorting when I tackled this problem, but this makes so much sense to sort it and compare only two strings instead of all. Loved the approach, shows me how much I still have to learn!

[@quantisticnumbers2633]

Definitely one of the easiest and best solutions that I came across. Thanks for the video!

[@ajaysaran6541]

When i saw the optimized solution, i really amazed. Thanks for helping in learning journey. 15mins completely worth it.

[@quartekstudio8127]

I was stucking this question for 4 days by watching many of videos but I can't find out the best way to answer this question. But you did and explain it really well. I know I'm not that good as problem solving but you are the one who can explain the concept from really hard to a really human can understanding really easy

[@officialdreamplayz]

how many questions u have conquered till now i m on 4th (easy)

[@quartekstudio8127]

@@officialdreamplayz just only 6-7. We are going slow because I think we are missing something or some concept to answer or design to answer the question. I think keep doing it will help us a lot. For me answering in Code war was really easy more than leetcode

[@nikoo28]

all the very best...and glad to see your progress

[@kunalkheeva]

your videos seem like bit longer, but its really very productive, I couldnot even skip it for a sec, just great work!! Thank you!

[@nikoo28]

I really try to go over the concept, rather than just giving away the code...the same you would be solving a problem in an interview. Thanks for your support. Much appreciated. 😄

[@englishpopop7694]

this breakdown made things much more clear and easy to comprehend

[@arjunmsit]

Very nicely explained. Your pace and explanation is super. I'm learning/revising DSA almost after 10 years and I should acknowledge that your videos are helping me a lot. Thank you very much for your efforts 🙏🙏

[@niteshpandey8207]

Esko dekh ye pta chala ki aise bhi question solve kr sakte hai.... keep posting

[@nikoo28]

😄

[@tashi7316]

So far one of the best solution I have seen

[@sumanthpai4587]

Very Articulately explained . with proper animations etc , really good job

[@HY-eu4ei]

tried a lot of videos and i would say this was the best explanation of all such a easy and elaborative explanation. thank you

[@nikoo28]

Glad you feel this way ♥️

[@enriquegrageda]

My bro, you are a great teacher!

[@sharcodes]

Love your Solution. Best Wishes for You. I Would Love to see your Channel Grow, why this is so Underrated.

[@nikoo28]

Thanks for your support and motivation. I really hope the youtube algorithm picks up the channel soon 😅

[@satyamgupta6030]

great teaching bhaiya. I always look for ur videos whenever I search for a solution of a leetcode question. No one comes close to you when it comes to explaining multiple approaches and finally solving with the most optimal approach . Thanks bhaiya.

[@nikoo28]

It's my pleasure

[@nginx.h]

bro i stuck at this problem and i dont understand how to solve this question but after watching some video i not getting it better but u solve this question with very simple approach. 👍👍 well done bro

[@subhamagarwal476]

addicted to your channel. You explain so simple and in easy way. Thanks a lot

[@candichiu7850]

Thank you! The BEST on KZbin. Super explanation.Keep up the great work.

[@kingop2679]

Your approach is really nice i didnt even think to sort array

[@Ishankulkarni]

The efficient solution isn't in fact the most optimal, sorting strings is not nlog(n) but the worst-case & average time is n^2log(n). Only in the best case where the first characters of the string do not match, the TC becomes O(long(n)).

[@nikoo28]

you are correct...happy to talk about an even more optimal solution if you have to bring up 🙂

[@sayednadirofficial]

This was asked to me in an interview

[@himadripaul7332]

excellent this channel will grow gradually ..no bakwas straight forward explanation ..

[@aryavardhan6807]

very nice explanation i am doing dsa in python but i am able to understand procedure very good coded own my own great 15 min worthy

[@CostaKazistov]

Great style of teaching algorithms. 10/10 👍

[@Prince-fv8mb]

loved the way you explained sir.....This channel is highly underrated

[@nikoo28]

Thanks for your support and motivation. I really hope the youtube algorithm picks up the channel soon 😅

[@PankajYadav-nz4gn]

Oh My GOD!!! I can't imagine that someone can be this good in explanation. Amazing and super thanks.

[@nikoo28]

Wow, thanks!

[@kirankutte7073]

Awesome explaination. I liked the slow and animated explainartion. Good job!

[@nikoo28]

Glad you liked it

[@kamleshkumarsingh5140]

SC will be O(n) right for using stringbuilder?? Great explanation though

[@nikoo28]

you are absolutely correct. My bad..the space complexity will be O(n)

[@nickjr9994]

Very good explanation brother. This solution was so precise and easy to understand. Thanks a ton

[@nikoo28]

glad you liked it.

[@egm5081]

man, what amazing explanation, thank you so much!

[@d54nirajthakare69]

The level of Explanation I saw today was amazing !!!

[@nikoo28]

So happy you felt that way

[@huquea]

OMG, you took teaching algorithm to next level

[@nikoo28]

So happy you feel this way 😄

[@spongebobsquarepants4576]

Interesting solution with sorting - but then you get n*log n. You can simply search the list for the shortest word and then compare it with all other words, thus making it O(n * k) where k is the length of the shortest word. Depending on the input you have in a real life scenario, your approach might be more efficient if the strings are very long, so if the length of the shoertest string k > n where n is number of values in the list, then your approach is more efficient. Anyway cool explanation thanks!

