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!
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
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!
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 :)
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.
@@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).
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.
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!)?
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.
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?
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
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 ?
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.
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!
No one cared to explain how recursion is working inside the for loop. But you did it thanks a lot man!!
That was the only video that helped me to really understand what happens in a backtracking solution.
Thank you sir
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
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
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!!!
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.
Thank andi guru garu, ee logic ardham kaaka oka roju antha chacchi poya!
This is one of the best explanation of recursion I have come across, please keep up the good work.
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~~~
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!
Wonderful recursive explanation !!! Thank you.
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 :)
Great solution and thinking, thanks!
1:22 algorithm - 1:29 recursion
1:54 go through the recursion tree
2:40
11:46 code
Time Complexity Infinity is a great concept , great channel and some brilliant work here
dude, u have the best explanation for this question
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.
way cleaner solution than whats on leetcode rn. good stuff man. thanks
This is it! I found the best recursion explanation. Good one 👍
Man you are a life saver !
One of the best videos on recursion. Thanks!
This video is very well done. Good Job. Man!
A very good explanation. I had to watch it 5 times though lol. Thank you!
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.
Please add more LeetCode challenges.. your explanations are awesome. Thank you.
Beautiful explanation! Been looking for something like this for a while now!!!
Brilliant explaination. Please do more videos
Thanks for the clear explanation. This is the best 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).
Explained in such an easy manner. Thanks!
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.
best explanation so far
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!)?
Awesome !
The great explanation about tree and recursion.
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.
Nice explanation!
you are best teacher so far
very helpful recursion tree visualization
great explanation !
Great video!
Best explanation!!
this guy is a god
Beautiful
Finally I get it!! thank you king
Love you bro
the best
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?
Thank you for your explanation! It is pretty useful!
heap’s algorithm is another way I believe.
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
What is the space complexity for this
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.
Great explanation. Thanks!
nice explanation,thanks man.
Very well explained video, thanks.
Would you mind explaining how this code will work in a stack frame?
dude amazing explanation thanks
Thanks :)
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.
I had to play 0.75 speed. But so far its okay, will update after I complete video.
Thank you.
Best explanation thanks :)
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 ?
num.filter has O(N) time complexity - your final complexity is even worse than O(N!). Be careful using this at interview!
Keep on teaching Man!
Thanks
best one
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....
good explanation
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.😂😂😂
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!
genius
Why do we have to do set.pop()
Can u please add java code?
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 somebody do this with java. ?
Make more videos
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
@@jagrit07 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.
😇
thanks m8
you lost me 3 minutes in lol...
great explanation!