Thank you for your patience and hard work for actually walking us through the recursive process. The recursion tree is clear and solid. You really know what you are teaching unlike most of the LeetCode RUclipsr out there who only copy and paste code. Thanks again!
I cannot thank you more. This is actually an interview question asked by Google. You made it look like very simple with your explanation. Tons of thank you, man!
You are actually explaining how to approach this problem unlike most of the leetcode youtuber( they just memorize the code and type it out). Thank you very much.
This is the best recursive tree video I have seen so far.Thank you so much SIR,now I can eat something,since I got it down.I was struggling to understand it for over a day.Thanks a million.
It's best ever!! Thank you so much for explaining step by step patiently lol. just hit subscribe but saw your latest update was 8 months ago, it would be great if you can keep going, your explanation is straightforward and super easy to understand, thank you!
this is the best illustration for recursion tree when called under a loop, every video just draw a two way or three way branch from beginning but this was exactly step by step , thnx
Today marks 3 years since you posted this video. Why have you stopped posting new videos? You are a very good teacher, Sir. You have a divine gift. Recursive thanks to you. PLEASE START POSTING NEW VIDEOS.
it was a piece of cake for you. And now for others too. Holy moly how well you explained, i cant describe it words, it was way to easy for us to understand what's going on. Thanks a ton mate. Cheers :)
@@mistercorea How does splice work internally? How do you say that splice is an O(1) algorithm? Splice itself cannot be O(1) but at the minimum has to be O(n).
Read the way of solving it using GeeksForGeeks, AlgoExpert, Leetcode prime solution explanation, couple of youtube videos. Didn't understand it from any, but you. Thank you.
Wow! You did a phenomenal job of explaining this in much simpler terms. Thank you so much for making this. Now, I understand this problem way better. I have one question though - Isn't the time complexity O(n * n!) from the tree diagram you illustrated? n times we are doing n! right, or can we reduce O(n * n!) to O(n!)?
You said we didn't use extra space in the video, but I think newNums is an extra space in each call. Space complexity will O(n*k) where n is the depth of tree and k is the extra space in each call. Please clarify.
we don't need to create a new list every time. We can use :- def permute(self, nums: List[int]) -> List[List[int]]: n = len(nums) ans = [] def rec(path): if not len(nums): ans.append(path[:]) return for i in nums[::-1]: nums.remove(i) path.append(i) rec(path) path.pop() nums.append(i) rec([]) return ans
I am really curios about what is really going on behind the scenes , for example when for the first time Permumation returns ,the next line set.pop is executed. But what was the next step? did it return back to the previous function call in the callStack( which seems it is) or did it just increased the i in for loop?
These things always make sense in hindsight once it's been explained, but do normal people actually come up with these crazy-ass solutions on their own?? Need to figure out what to take away from this conceptually so I actually learn something instead of just being able to do this exact problem if I see it again.
I guess this is how it works. Problem solving is all about reading the pattern, the more problems you solve the faster you can come up with an idea what algorithm you need to apply. You can say that we are memorizing it, but who isn't? That's why "Practice makes perfect" is a thing. I know I probably never gonna use it again when really working, unless I am on an algorithm research team, but this is how the game is played.
Hi. I've tried to make sense of this for a long time but I am still confused. Because the base case (the if statement) is checked for before the nums array is filtered, shouldn't there be one more step of the tree for just pushing the array into the answers. In the tree showed in the video, the last step does not meet the base case because nums is not empty when the base case is checked (nums === [3]). Could someone please clarify?
IN line 6 why are we doing this : answers.push([...set]) and why not simply this : answers.push(set)? set is an array and we can simply push the array into the answers array ?
I used Java to write this code. But I get something wrong, the result would be [1,2,3] only. What's the problem with my code here? And how to write it correctly in Java? public List permute(int[] nums) { if (nums == null || nums.length == 0) return new ArrayList(); List result = new ArrayList(); List list = new ArrayList(); for (int num: nums) list.add(num); List todo = new ArrayList(); backtrack(list, todo, result); return result; } private void backtrack(List list, List todo, List result) { if (list.size() == 0) { result.add(new ArrayList(todo)); } for (int i = 0; i < list.size(); i++) { todo.add(list.get(i)); list.remove(i); backtrack(list,todo,result); todo.remove(todo.size() - 1); } } Thanks a lot!
Good explanation but im not sure if its the best implementation in code? I've seen this problem solved using 2 functions with different # of parameters, what are the pros and cons of both implementations?
