Rotate Matrix/Image by 90 Degrees | Brute - Optimal

Problem Link: bit.ly/3Qk14gY
Notes/C++/Java/Python codes: takeuforward.org/data-structu...
We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.
Full Course: bit.ly/tufA2ZYt
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00:49 Problem statement
2:06 Observation
2:24 Brute force approach
6:25 Brute force code
7:49 Optimized approach
14:00 pseudo code for transposition
15:01 Optimized approach code

[@takeUforward]

Please watch our new video on the same topic: kzbin.info/www/bejne/kGG1Y6hsnMlmfbc

[@rainyvideo6936]

It's a loop same video link

[@rudraprasad8912]

The way you approach the problem.......It gives me the confidence that i can also do it!!!!!💀

[@takeUforward]

Timestamps pleaseee. Let's march ahead, and create an unmatchable DSA course! ❤ Use the problem links in the description.

[@iamvaibhav_10]

notes link is not visible

[@shra1]

Love you brother, marching ahead consistently.

[@aman_singh__]

TIMESTAMPS 00:49 Problem statement 2:06 Observation 2:24 Brute force approach 6:25 Brute force code 7:49 Optimized approach 14:00 pseudo code for transposition 15:01 Optimized approach code

[@riyadhossain1706]

How you elaborate on the problem and solution is unique to any other free content I have gone through. I'll surely gonna recommend your channel if somebody asks.

[@chinmay6152]

Understood. Thank you for this amazing content. I have tried many lectures but the way you approach the problem it seems extremely easy.

[@alessandrocamilleri1239]

Great explanation. I tried to come up with my solution prior to watching the video. I used the intuition of concentric squares within the matrix. I traverse one side of each concentric square and perform three swaps for each element . Since only one side of each concentric square is traversed, the number of elements traversed is approximately 1/4(m*n) and since there are 3 swaps for each element the time complexity will be O(3/4 m*n). Using a swap counter, Striver's solution is very close to O(m*n). However, unlike Striver, I do use some extra auxiliary constant space in the form of 4 pairs of co-ordinates which I use to determine the correct placing of the elements during swapping. void rotate(vector& m) { int l = 0; int h = m.size() - 1; pair a, b, c, d; while (l < h) { a = {l,l}; b = {l,h}; c = {h,h}; d = {h,l}; for (int i = l; i < h; i++) { swap (m[a.first][a.second], m[b.first][b.second]); swap (m[a.first][a.second], m[c.first][c.second]); swap (m[a.first][a.second], m[d.first][d.second]); a.second++; b.first++; c.second--; d.first--; } l++; h--; } }

[@Manishgupta200]

Best optimal explaination with in depth time complexity. Great. I do by myself with rotated by anti-clockwise and clockwise both. THankyou

[@umeshkaushik710]

Thanks a lot bhaiya. This time I must say you are on fire. Your explaining capability is next level, bez I had problems in understanding the matrix(index and all). But Now super clear. OP Striver Guru 🔥🔥🔥🔥

[@cinime]

Understood! Amazing explanation as always, thank you very much for your effort!!

[@mlkgpta2869]

Nice explaination, the best part is that you teach how to build your mind to think in that way.....

[@Krishnayadav-fu3uv]

very well understood, thank you for the great content ❤

[@nitinpatel9259]

Brother you are awesome. The way you give the solutions of the problems and it very helpful for me to explain whole code(dry and run).🙏🙏

[@ryanmathew6397]

so amazing watching your videos and getting to know how one should change there mind to observe the problem.

[@mariaonokhina4493]

Thank you for work you do. Really helpful!

[@primespeed121]

best DSA sheet ever you are the god of DSA really

[@abhijeetmishra3804]

Bhaiya ur amazing . How can one explain with soo much perfection man. Live long and keep making videos for us . Hope to meet you soon .

[@neilkapadia5407]

Understood! Amazing explanation!

[@codedByAyush]

Bhaiya, apka har solutions are just too OP and easy to understand 🔥🔥

[@jatinsharma1595]

Understood. Thank you Striver

[@sayantandey4708]

I did this optimal solution on my own, then came to see the solution video, this sheet building my confidence and skills little by little. (Rikon was my childhood friend. He worked for you some days back. No wonder why he praised you so much.)

[@utsavseth6573]

Beautifully explained.

[@ManognasaiSurineniManu]

Understood Thank you for this amazing lecture sir.

[@torishi82]

Samaj aa gaya bhai. Thank you.

[@rishipandey123]

Wow sir amazing and super easy explanation ❤❤❤😊

[@nikhilrajput5820]

thanks you sir, i easily understand how to transpose matrix inplace.

[@technicaldoubts5227]

understood very well !

