Rotate Image - Matrix - Leetcode 48

[@NeetCode]

🚀 neetcode.io/ - A better way to prepare for Coding Interviews

[@meowmaple]

This is a clear explanation, but definitely still not the simplest. For me, the most straightforward method is to transpose the matrix and reverse each row. The code is simple and short. #transpose for row in range(len(matrix)): for col in range(row,len(matrix)): temp = matrix[row][col] matrix[row][col] = matrix[col][row] matrix[col][row] = temp #reverse for row in matrix: row.reverse() Accepted by leetcode.

[@almasmirzakhmetov8590]

excellent solution. By the way your solution is based on rotation matrix, right? For 90 degree, we have (x,y) -> (y,x) en.wikipedia.org/wiki/Rotation_matrix

[@Rahul-pr1zr]

How do you even get the idea to transpose and then reverse? I agree that implementation is easier but the idea doesn't seem that simple.

[@333jjjjjj]

@@Rahul-pr1zr You need to recall it from your linear algebra class. Good luck if that was more than a few months ago or never.

[@peterpace3379]

@@333jjjjjj for me it was a whole year ago lmao

[@przemysawpobrotyn1195]

I have another solution in similar vein. I came up with it by eyeballing the leetcode provided input/output samples and noticing that the firs element of the last row of the input is the first element of the first row of the output, the second element of the last row of the input is the first element of the second row of the output etc. thus n = len(matrix) for row in matrix[::-1]: for i in range(n): element = row.pop(0) matrix[i].append(element) also does the job ;)

[@srinadhp]

This has been one of the toughest problems for me. Very hard to visualize and always used to make mistakes even after multiple attempts. The way that you explained the approach is THE BEST. You made it so crystal clear in visualizing the solution. Thank you so much!

[@sidkapoor9085]

I found it way easier, almost trivial when I stopped looking at the "2D matrix" and just at the input and output lists.

[@caniaccombo123]

@@sidkapoor9085 mind blown

[@markomekjavic]

I honestly think this is a Hard problem when it comes to implementation.. you can see the idea but coming up with the double pointer approach and a loop, thats a different story!

[@huansir1922]

@@markomekjavic yes,coming up with the double pointer approach , it seems hard

[@princeanthony8525]

Same here.

[@doublegdog]

Just got an offer at amazon. Your videos rock and helped me out so much!

[@NeetCode]

Congratulations 🎉

[@JustinBieber38524]

Hi hope you are doing good Can you please guide me how's your preparation looks like

[@xqfang3171]

This is the best explanation for this problem. Crystal clear visualization, elegant code. Great job. Thank you so much for posting!

[@NeetCode]

Happy it was helpful! :)

[@d1rtyharry378]

Bro what an structured approach . Really loved your way of teaching man! You made it look so easy.

[@NeetCode]

Thanks!

[@parthpatel8532]

Although this is a good way to do it, I found my way to be a bit simpler once you understand matrix manipulation. Rotating a matrix by 90⁰ is equivalent to flipping the matrix diagonally and then flipping it vertically. First try it out with paper, and once u get it, it's really easy. It doesn't save runtime or anything, but I find it easier in terms of code than to move 4 things at a time layer by layer.

[@jim5621]

Brilliant idea. But this solution takes 2x time since you need to loop through the matrix twice. But the time complexity is still O(n) though. Good thinking!

[@brainmaxxing1]

​@@jim5621 "This solution takes 2x time" isn't actually true. Because of things like cache locality, where the elements that are in the same row will be closer to operate on for the CPU, it's not possible to say that the element-wise method is faster. The profiling method actually worked about 25% faster from tests on my computer!

[@sanidhyax]

This is so elegant. Have solved multiple of 2d array problems but never thought of accessing the rows literally by [bottom[[R] and [top][L]

[@MsSkip60]

Thanks a lot for the content mate! No offence to others but I really like your clear accent and structured material which is easy to follow. Hope you keep up posting!

[@lottexy]

I gotta say your videos are amazing. I've been grinding LC for the past 3 weeks, I went from struggling to solve even 1 question on the weekly leetcode contest to solving 2 - 3 questions each week. Thank you so much. I've also and will always share your videos and excel sheet on reddit whenever people ask for leetcode tips. Oh and its abit late but congrats on the Google offer! I hope to one day get into google as well or any other company tbh ( my current tech job kinda blows ) ...

[@suraj8092]

Good luck!