I used general backtracking logic to implement this. I don't know if its similar to what he is implementing as I only watched the logic. class Solution { List list=new ArrayList(); public List permute(int[] nums) { backtrack(new ArrayList(),nums); return list; } public void backtrack(List curr,int[] nums) { if(curr.size()==nums.length) { list.add(new ArrayList(curr)); } for(int i=0;i
This is the best recursive tree explanation I ever seen. Thank you
Thank you for your patience and hard work for actually walking us through the recursive process. The recursion tree is clear and solid. You really know what you are teaching unlike most of the LeetCode RUclipsr out there who only copy and paste code. Thanks again!
I cannot thank you more. This is actually an interview question asked by Google. You made it look like very simple with your explanation. Tons of thank you, man!
everything is simple with good approach
No one cared to explain how recursion is working inside the for loop. But you did it thanks a lot man!!
This was the first explanation of the Permutations problem I've found that makes sense. Thank you.
answers.push([...set]) instead of this : answers.push(set) solves the problem.
Can’t agree more!!!
That was the only video that helped me to really understand what happens in a backtracking solution.
Thank you sir
You are actually explaining how to approach this problem unlike most of the leetcode youtuber( they just memorize the code and type it out). Thank you very much.
This is the best recursive tree video I have seen so far.Thank you so much SIR,now I can eat something,since I got it down.I was struggling to understand it for over a day.Thanks a million.
It's best ever!! Thank you so much for explaining step by step patiently lol. just hit subscribe but saw your latest update was 8 months ago, it would be great if you can keep going, your explanation is straightforward and super easy to understand, thank you!
please start posting new videos, I watched loads of algothrim videos, yours are the ones that make more senses. and easier to understand, Thank you~~~
This is one of the best explanation of recursion I have come across, please keep up the good work.
this is the best illustration for recursion tree when called under a loop, every video just draw a two way or three way branch from beginning but this was exactly step by step , thnx
Thank andi guru garu, ee logic ardham kaaka oka roju antha chacchi poya!
Great solution and thinking, thanks!
Time Complexity Infinity is a great concept , great channel and some brilliant work here
Today marks 3 years since you posted this video. Why have you stopped posting new videos? You are a very good teacher, Sir. You have a divine gift. Recursive thanks to you.
PLEASE START POSTING NEW VIDEOS.
it was a piece of cake for you.
And now for others too. Holy moly how well you explained, i cant describe it words, it was way to easy for us to understand what's going on.
Thanks a ton mate. Cheers :)
This is it! I found the best recursion explanation. Good one 👍
dude, u have the best explanation for this question
Your use of filter to remove an element at an index is awesome. I've been using splice() with concat() but filter() is so much more elegant.
splice should be faster than the filter.
Wonderful recursive explanation !!! Thank you.
One of the best videos on recursion. Thanks!
A very good explanation. I had to watch it 5 times though lol. Thank you!
Man you are a life saver !
Beautiful explanation! Been looking for something like this for a while now!!!
way cleaner solution than whats on leetcode rn. good stuff man. thanks
Best explanation!!
Explained in such an easy manner. Thanks!
great explanation !
Nice explanation!
the removal of one element from an array itself is an O(n) complexity. That turns your solution into O(n*n*(n!))
What if you use a splice() function that will be O(1)? can you explain why it is O(n^2 * n!)?
@@mistercorea How does splice work internally? How do you say that splice is an O(1) algorithm? Splice itself cannot be O(1) but at the minimum has to be O(n).
Please add more LeetCode challenges.. your explanations are awesome. Thank you.
Read the way of solving it using GeeksForGeeks, AlgoExpert, Leetcode prime solution explanation, couple of youtube videos.
Didn't understand it from any, but you.
Thank you.
This video is very well done. Good Job. Man!
Thanks for the clear explanation. This is the best explanation.
The great explanation about tree and recursion.
Wow! You did a phenomenal job of explaining this in much simpler terms. Thank you so much for making this. Now, I understand this problem way better.
I have one question though - Isn't the time complexity O(n * n!) from the tree diagram you illustrated? n times we are doing n! right, or can we reduce O(n * n!) to O(n!)?
Love you bro
best explanation so far
Brilliant explaination. Please do more videos
Thank you for your explanation! It is pretty useful!
the best
You said we didn't use extra space in the video, but I think newNums is an extra space in each call. Space complexity will O(n*k) where n is the depth of tree and k is the extra space in each call. Please clarify.