[@satyasegu3566]

striver so happy to learn with youuh

[@harshilsutariya1793]

your dedication 🙌🙌

[@sarangkumarsingh7901]

Awesome Lecture Sir.................

[@konankikeerthi]

Understood bro. Thank you

[@_hulk748]

Great Explanation❤🙇‍♂✨🙏

[@tamilmukbang3789]

understood. thank you so much bro

[@NazeerBashaShaik]

Understood, thank you.

[@samuelfrank1369]

Understood. Thanks a lot

[@suyashshinde2971]

SDE Sheet Day 2 Problem 1 Done!

[@rahuljmd]

Understood, thanks💚

[@DesiGodOfWar]

great explanation

[@DeepakKumar-oz5ky]

Thank u Bhaiya Very helpFul video

[@HARSHA_27]

understood!!🙇‍♂

[@curs3m4rk]

I never thought, this question was that simple :( Understood Striver, Thanks

[@rajatpatra2593]

Understood very well

[@computer_tech98]

Thank you

[@RituSingh-ne1mk]

Understood!

[@infernogamer52]

Understood bhaiya!

[@sayanibiswas3741]

Thank you 🙏

[@mind9889]

I also came up with the transpose approach very happy 😁😁

[@CodingEnv]

I wish , I would have seen this video before makemytrip interview.. Thank you for great content.

[@aryanmandi7748]

they ask to complete the func in interview or just write pseudo code

[@mr_weird3680]

Thank you very much brother🙇

[@user-dv1ts9db8i]

UNDERSTOOD SIR

[@PankajSingh-pb4vs]

Understood ❤

[@ddevarapaga5134]

Superb UNderstood

[@heyOrca2711]

Understood! Sir

[@vidushibhardwaj1415]

thankyou so muxh for these videos❤❤... and the problem link added is a different question, but in your dsa sheet its same

[@anuragprasad6116]

I used the idea of concentric squares to solve the problem. Say, n = 6. Now the square will be of 3X3 size. You can draw a matrix to see how the outer square is of length = 6, inside it there's a square of side = 4 and inside it there's another square of side = 2. Basically each inside square is of 2 units lesser length then its outer square. We traverse from outside to inside and rotate each square one by one. For rotation, we traverse the upper side of the square and use 3 swaps for each grid. Also, the traversal is done till 2nd last grid because if you do the dry run, you'll notice that the last grid is already swapped in the first step, i.e., the corners are common between 2 given sides. The most difficult part is to deduce the co-ordinates for the replacing element. Imagine a square which you're traversing on its top side. Now, the top left element will be replaced by bottom left, bottom left by bottom right, bottom right by top right and top right by top left. It's hard to explain in a comment how I arrived at the co-ordinates but if someone wants to try this out, instead of swapping element by element, first try swapping row by row. I've attached codes for both. Basically, store the upper side of square in a temp array then replace top row with left column, replace left column with bottom row and so on. Once you understand how that's working, the co-ordinates for element by element swap is same but using lesser extra space. If someone needs a video explanation, do reply and I'll try to post a video explaining the same. // Swapping row by row: for (int i = 0; i < n/2; i++) { vector temp; for (int j = i; j < m-i; j++) temp.push_back(mat[i][j]); for (int j = m-i-1; j >= i; j--) mat[i][j] = mat[n-1-j][i]; for (int j = m-i-1; j >= i; j--) mat[n-1-j][i] = mat[n-1-i][m-1-j]; for (int j = m-i-1; j >= i; j--) mat[n-1-i][m-1-j] = mat[j][m-1-i]; for (int j = m-i-1; j >= i; j--) mat[j][m-1-i] = temp[j-i]; } // Swapping element by element int len = mat.size(); for (int i = 0; i < len/2; i++) { for (int j = i; j < (len-i-1); j++) { int temp = mat[i][j]; mat[i][j] = mat[len-1-j][i]; mat[len-1-j][i] = mat[len-1-i][len-1-j]; mat[len-1-i][len-1-j] = mat[j][len-1-i]; mat[j][len-1-i] = temp; } }

[@SatendraChauhan-ke5yr]

thankyou so much sir

[@user-tk2vg5jt3l]

Thank you bhaiya

[@AbhishekKumar-cv1dh]

Understoooood 🔥😁

[@shobhitsrivastava1223]

Understood❤

[@kushagramishra5638]

understood!

[@itsabhinav04]

Understood, thanks :)

[@per.seus._]

understood❤

[@rashiqajameel7822]

excellent!

[@aarishfaiz7880]

Sir app bhut accha Padhte hoo.

[@brajeshmohanty2558]

now I understood why bhai chose c++ over java because u have to write so many function in java but in c++ u have stl :( . But bro i understood the question thanku :b

[@user-is6ky7pp2n]

Understood !!