[@mykytapiskarov7291]

Amazing solution and explanaition! I spent about 2 hours trying to understand leetcode "Rotate group of 4" solution - but no luck. Here 15 minutes - and it's clear.

[@vaibhavkhanna2922]

Great approach! Here you can see if you are confused with variable naming I have used some easy to understand names.Approach is still the same. void rotate(vector &matrix) { int size = matrix.size(); int startRow = 0; int startColumn = 0; int endRow = matrix.size() - 1; int endColumn = matrix.size() - 1; while (startRow < endRow && startColumn < endColumn) { int current_column_for_start_row = startColumn; int current_row_for_end_Column = startRow; int current_column_for_end_row = endColumn; int current_row_for_start_column = endRow; int current_size = endColumn - startColumn; for (int i = 0; i < current_size; i++) { int temp = matrix[startRow][current_column_for_start_row]; matrix[startRow][current_column_for_start_row] = matrix[current_row_for_start_column][startColumn]; matrix[current_row_for_start_column][startColumn] = matrix[endRow][current_column_for_end_row]; matrix[endRow][current_column_for_end_row] = matrix[current_row_for_end_Column][endColumn]; matrix[current_row_for_end_Column][endColumn] = temp; current_column_for_start_row++; current_row_for_end_Column++; current_column_for_end_row--; current_row_for_start_column--; } startRow++; startColumn++; endRow--; endColumn--; } }

[@suhasdon1]

Thank you. Best explanation without having to deal with 2 for loops with i, j or recursion and all other BS to be worried about.

[@emanawel9224]

This code makes the problem look way easier than it is! Love the code and explanation.

[@ArunRampure]

Have been struggling with coming up with an notion of using boundaries and using that i variable. Neet explanations for the neet code to write. Thanks for adding this.

[@mhdzahir423]

Broo! your explanation is fantastic! as a beginner, this whole and sole problem made me understand how to play with 2D matrix in any situation!

[@yabgdouglas6032]

my ego has made me attempt this problem almost 3 hours - thank you for this clean explanation!

[@wij8044]

Easier solution for nXn matrix! Neetcode solution may be better suited for nXm matrix. function rotate(matrix: number[][]): void { const n = matrix.length; // Transpose the matrix, starting from i = 1 for (let i = 1; i < n; i++) { for (let j = i; j < n; j++) { [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]]; } } // Reverse each row for (let i = 0; i < n; i++) { matrix[i].reverse(); } }

[@himanshubansal7040]

This is a beautiful way to write the code for this tricky problem. Kudos!!

[@DanielTruongDev]

Great explanation, however if you recalled from Linear Algebra, this is basically transpose the matrix so (row,col) becomes (col,row). So here's a shorter solution in Python rows = len(matrix) matrix.reverse() #Reverse the matrix for r in range(rows): for c in range(r,rows): matrix[r][c], matrix[c][r] = matrix[c][r], matrix[r][c] #Transpose

[@nayanachandran7072]

I must've have tried to understand this problem atleast 10 times and always failing to remember it. I now know I will never forget it! Thanks!

[@codewithtejesh1284]

thanks a lot, great explanation! i stopped my leetcode subscription , now i am just watching your videos and solving question.

[@ishaanjain4973]

I cant explain you how much this channel helps me !! Other channels just tell the transpose method which is not so intuitive, you always tell solutions which I can think in future in real interviews and exams. Thanks a lot Neetcode !! Keep up the good work man

[@sureshgarine]

I really like the way u handled minimizing the temp variable swap. very well explained. Thank you so much.

[@huimingli9207]

This is really a pure math problem. rotating a cell 90 degree, the index/coordinate change is from (x,y) -> (y, n-1-x).

[@jananiadanurrammohan9795]

Best explanation one could ever give for a problem!!!. Thank you for the effort and time you are putting into making all these videos.

[@krishnateja6428]

Cleanest explanation I have ever seen. Thank you!

[@josecarlosfontanesikling306]

It's possible to do this without any extra memory at all. Suppose you have variables x and y. You can swap them like this y=y+x x=y-x y=y-x This is all you need to transpose a matrix and to swap columns or rows. This rotation is just a transposition followed by inverting the order of the rows.

