BS-24. Search in a 2D Matrix - I | Binary Search of 2D

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[@takeUforward]

Please watch our new video on the same topic: kzbin.info/www/bejne/gIm4ZXShm9lqr80

[@AbhishekSharma-ip4df]

bhaiya..kya approach..kya intuition..kya energy..explain krne ka tareeka..legit sb awsm hota h aapka..u make us feel dsa🔥🔥

[@prachigoyal2510]

A huge thanks Striver for putting all these efforts 🙏

[@princebhanushali6255]

Best DSA questions and approaches among all the DSA tutorials on KZbin. Big thankss❤❤

[@devkirandass7930]

Striver is the best...Feel like worshipping him🥰

[@SandeepM0ttan]

Try kunal kushwaha then

[@devkirandass7930]

@@SandeepM0ttanyeah I know him....idk but I felt like he is too arrogant ....and he teaches JAVA and I am into C++

[@cinime]

Understood! Super wonderful explanation as always, thank you very very much!!

[@sairam3351]

You are excellent teacher bro,no professor comes close, 💖

[@poonam-kamboj]

very nice and clearly explained , thanks for the good explanation!!

[@jhanavikumari274]

you are definitely doing some sorcery out there like how can you make learning this interesting .

[@hardikpatel352]

Understood! Such a great explanation as always, thank you very much for your effort!!

[@Gagansharma3002]

Bhai babbar ke to bus hype h asli udham to tumne machai h student ke learning me❤

[@aswinsrikanth408]

A great explanation as a whole bruh

[@harshhwardhanrai3716]

at 10:15 I literally started searching for the intuition for it but thanks for explaining it :)

[@theyashsawarkar]

very very helpfull explaination

[@hareshnayak7302]

Understood,thanks striver for this amazing video.

[@md.imrankhan6912]

Thank You Mr Legend

[@rushidesai2836]

Amazing solution

[@mrminus8590]

Thanks for everything brother

[@ayushagarwal5271]

coded it down myself! thanks!!

[@ashishpradhan6250]

amazing..thanku

[@iampatelajeet]

Thanks Bhaiya 💯❤

[@iceyyeah]

You go crazy love that dude

[@NazeerBashaShaik]

Understood, thank you.

[@ArpanChakraborty-do6yz]

done and dusted❤‍🔥❤‍🔥just awesome

[@shawshak2148]

understood thanks

[@user-tk2vg5jt3l]

Thank you Bhaiya

[@YourCodeVerse]

Understood✅🔥🔥

[@EC20022ELAKKIYAC]

Understood thank you

[@harshitpatil6138]

Understoood✌✌👍👍

[@ankanmalik409]

dada make a video on - Divide two integers without using multiplication, division and mod operator

[@chiragsharma8905]

My code with slightly more intuitive approach(according to me). O(log(n)+log(m)). Finding correct row first, and then finding target in that row. class Solution { public boolean searchMatrix(int[][] mat, int target) { int m = mat.length; if(m==0) return false; int lo = 0; int hi = m-1; while(lo

[@saswatrath4646]

great bhai same as my approach

[@srujangajjala2658]

log n + log m = log nm so time complexity is still same

[@RajNamdev_19]

@@srujangajjala2658 thanks for reminding me log formula I just forgot about it

[@arihantjammar8888]

understood

[@culeforever5408]

understood 😇

[@KartikeyTT]

tysm sir

[@senseiAree]

Understood ❤

[@yatintripathi5638]

Hey Striver , I somehow find this question and Median in a row-wise sorted Matrix a bit similar , but my approach gets wrong when i try to flatten the matrix to find the median. Is my approach wrong towards the question?

[@PrabhaKaran-36]

Understood❤

[@RituSingh-ne1mk]

Understood!

[@nikhilkumarjamwal5322]

It can also be done in O(n).

[@Ajeetkumar-uo4hf]

Striver Energy OP !!

[@artifice_abhi]

but its not necessary that if my target lies within the range it will surely be there for eg if an array is: 3 6 7 8 and target is 5 so now I would apply binary search as it is within my range but actually its not present. And this can happen for the rows where every time target is within the range but not present there. So in worst case time complexity would be: N*log(M)

[@harshilpatel3205]

No brother time complexity in the worst case stays as 0(n) + 0(log2(m)) ... Because once we find the answer we are returning the answer so it'll never go through each row...

[@ShahNawaz-cx3pi]

One more solution can be , first try to find out the correct row from 0th column (element in 0th column

[@kundann_n9989]

Can we not do a binary srach on i to look for the row that contains the target the do binary search on that row to look for target?