[@nagame859]

Understood 👍

[@anantsingh2004]

Understood 😊

[@user-ik3qu5uy5e]

UNDERSTOOD

[@karanhaldar755]

great job bro

[@her_soulmate]

Understood 🎉

[@user-uv5or5bm2c]

Understood.

[@sarthak4989]

understood ❤💘

[@arjit1495]

Thanks. We can further improve by not using loop for reversing of row in optimal instead just use reverse after j loop is finished like this: for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { swap(mat[i][j], mat[j][i]); } reverse(mat[i].begin(), mat[i].end()); }

[@nishant1456]

Understand❤️

[@rashi1662]

this is how I wrote the transpose code for(int i=0; i< n ; i++){ for(int j=i; j < n; j++){ swap(matrix[i][j], matrix[j][i]); } } which is less confusing

[@ast_karan128]

bro this code is not optimal, because it this code u will traverse through all elements, and the code in the video is not traversing the diagonal element which reduce the time complexity

[@rashi1662]

@@ast_karan128 thanks for pointing that out bro 🤜

[@culeforever5408]

understood 🚴‍♂

[@AzizurRahaman-zq4io]

Understood

[@Mohammad46552]

understood

[@DSAMADESIMPLE]

understood sir

[@shaheenparween4326]

understood✔✔

[@yamini436]

love u bro :)

[@samitkumar18]

Understand

[@ghayoorhussain8930]

Java Code: ```class Solution { public void rotate(int[][] matrix) { int n = matrix.length; // Transpose of Matrix for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int temp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = temp; } } // Reverse each row for (int i = 0; i < n; i++) { int left = 0, right = n - 1; while (left < right) { int temp = matrix[i][left]; matrix[i][left] = matrix[i][right]; matrix[i][right] = temp; left++; right--; } } } }```

[@ishanjindal9001]

thanks

[@khanra17]

9:54 We do transpose not because we need to convert rows into column. If we have a another matrix to store then we can do it directly instead of two steps. we do it so that we can swap elements. you can't do it directly. so do the extra step

[@anuragroy9045]

understood...

[@ksankethkumar7223]

TC for the first 2 nested for loops would be O(N*N/2) in my point of view?!

[@yhbarve]

Hey Striver, the problem link and the video link doesn't match in the SDE sheet, please get it updated...

[@rakshitrabugotra8354]

Hi bro! I came across this problem and the first two operations on my mind were transpose and reverse. As the problem requried it to be solved in O(1) space, I carefully examined if the sequence of these operations made any signifcant changes to the performance. What I did was to reverse the matrix (row-wise) first, then take a transpose. The first operation used n/2 iterations (for optimal reversing). The second operation used n*(n+1)/2 iterations (for optimal transpose). So the total number of iterations with optimization: (n(n + 1) + n)/2 = O(n^2 + n) = O(n^2) With transpose first and then reverse each row: (n(n+1) + n^2)/2 = O(n^2 + n^2) = O(n^2) It doesn't make a difference as our PCs are blazzingly fast, but I found it neat :) Thanks a lot! This series is amazing!♥♥

[@milanthakur4975]

❤

[@roshanparajuli2172]

@takeUforward, Does this works for non squared- matrix as well ??

[@ayushdhiman9378]

i think the time complexity of the optimal should be O(N) * O(N/2) + O(N) * O(N/2)

[@thelightoffight7881]

How is N/2 brother i don't understand can you explain

[@theornament]

I thought that too. For the first loop, we still have to go through all of the rows except last one, which would be O(n). Now, for each row you go through, you will go through n - (i + 1) columns, which theoretically means you go through the later half of our matrix. This is why it is calculated as n/2. Then, when we loop through the rows and reverse them, we say we go through n rows and for each row we have to go loop through it to reverse them, which with the algorithm provided by Striver, it takes n/2. So, the answer would be in fact O(n * n/2) + O(n * n/2), which in simpler terms is O(n^2/2).

[@HarchitGulati]

one doubt if we have to space optimise why are we not swapping the elements in the same brute force aprroach instead of creating extra dp

[@computerscience68]

Please upload soon next video

[@mdanashkhan5144]

at 14.33vu said j loop goes till n-1 and in code at 15 .26 u have wriiten j

[@abhisheksinghmehra9576]

Understood🔥🔥🔥

[@elmo4672]

trying to pass first year of college : ( thank you : )

[@AkOp-bf9vm]

i think swap part of optimal approach take time complexity of O(N * N/2) bcz first loop is running for n times and second N/2 times

[@nahidfaraji5069]

Pure Platinum.

[@divyanshthakur2026]

Happy Ram Navami Everyone 🚩