[@annieonee]

This is the best explanation of this problem so I've found. Thank you so much for the content! Keep up the good work 👏

[@anxonpues6018]

Good, explanation, improoving steps ... really golden pedagogical

[@AlfredPros]

While the LeetCode problem doesn't want the cheat solution, I tried it anyway and do the following code. def rotate(self, matrix: List[List[int]]) -> None: mat = list(zip(*matrix[::-1])) for j in range(len(matrix)): for i in range(len(matrix[0])): matrix[j][i] = mat[j][i] This code performs almost as fast as the solution with a caveat that it uses more memory.

[@dineshraj5957]

The best explanation by far of the layer rotation method. Damn it. The best!!

[@dr.merlot1532]

The guys talking about transpose mean this: Geometrically, rotation of the plane by 90 degrees is equivalent to flipping about the diagonal line y=-x and then flipping about the y-axis. That's why this works.

[@Cruzylife]

wow this explanation was so clear and the code was so clean!!

[@bruce716]

Thanks for the details and it is really easier to understand the concept with your good variable naming convention.

[@kwaminaessuahmensah8920]

Mans went into god mode swapping the elements in reverse

[@yasharma2301]

Knew the O(2*n^2) solution using transpose followed by inversion, thanks for this great one pass solution.

[@zekonja90]

C# solution: public void Rotate(int[][] matrix) { int n = matrix.Length; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int tmp = matrix[i][j]; matrix[i][j] = matrix[j][i]; matrix[j][i] = tmp; } Array.Reverse(matrix[i]); } } Cheers!

[@sumosam1980]

Your videos are excellent. You do a great job of being super clear! I often come here to Neetcode to see if you have the solution as it is better than the official explanation. Keep up the great work!

[@calvin_713]

Super duper helpful! This math question is tricky as hell...

[@bilalahmedkhan9518]

I thought this was a very though problem but you made it so easy for me. Thank you!

[@dollyvishwakarma2]

No doubt I love your simple algos but this one can be done in a much simpler way which is to reverse the matrix row wise and then swapping the elements like we do for a transpose. Kudos to the great work you do :)

[@user-xg2wj4dy5f]

But to do that you will have to create another matrix which is the copy of the Matrix which is to be transposed and that is against the constraints of the question you have to work in the same Matrix

[@welcomb]

There a very easy way to do this. First transpose the matrix, which is just flipping along a diagonal in-place. Then reflect vertically, which is another simple in-place swap.

[@Qxismylife]

I am thinking about the same. But the problem is we don't know if this is allowed, at least for the purpose of what this question is examining. If this is allowed then you can also use the 2D rotation matrix too.

[@Qxismylife]

Just saw other comments. This is allowed. I am happy now.

[@welcomb]

@@Qxismylife using a 2D rotation matrix may not be inplace as you need extra space for the matrix. Transpose and flip can both be done inplace

[@nayandhabarde]

Good work with comments and simplicity!

[@burburchacha]

I hope you realise that you have a gift in explaining difficult concepts

[@seungjunlee00]

Thank you so much for the straightforward and clear answer!

[@ElijahGeorge]

This is a very intuitive explanation. Thanks so much!!

[@jimmycheong7970]

This is a freaking amazing explanation. Thank you so much for sharing!

[@Ilmard99]

1) Invert the order of the rows 2) Transpose

[@KarthikChintalaOfficial]

Like the approach. But it's tedious to really remember and do it. I like the transposing and reversing using a 2 ptr approach to do this.

[@IncrementalNova]

Great content Bro :) Solution in Java: class Solution { public void rotate(int[][] matrix) { int n = matrix.length; int right=n-1, left=0; //neetcode solution video while(left

[@sebastian_tec]

@theraplounge because as you enter in the inner loops the offset, or the amout that you have to add/subtract for the rotation decreases. in the video for the outer matrix the offset can be as much as 2, but in the inner matrix that offset is 0.

[@inspiredomkar1239]

Thanks. I was looking for it.

[@pampanasubrahmanyam5693]

small correction, in the for loop, at the range function it should be range(r) not range(r-1) for python3

[@rohatgiy]

bro it’s an L

[@nikiforovsansanich]

Perfect explanation! Thank you!

[@leonfeng4006]

please continue to make more videos, this channel is pure gold

[@alinisar87]

Beautiful solution, great explanation. Thank you so much.

[@mr.anonymous6098]

Perfect Solution. When I first read the solution in the Cracking the coding interview book, I spent quite a lot of time and still could not understand it. You really simplified the logic which is so much easier to follow! Great Job, man

[@sanketkoli8641]

Very nice explanation. Thanks NeetCode!

