Rotate Image - Matrix - Leetcode 48
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- čas přidán 8. 07. 2024
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Problem Link: neetcode.io/problems/rotate-m...
0:00 - Read the problem
3:25 - Drawing explanation
9:55 - Coding explanation
leetcode 48
This question was identified as a microsoft interview question from here: github.com/xizhengszhang/Leet...
#microsoft #matrix #python
Disclosure: Some of the links above may be affiliate links, from which I may earn a small commission. - Věda a technologie
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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.
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
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.
@@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.
@@333jjjjjj for me it was a whole year ago lmao
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 ;)
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!
I found it way easier, almost trivial when I stopped looking at the "2D matrix" and just at the input and output lists.
@@sidkapoor9085 mind blown
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!
@@markomekjavic yes,coming up with the double pointer approach , it seems hard
Same here.
Just got an offer at amazon. Your videos rock and helped me out so much!
Congratulations 🎉
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]
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!
This is the best explanation for this problem. Crystal clear visualization, elegant code. Great job. Thank you so much for posting!
Happy it was helpful! :)
Bro what an structured approach . Really loved your way of teaching man! You made it look so easy.
Thanks!
This is the best explanation of this problem so I've found. Thank you so much for the content! Keep up the good work 👏
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 ) ...
Good luck!
Best explanation one could ever give for a problem!!!. Thank you for the effort and time you are putting into making all these videos.
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!
I really like the way u handled minimizing the temp variable swap. very well explained. Thank you so much.
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
This code makes the problem look way easier than it is! Love the code and explanation.
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.
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!
@@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!
These videos are great, this one in particular is perfect. I struggled with this a lot until I checked out this video. Awesome stuff!
This is a beautiful way to write the code for this tricky problem. Kudos!!
Thank you so much for the straightforward and clear answer!
thanks a lot, great explanation!
i stopped my leetcode subscription , now i am just watching your videos and solving question.
This is a freaking amazing explanation. Thank you so much for sharing!
This is a very intuitive explanation. Thanks so much!!
Thanks for the details and it is really easier to understand the concept with your good variable naming convention.
wow this explanation was so clear and the code was so clean!!
Beautiful solution, great explanation. Thank you so much.
This is super useful, thanks for the great work!
Good work with comments and simplicity!
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.
Cleanest explanation I have ever seen. Thank you!
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!
Awesome Explanation. Just loved it. Keep up the good work :)
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.
Excellent article. Keep'em coming!
Amazing approach - very structured! and of course backed by great visualization!! You've got a subscriber, just based on this!! :)
The best explanation by far of the layer rotation method. Damn it. The best!!
Thanks for posting, this is great!
please continue to make more videos, this channel is pure gold
I thought this was a very though problem but you made it so easy for me. Thank you!
Perfect explanation! Thank you!
Very nice explanation. Thanks NeetCode!
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
love it. thank you!
After playing around with matrices, this question was a cake walk.
Good, explanation, improoving steps ... really golden pedagogical
Wow what an amazing explanation. Thank you !
made it look like so simple, great explanation
my ego has made me attempt this problem almost 3 hours - thank you for this clean explanation!
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--;
}
}
Your code was really elegant. Well done
Very clearly explained, thank you
This is the best explanation!!
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();
}
}
Thank you so much. You're the best.
Mans went into god mode swapping the elements in reverse
Thanks for such a clear explanation.
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 :)
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
Another great explanation my dude !
Knew the O(2*n^2) solution using transpose followed by inversion, thanks for this great one pass solution.
WOW, really neat!!
This is really a pure math problem. rotating a cell 90 degree, the index/coordinate change is from (x,y) -> (y, n-1-x).
The explanation was awesome and helpful to understand the problem effectively. I was waiting to see the final input output but nevertheless the code is correct so i guess i will write and check myself 😁😊
You really are the Bob Ross of LeetCode XD
Great explanation!!!
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
"This is still a square if you tilt your head enough" - that got me laughing harder than it should have
Best of the best solution!
Your explanation is so good
Here is a js solution
const rotate = (matrix) => {
let left = 0;
// number of columns - 1,
// also think, actual position of right
let right = matrix.length - 1;
while (left < right) {
for (let i = 0; i < right - left; i++) {
let top = left;
let bottom = right;
let 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--;
}
};
yes I know, py and JS are about as close to languages as you can get but for anyone new or may not understand the small differences between py and JS, here is your answer. NeetCode, thank you, your explanation was thorough and also less than 15 minutes which is great.
Great explanation
Thanks for this explanation. Really liked it.
Thank you so much!! Glad it was helpful!
I really wish I had such clear approach
Awesome !!!!
Nice, thanks for that
Beautiful code~
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
@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.
Thanks. I was looking for it.
fantastic teacher!
very intuitive
You are so genius!!
Thank You Brother for this amazing video.............🙏🙏🙏🙏🙏🙏
Nice, explanation. I thought the same thing but got thrown off on how to get the indexes
This shows me how many people in the comments don't even try the solution themselves.
I tried the solution in this video on leetcode and it doesn't pass the test cases.
I tried it, it passes.
Thanks for the clear explanation. I can understand it while I am in a food coma. lol
Great video. Just one question - why are the elements replaced in a counterclockwise manner? What benefit does it have over the more intuitive way of just replacing it clockwise?
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.
amazing !
Cool beans dude!
video is so helpful.
I hope you realise that you have a gift in explaining difficult concepts
This video is so so so (X100) much better than the leetcode's solutions on this problem.
Hey. Thank you for the explanation. Query: In your drawing, you portray the bottom pointer moving up and the top pointer moving down, so it's kinda confusing for me because in your code you write 'bottom - i' and 'top + i'. Mind explaining this distinction?
you are awesome!
There is no way I could have figured this out myself. Not even if they give me a million years time.
Thanks!
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.
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.
Just saw other comments. This is allowed. I am happy now.
@@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
I did this in one line 😎 which beats 97%
matrix[:] = list(zip(*matrix[::-1]))
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.
as usual, you made it easy man
awesome explanation.. thanks (Y)
beautiful code
You don't need 'top', 'bottom', 'left' and 'right' mambo jumbo!
Simplified transformation for clockwise rotation of [ X, Y ] is:
[ X, Y ] => [ Y, abs(X - len(matrix) - 1) ]
...and for counterclockwise:
[ X, Y ] => [ abs(Y - len(matrix) - 1), X ]
i.e. (switch coordinates and replace one with the absolute value of itself minus the length of the matrix minus 1)
Do all transformations from [ 0, 0 ] using TEMP variable, then walk diagonally toward the center... [ 1, 1 ], [ 2, 2 ], ...
Just to point out:
whenever you see a 2D, 3D or nD rotation or translation challenge, you know there is ONE rule that works for all the nodes. There are no special cases in rotation nor in translation. Always try to find the rule instead of breaking the matrix into small pieces -> i.e. more challenges.