[@ronakpatel2826]

string longestCommonPrefix(vector& strs) { int mini=INT_MAX; int ind=-1; for(int i=0;istrs[i].size()){ mini=strs[i].size(); ind=i; } } string ans=strs[ind]; int j=mini-1; while(j>=0){ int k=0; //bool isOk=true; //cout

[@tejas_2540]

Awesome explanation

[@LuisGuarin-y4z]

You're such a good teacher. Thank you!

[@nikoo28]

Glad you think so!

[@mdrabby1259]

Very good explanation ❤😊

[@sandeepkarne8407]

iam ur new subscriber today nice that i found u in my early prep of time

[@nikoo28]

Hope you find the content helpful!

[@mohammadaftabansari6882]

Thanks for your guidance. Much appreciated !

[@ShiraBoruchovich]

Great explanation, but I would add a check for potential issue where an Index error can occur if the first and last strings in the strs array have different lengths. Like this: min_length = min(len(first), len(last)) for i in range(min_length)

[@nikoo28]

Sure, you can do that

[@erenyeager4452]

Hey, found a similar problem but no one had a good solution online. The idea is to find the longest palindromic prefix. Would help a lot if you can make a video on this. thanks for the videos. love them.

[@zack176]

Wow so you are finally tired of destroying world

[@raghavkaushal7561]

Thank you so much brother!!! Your explanation is extremely simple to understand and follow. Wish I had a teacher like you.

[@s.harishkumar3087]

I understand this problem with your explanation Easily❤

[@yokeshbl4133]

wonderful explanation

[@gopro12378]

Crystal clear concept thank you sir!

[@CodingJourney-sz3og]

i have a query , why we are comparing rhe first and the last string in the efficient solution after sorting the strings .. after sorting even first two strings can give the solution

[@officialdreamplayz]

clever clue deliver now compare 1st and 2nd 😂

[@rishidangi2978]

first and the last string will have the maximum possible difference, while first and second will have the minimum difference as they're sorted

[@ankanhaldar1130]

if your string str=["clue","clue","clear","clumsy "] then??

[@bivashroy2829]

If the string contain a mixture of capital and small letter then sorting can't be the solution... string commonPrefix(vector& strs,int n){ string s1 = strs[0]; int ans_length = s1.length(); for(int i = 0 ; i< n ; i++){ int j = 0 ; while(j

[@nafiskamran6900]

really good explanation

[@chakri2622]

this video made to subscriber to never loss hope and don't give up u will grow

[@saiyamjain6099]

Excellent explanation 🔥 keep making more videos loved it

[@rishidangi2978]

don't you think that for loop should also have the condition "&& i < last.lenght()" as if the first array has length greater, then it would also check that index char for the last array, which would result in "IndexArrayOutOfBonds" exception like the example: canada, car, czech (sorted) here the first element would have the length 6, so the for loop would also run for the value i = 5, but the last[5] doesn't exist

[@rishidangi2978]

ok so I came up with the answer, its because the traversal of the string would just stop before it hits index more than the last element, correct me if i'm wrong

[@nikoo28]

absolutely correct

[@ChaitanyaPimparkar]

keep them coming

[@jatinkumar4410]

Hello... Nice and clear explanation. But i have a doubt. Shouldn't the time complexity of efficient approach be: O(n*s*log(n)) (where n: length of array and s: average length of string) Instead of O(nlogn) (13:52) . If that is the case, then time complexity of brute force would be better than this approach. Isn't it? Also aren't we going to consider space complexity taken by sorting algo internally?

[@sasidharnaidu4507]

Yeah, you are right. The brute force approach is even less time consuming when the initial chars of the words won't match.

[@chocochips-as]

Hey Bro really thanks, wanna give a suggestion, please use dark color in ppts.

[@nikoo28]

i prefer while, as it gives an experience with whiteboarding as well. You will be doing a lot of whiteboarding in any organization. 😄

[@bxp_bass]

That optimization only works in python or other interpreted languages, where sorting is much cheaper than a loop. In golang or c or c++ it's much faster to do loops, so you don't need to spend O(n log n) on sorting

[@AvinashKumar-ps4tw]

Very Crisp and interesting explanation

[@aradhyashukla2813]

Explained a complex problem very easily 💯👏

[@keshavkaushlya8355]

if possible try to show the code for brute force approach also, this helps in understandig the problem in a much better way,btw doing great work,appreciated.

[@nikoo28]

thanks for your feedback and support :)...but usually coding for brute force is not desirable...i only do that in scenarios when you tweak a brute force approach to convert it to an optimal solution.

[@saurabhsaxena8470]

In actual interviews, should we just explain the brute force solution and code only the optimal solution or try to code both?@@nikoo28

[@aparna5778]

Thanks for the explanation. Was easy to follow.