[@chichiem2397]

These videos are great, this one in particular is perfect. I struggled with this a lot until I checked out this video. Awesome stuff!

[@Michael-zh3op]

After playing around with matrices, this question was a cake walk.

[@shivaacharya7247]

Great explanation! For some weird reason, for loop with range(l, r) does not pass all test cases!

[@markvaldez526]

My code isn't passing either...

[@shivaacharya7247]

@@markvaldez526 range(l, r) doesn't work because we are iterating r - l times, not from l to r. For example, if l = 5, and r = 10, we need to run our for loop 5 times starting index from 0, 1, 2, 3, and 4 NOT from 5, 6, 7, 8, 9 which would be the case if we do range(5, 10).

[@ugochukwustephenmbonu7974]

@@shivaacharya7247 thanks for explaining. Cleared it up for me!

[@siddharthgupta6162]

This video is so so so (X100) much better than the leetcode's solutions on this problem.

[@pengmick2046]

Your explanation is so good

[@itspete2444]

"This is still a square if you tilt your head enough" - that got me laughing harder than it should have

[@eltaylor778]

Great explanation, does anyone know why the range has to be range(right- left) instead of range(left,right)? The range(left,right) works on many of the smaller test cases but does not work on many of the harder ones. Help!

[@kritmok1875]

that's because we always want to start the i with 0. let's take (2, 4) as an example range(2,4) return 2,3 but range(4-2) return 0, 1

[@ShivangiSingh-wc3gk]

Nice, explanation. I thought the same thing but got thrown off on how to get the indexes

[@voxel9470]

Apparently you can rotate by transposing the matrix and then flipping each row. I've seen that solution be better than the way you illustrated here I'm assuming because it's more cache friendly in the flip phase.

[@Kenny-st8cg]

Since its python you can just do def rotate(self, matrix: List[List[int]]) -> None: n = len(matrix) l, r = 0, n - 1 while l < r: top, bottom = l, r for i in range(r - l): ( matrix[top + i][r], matrix[top][l + i], matrix[bottom - i][l], matrix[bottom][r - i] ) = ( matrix[top][l + i], matrix[bottom - i][l], matrix[bottom][r - i], matrix[top + i][r] ) r -= 1 l += 1 So theres no need for a temp variable, in that case we also wouldnt have to care about doing it in reverse. But I think readablitiy suffers a little

[@brainmaxxing1]

An even better solution to consider: n = len(matrix) # Transpose Matrix for i in range(n): for j in range(i, n): matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j] # Reverse Matrix for row in matrix: row.reverse()

[@Kenny-st8cg]

@@brainmaxxing1 Sure, but the TC of that solution is obviously much worse. For that solution theres no need for the second for loop either: n = len(matrix) for row in range(n): for col in range(row + 1, n): matrix[col][row], matrix[row][col] = matrix[row][col], matrix[col][row] matrix[row].reverse()

[@brainmaxxing1]

The time complexity of both solutions is the same, O(n^2). You do more 'operations' but the performance is actually better in my testing (somehow because of how the computer handles the more common operations of reversing a row and transposing a matrix) My notebook testing all 3 colab.research.google.com/drive/1laaMa1XLalAsShcoqENbbD5peb-PdCHo#scrollTo=QSThsieXZiCO Also nice on removing the second loop, that does speed it up.

[@ChocolateMilkCultLeader]

Your code was really elegant. Well done

[@UmaAbhinavHindu]

Great solution. Thanks

[@navenkumarduraisamy6260]

I really wish I had such clear approach

[@ashinshenoy84]

Very well explained, but i spotted one line which could be shifted up. Line 10 can be put before the for loop isnt it ? no need to initialize it on every loop

[@KaushikChavali]

I like the intuitively named variables.

[@akshaibaruah1720]

literally my go to channel..well was it really asked by microsoft?...I solved it by taking its transpose and then reversing it which is in place but needs two traversal

[@Tensor08]

I did this in one line 😎 which beats 97% matrix[:] = list(zip(*matrix[::-1]))

[@reaiswaryaa]

Wow what an amazing explanation. Thank you !

[@TheMzbac]

Very clearly explained, thank you

[@SarveshBhatnagar1214]

I found more two approaches to this, First one is to reverse the image and then transpose the resulting matrix (Easier). Other approach is to do this matrix[:] = zip(*matrix[::-1]) I don't really understand what's happening here so if someone understands, please do tell :) Thanks!

