LeetCode Spiral Matrix Solution Explained - Java
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- čas přidán 23. 07. 2024
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Great.... Thumbs up to the clearity of your explanations!
Thanks this is great explanation and variable naming!
great job dood. keep doing this
very well explained especially descriptive variable names make it easier
Good one! It made it very easy! Thank you
Now I understand where those ifs come from. Thank you!
Same Here. THUMBS UP!
Of all the videos I watched so far, this one is well explained.
true ;)
thanks for helping me understand how to solve this problem!!
Great solution! Very intuitive.
I live how you explain so thoroughly and use brute force thats understandable and not too optimal to get! Thanks!!
you all probably dont give a shit but does anyone know a way to get back into an Instagram account..?
I was stupid lost my login password. I would love any tricks you can offer me.
@Brentley Randy instablaster :)
@Kole Fabian I really appreciate your reply. I found the site thru google and im trying it out now.
Takes quite some time so I will reply here later with my results.
@Kole Fabian It worked and I finally got access to my account again. I am so happy!
Thanks so much, you really help me out :D
@Brentley Randy You are welcome :D
So cool idea!
Thank you a lot for your efforts!!!
As always great work sir 👍
Awesome explanation ! Thanks
Great way to understand! Thank you!
what a nice explaination thanks a ton brother. please keep making such videos.
I dont know anything about coding but the coco powder almonds are really good! Thanks for the suggestion Nick!!
That was very understandable.. Thank you
Great Explanation ! dude
Great video! Thank you so much!!!
It's a fun problem. Thanks Nick 😊
Nick is the best!
Best explaination..thanx man
You are awesome bro.!! Love from India
I attempted to do it in a single loop, with some logic to decide what direction the spiral is moving. It works but took me a while and doesn't perform as well as the above.
Really great video. Helped a lot to understand the problem and look for a solution. I was able to solve it using the approach you showed in rotating the matrix.
class Solution {
fun spiralOrder(matrix: Array): List {
val result = mutableListOf()
var l = 0
var r = matrix.first().size - 1
val totalSize = matrix.size * matrix.first().size
while (l
but looks like your solution is much faster
I like how you use descriptive variable names instead of i and j
awesome solution
What is the time and space complexity for this solution?
Feel much better to solve problems while watching your video
can u explain the if check inside whike loop
woooow, it is very simple using boundaries
Can anyone, please explain the time complexity for this algorithm?
love it!
better than that in geeksforgeeks(where I had left comment to use self descriptive var names would do well) which was better than that of code school where I disliked the logic & var names
Thanks for this. Also, +1 on chocolate covered almond things.
It would be nice if you can post the solution in Github.
great work, what's the time complexity?
O(N) - N is number of elements the matrix
Gooooood!
oh my god.. I am just typing the code continuously after his explanation. It was a nightmare before watching this video. +1 for variable naming. That simply made to write flaw less code.
Thanks
I came back to thank you again orignally I was looking for a solution for spiral matrix 2 and your explanation was so good that I solved a different question with the help of your explanation in this video.
hwo to treaverse this spiral matrix in O(logn) time?
If you don't want to type the code, you can get it from here github.com/eMahtab/spiral-matrix
please give a solution in O(logn) time
please please
A similar and better solution
class Solution {
public List spiralOrder(int[][] matrix) {
List res = new ArrayList();
if(matrix.length == 0 || matrix[0].length == 0) return res;
int top = 0;
int bottom = matrix.length-1;
int left = 0;
int right = matrix[0].length-1;
while(true){
for(int i = left; i right || top > bottom) break;
for(int i = top; i right || top > bottom) break;
for(int i = right; i >= left; i--) res.add(matrix[bottom][i]);
bottom--;
if(left > right || top > bottom) break;
for(int i = bottom; i >= top; i--) res.add(matrix[i][left]);
left++;
if(left > right || top > bottom) break;
}
return res;
}
}