[@animationTip]

I love the way of explanation.Thank u so much making such a video bhaiya .

[@shivanijain2192]

what is the time complexity of brute force. And Sorting must be also compare all the characters internally then just wondering how it is optimized solution

[@deepti8787]

Nice explanation

[@gokul6120]

But i thought Array.sort() will take time complexity of O( n k log(n) ) where n is size of arrays and k is the length of longest string in arrays to sort lexicographical order ... So may it would be too expensive to use it. Now i could store our data in tries format instead of array format. but if we have still the array format then we should use bruteforce approche couse it will take only O(n * k)

[@ankanhaldar1130]

Yess!! good observation !!

[@amolgawai010]

Great Explanation, small correction: Space complexity is O(n)

[@nikoo28]

we aren't using any extra space to arrive at the solution

[@aryasharma69]

@@nikoo28 but we are using string builder it has the property to store n elements so the overall space complexity will be O(n)

[@eyerahulk]

Nice explanation, thanks man.

[@VaibhavSingh_code]

Maza aayega ❤

[@debojitmandal8670]

But I think we can also do without sorting bcs ur comparing first and last word of the list and also how is the space complexity o(1) when ur appending Plus sorting itself will take o(n) Plus iteration of first word will take o(length of the word)

[@amitarajput1153]

Best Teacher. Please upload more videos. Super thanks

[@Chhooso]

Finally was able to understand it😭👍

[@mohdsameer2445]

very nicely explained bro

[@nikoo28]

🤝

[@SaumyaSharma007]

Thank you brother 🙌

[@LuckyAndDad]

Excellent narration

[@shivenderkumar4521]

Very Simple and clear explaination , Thank you ❤

[@DalisYn]

Thank you so much this one was a bit confusing

[@curiousvaita]

Amazing Solution. Thanks a lot.

[@muskancreativity7]

Best explanation 💜

[@nikoo28]

Glad that you feel this way :)

[@PranitKothari]

You explain nicely!

[@SMARTRIXDbztttmods]

*Excellent Work 🔥Thanks Sir*

[@denisgordiy5718]

Thank you for great explanation. IMHO more efficiently is not to use StringBulider but just return first.substring(0, index)

[@nikoo28]

The difference will be very minute. :)

[@palashbharati8818]

string longestCommonPrefix(vector& strs) { sort(strs.begin(), strs.end()); string st=strs[0], end=strs[strs.size()-1]; string ans; for(int i=0;i

[@oncloudnine4334]

bro can you explain why we go with first and last and explain why can't we go with first and second string

[@SouravCheck]

2:07, A is also a counted in prefix?

[@parulsaraogi6770]

Woww explained so well.. just loved it 🥰❤️

[@gozebe]

This is a great solution.

[@rishabhjain7651]

love to way of explaination 😍

[@kidoo1567]

Asusal nikil is on fire❤

[@rohanyadav5246]

Hi, great video. But after sorting, can't we compare the first and 2nd string in the array. That will also work ? or it will break in any case ?

[@nikoo28]

the first and second string will not give the longest common prefix.

[@likithganesh9598]

​@@nikoo28why it won't give? If it's not present in 2nd string it isn't considered for prefix right?? Could you pls explain

[@nikoo28]

Because you need the longest common prefix. Just the first and second gives you the longest prefix. We want a common prefix that is in all the strings.

[@salmarafiq7695]

Amazing explanation!!!

[@sahilbait2981]

Please make a video on Repeated String Match question

[@punamroy6639]

great explaination. Thank you

[@MaksymKondratiuk]

Thanks for the video. But in the end, it looks a bit confusing. The time complexity of the "brute force" approach should O(n * m) where m is the length of the string and in the case of the "efficient approach" we will have O(n * m * log n) which is slower than the previous one. Is it a case or did I miss something?

[@Ishankulkarni]

You are correct

[@vm1662]

Great explanation! Thank you. :)

[@DebsDigitalDiary]

You are amazing. ❤

[@nikoo28]

You're the best!

[@ahmed91372]

What is the time complexity of the brute force approach?

[@anshikagarg9485]

I can't think... Thanku for this optimal approch

[@munzirahmed6424]

#this code for #python. def pre(a): a.sort() first=a[0] last=a[-1] out="" for i in range(len(first)): if first[i] == last[i]: out+=first[i] else: break return out print(pre(["flower","flow","flight"]))

[@hemant8762]

Thank you so much sir

[@Abhishek-ji6qj]

#python:- x = ["flower", "flosing", "flowing"] prefix = "" for i in range(len(x[0])): flag = 0 temp = x[0][i] for j in range(1, len(x)): if not x[j][i] == temp: flag = 0 break else: flag = 1 if flag ==1: prefix+=temp else: pass print(prefix)

[@TemuriChitashvili]

Really good.😉

[@richieabbugamingff498]

in the time complexity y u dont count the complexity for the loop

[@ashish4k07]

Can O(n) be possible like finding the max lengthed word and least lengthed word and compare them just the doubt

[@nikoo28]

Try writing a pseudo code for it