[@suryasaimaheswar8636]

understood :)

[@abhishekprasad010]

Understood

[@sarangkumarsingh7901]

Understood...............

[@harikakesharaju8799]

Same code u submitted is giving me TLE error for me sir

[@manikantachegu2182]

How about finding the row in which the element will be (by binary search ) and then applying binary search on the row can result in logn + logm crct me if im wrong because it worked for me in coding ninjas

[@aradhyapandey1489]

bhaiya...kitna bhayankar padhate ho🙏

[@graviton001]

Solved myself ❤

[@piyushmahale9024]

❤❤❤❤

[@lakshmanprabhu6722]

Which is lesser time complexity, log(m) + log(n) , or log(m*n) ?

[@rohitsai806]

both are same... log a + log b = log ab

[@dayashankarlakhotia4943]

Optimal approach. Public class solution { static boolean Search in matrix( Array list mat,int target){ int m=mat.size(),n=mat.get(0).size (); int low=0, high =m*n -1; while (low

[@sayantanpoddar5428]

Understood.............Understood!!!!! G.O.A.T

[@shashankgsharma0901]

Understood, but please take m x n instead of n x m, took i / m the whole time taking m x n, so got segmentation fault the entire time.

[@studyah]

bhaiya why did you take log of base 2?

[@saswatrath4646]

search space is reducing by half everytime we are doing mid = (low +high)/2. Anytime if you are reducing your working space by diving something your time complexity becomes log base to diviser. Had we been doing mid = (low + high)/5, just imagine it, our time complexity would have been log base 5. Hope it helps!

[@studyah]

@@saswatrath4646 thanks for explaining:) But I can't grasp why it happens...why the time complexity becomes log of base of the divisor

[@saswatrath4646]

@@studyah Search space is reducing from N to N/2 to N/4 to N/8 and so on.. why so ? bcz we are doing eliminating one half either left of right from the mid according to our situation. before you read any further 2^k means 2 raised to the power k. Here is the detailed explanation buddy. You can also write N, N/2, N/4, N/8..... as N/2^0 , N/2^1, N/2^2, N/2^3.....let's say it goes till 2/2^K. N/2^K = 1 hoga right ? kyuki last me 1 element == target hoga. Solve the equation, N = 2^k hog. Recall the formula of log and exponential or search it on google, if N = 2 ^ K then K = log base 2 of N. Now if you observer we are going from 1 to K right ? when we were at 1 we considered the whole array, when we were at k we comsidered only a single element. So, we moved from 1, 2, 3,...K. K number of iterations, now what is K that we found out above ? log base 2 of N. Time complexity will be O(k) =O(log base 2 of N) .

[@CodingJoySoul]

Please help Indians become candidate masters on codeforces.

[@rahulhembram4519]

UnderStood

[@aniketdas4358]

why am i getting runtime error with this code bool find(vector &arr,int col,int target) { int lo=0; int hi=col-1; while(lotarget) { hi=mid-1; } else if(arr[mid]

[@saswatrath4646]

when u r asking out do mention type or name of the error, it wil assist the person trying to help u out.

[@saswatrath4646]

change col to mat[i].size() and row to mat.size(), and write the col inside the for loop before calling bool function it will work perfectly fine.

[@nakulchauhan6713]

Your content and amount of effort is great. Huge respect for this course and you. But i feel you are not poking students to think, it appears as if you already know answer and here you are explaining it. Thought to process to come to a point should also be recorded. Lets say live solving DSA problems.

[@Fe-ironman]

You should watch this video only after going through the problems yourself

[@dayashankarlakhotia4943]

BF approach. Public class { static boolean search matrix (Array list mat,int target){ int m=mat.size(), n=mat.get(0).size (); for(int i=0;i

[@cenacr007]

us

[@pulamollavenkatreddy8162]

Can we do this O( log(n)+ log(m)) by eliminating rows

[@abhisheknag1929]

yes

[@rohitsai806]

log(m) + log(n) = log(m*n) as per logarithms.. So both time complexities are same

[@nayankhuman1043]

Understood ❤

[@abhinanda7049]

understood

[@jasmeenjz]

Understood

[@chiragbansod8252]

understood

[@KAMLESHSINGH-vr3bl]

understood

[@shaiksoofi3741]

understood

[@shashankvashishtha4454]

understood

[@arnabkundu1648]

understood

[@dewanandkumar8589]

Understood

[@sandeshgupta7950]

Understood

[@ganeshvhatkar9253]

Understood

[@RajNamdev_19]

Understood

[@yrzwisdom]

Understood

[@user-nb1ye5tf1r]

Understood

[@NARUTOUZUMAKI-bk4nx]

Understood