[@anandsrikumar007]

I took a temp array, copied all columns into the rows of the temp array but in reverse order, finally, I iterated the temp array and replaced matrix[row] = temp_array[row]

[@johnlocke4695]

There is no way I could have figured this out myself. Not even if they give me a million years time.

[@Bingo901]

You can transpose the matrix in place and reverse each row to achieve the same result

[@akifozkan5065]

made it look like so simple, great explanation

[@peterosi192]

Now this... I like!

[@maxwong4657]

I use JS. The same code but doesnt work! can't pass the case 6. I checked many times and asked GPT, GPT gave the correct answer, but leetcode didnt. Can anyone pass the case 6 with JS? var rotate = function (matrix) { let left = 0 let right = matrix.length - 1 while (left < right) { for (let i = left; i < right; i++) { let top = left let bottom = right const topLeft = matrix[top][left + i] matrix[top][left + i] = matrix[bottom - i][left] matrix[bottom - i][left] = matrix[bottom][right - i] matrix[bottom][right - i] = matrix[top + i][right] matrix[top + i][right] = topLeft } left++ right-- } };

[@AshishSarin1]

Thanks for this explanation. Really liked it.

[@NeetCode]

Thank you so much!! Glad it was helpful!

[@harishsn4866]

Thank you so much. You're the best.

[@alexeymelezhik6473]

"top" and "bottom" variables are not necessary, one could replace them by "l" and "r"

[@brainmaxxing1]

Sure, but why sacrifice readability? The difference is very small in terms of memory usage

[@rohitkumaram]

still your logic a more pythonic approach: class Solution: def rotate(self, matrix: List[List[int]]) -> None: l = 0 r = len(matrix) -1 while l

[@stith_pragya]

Thank You Brother for this amazing video.............🙏🙏🙏🙏🙏🙏

[@zekaiimran2464]

Best solution! thanks.

[@hunglam2680]

thank you. this problem has been bugging me, i figured out how to rotate the image but couldn't implement it properly.

[@franly656]

hi neetcode, i have a question, how to rotate matrix 45 degrees so the matrix size will increase, i trying to make the solution and now i want to give up (i have tried straight 5 hours btw). Can you make this solution a video pls.

[@preetirani8272]

List2 = [] for x in range(len(matrix[0])): List1=[] for y in matrix[::-1]: list1.append(y[×]) List2.append(list1) return list2 Why this code is giving wrong output in leetcode but right output in local jupyter notebook

[@nilkamalthakuria4030]

Great explanation

[@eltonlobo8697]

There's another approach to this problem: -Swap the non-diagonal elements (Because we want to make rows columns and vice-versa) -Use two pointers, one at the first column and the other at the last column, swap column elements -Do the above step until the first pointer is no more lesser than the second pointer (Last 2 steps are just to swap columns) Code: class Solution: def rotate(self, matrix: List[List[int]]) -> None: """ Do not return anything, modify matrix in-place instead. """ n=len(matrix) r=0 while (r

[@plotfi1]

I find the standard rotation technique to be pretty frustrating to think about. Instead ive found its easier to invert the values on the diagonal then reverse the rows.

[@samuelrobert2927]

Transpose and reverse in each row

[@khatharrmalkavian3306]

You know you can just swap instead of using a temporary. Walk through it: Swap TL with TR Swap TL with BR Swap TL with BL Shazam, the corners are rotated and aux space is O(0).

[@trongthuong9581]

How can u perform swap without temp variable xD

[@khatharrmalkavian3306]

@@trongthuong9581 There are two common methods and one other method that is CPU dependent: A) XOR swap: x ^= y y ^= x x ^= y B) Mathematical swap: x += y y = x - y x -= y C) x86 and x64 CPUs have the XCHG instruction, but it's likely to have the same performance as the other methods. Theoretically it would be faster since it can target a memory value directly, but doing so causes the CPU to lock all cores for synchronization, which is a massive performance hit. Operating just on registers, it has about 3 cycles of latency, which is similar to the XOR and mathematical methods.

[@karthikjayaraman9646]

Excellent article. Keep'em coming!

[@deewademai]

10:28 confuses me. range (L, r -1) does not the same as range (r -L)

[@jagrutitiwari2551]

The explanation was amazing. I find subtracting i a little confusing. Though dry run help me understand how it works. How do I remember it when I code?

[@Yougottacryforthis]

pen and paper