Beautiful
Thanks :)
Great video!
you are best teacher so far
very helpful recursion tree visualization
Thanks
Finally I get it!! thank you king
this guy is a god
Awesome !
Great explanation. Thanks!
Very well explained video, thanks.
nice explanation,thanks man.
dude amazing explanation thanks
Best explanation thanks :)
Keep on teaching Man!
Thank you.
1:22 algorithm - 1:29 recursion
1:54 go through the recursion tree
2:40
11:46 code
best one
heap’s algorithm is another way I believe.
Cracking the code interview book says the time complexity of this algorithm is not n!
The book is right, in the video he didn't take into consideration the additional time required to copy into a new array while removing the item.
good explanation
we don't need to create a new list every time.
We can use :-
def permute(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
ans = []
def rec(path):
if not len(nums):
ans.append(path[:])
return
for i in nums[::-1]:
nums.remove(i)
path.append(i)
rec(path)
path.pop()
nums.append(i)
rec([])
return ans
I am really curios about what is really going on behind the scenes , for example when for the first time Permumation returns ,the next line set.pop is executed. But what was the next step? did it return back to the previous function call in the callStack( which seems it is) or did it just increased the i in for loop?
genius
These things always make sense in hindsight once it's been explained, but do normal people actually come up with these crazy-ass solutions on their own?? Need to figure out what to take away from this conceptually so I actually learn something instead of just being able to do this exact problem if I see it again.
I guess this is how it works. Problem solving is all about reading the pattern, the more problems you solve the faster you can come up with an idea what algorithm you need to apply. You can say that we are memorizing it, but who isn't? That's why "Practice makes perfect" is a thing. I know I probably never gonna use it again when really working, unless I am on an algorithm research team, but this is how the game is played.
Exactly What I've been thinking.how on Earth do these guys come up with such solutions.😂😂😂
Hi. I've tried to make sense of this for a long time but I am still confused. Because the base case (the if statement) is checked for before the nums array is filtered, shouldn't there be one more step of the tree for just pushing the array into the answers. In the tree showed in the video, the last step does not meet the base case because nums is not empty when the base case is checked (nums === [3]). Could someone please clarify?
Anyone know the space complexity?
Thanks in advance!
I believe O(n!) as well because the array containing the answers will have that many elements
Its O(n!). The amount of permutations that you make is n!. your array then is n! large.
thanks m8
num.filter has O(N) time complexity - your final complexity is even worse than O(N!). Be careful using this at interview!
IN line 6 why are we doing this : answers.push([...set]) and why not simply this : answers.push(set)? set is an array and we can simply push the array into the answers array ?
What is the space complexity for this
Would you mind explaining how this code will work in a stack frame?
😇
I had to play 0.75 speed. But so far its okay, will update after I complete video.
Very nice explanation. Can someone show how to do write this in Python?
is it allowed to add function parameters? not leaving just the "nums" one? I thought you are not allowed to do that....
I used Java to write this code. But I get something wrong, the result would be [1,2,3] only. What's the problem with my code here? And how to write it correctly in Java?
public List permute(int[] nums) {
if (nums == null || nums.length == 0) return new ArrayList();
List result = new ArrayList();
List list = new ArrayList();
for (int num: nums) list.add(num);
List todo = new ArrayList();
backtrack(list, todo, result);
return result;
}
private void backtrack(List list, List todo, List result) {
if (list.size() == 0) {
result.add(new ArrayList(todo));
}
for (int i = 0; i < list.size(); i++) {
todo.add(list.get(i));
list.remove(i);
backtrack(list,todo,result);
todo.remove(todo.size() - 1);
}
}
Thanks a lot!
Make more videos
Can somebody do this with java. ?
Why do we have to do set.pop()
Good explanation but im not sure if its the best implementation in code? I've seen this problem solved using 2 functions with different # of parameters, what are the pros and cons of both implementations?
Can someone give the java version of this?
Yes
Can any one please share code for Java ?
I used general backtracking logic to implement this. I don't know if its similar to what he is implementing as I only watched the logic.
class Solution {
List list=new ArrayList();
public List permute(int[] nums) {
backtrack(new ArrayList(),nums);
return list;
}
public void backtrack(List curr,int[] nums)
{
if(curr.size()==nums.length)
{
list.add(new ArrayList(curr));
}
for(int i=0;i
@@jagritbhupal5836 hey, can you please explain the code by chance?
Why poke yourself in the eyes and into the brain when learning a new concept whilst coding with Java will do it for you.
Can u please add java code?
you lost me 3 minutes in lol...