Відео

Awesome solution! Thanks for sharing~ Very curious how did you come up with this solution and why we can get the *minimum* swaps in this way.🤔

when using this solution my output is just [2]. Here is the solution I submitted on leetcode: class Solution { public int[] intersect(int[] nums1, int[] nums2) { HashSet<Integer> set1 = new HashSet<>(); HashSet<Integer> intersect = new HashSet<>(); for(Integer i : nums1){ set1.add(i); } for(Integer i : nums2){ if(set1.contains(i)){ intersect.add(i); } } int size = intersect.size(); int[] ans = new int[size]; int index = 0; for(Integer i : intersect){ ans[index++] = i; } return ans; } }

Four lines of code makes this hard question looks surprisingly easy

The debugger really helped.... Thanks for showing the code through the debugger

I was hooked as soon as I heard shook ones pt. 1 by mob deep. I am tired of watching neetcode.

Very nice technique to explain sir

man you are good. i like the way you explain thanks

The solution is incorrect

It doesn't work with input [0,10,10,10,10,10,10,10,10,10,-10,10,10,10,10,10,10,10,10,10,0]

Decent explanation, but I am not passing one test case on LeetCode

Well explained

You missed explaining the part where the actual explanation was needed.

We don't count the answer as part of the space complexity. We use The extra space here for ans and the recursion call stack. The depth of the call stack is equal to the length of ans, which is limited to k. Therefore: Space complexity: O(k)

Gee what kind of person can come up with this idea? I was able to make O(Qn) algorithm with time limit exceeds for some test cases.

why (0,1) map.put(0, 1)

I don't think this is a nice explanation. Realistically, you didn't even know how to solve a problem without looking at the cheat sheet, which means you don't actually fully understand the solution. And since you do not understand the solution it is impossible to explain it properly. My recommendation is to study the solution deeply before creating any future videos.

waw its really simple and awesome

So, does anyone understands this problem at all, or are we all just copy pasting the same exact solution without actually understanding what's going on? I can tell by the way Xavier explained it, that he doesn't get it at all. He just memorized the code. And the explanation was done on a [1,1,1] array. Imagine if it was [7,3,2,1,8,5,4,9], int k = 6. He would go into an infinite loop of confusion.

I've gotten stuck on this problem for a few days. Your recursion is neat and clearly shows the algorithm. Thanks guy!

You just gave out the answer on the whiteboard. That's not an explanation.

Coming here after a long time just to thank you Xavier, for the comment in the beginning of the video. About not knowing how could someone come up with this on their own. I've struggled a lot with the idea that some people are just so incredibly much smarter and better at this than I am. And thats ok that they are. But at the same time some people are just more educated, have more study, have more contact with these challenges and patterns, as I have more contact with other stuff, and remembering that is very good.

video editing is epic

I dislike this question immensely. The problem description leads you to think you gotta replace all those individual indexes. I personally feel like the problem intentionally misleads you.

Thank you no one else explained how the removal part working.Great video!! Keep Uploading

Amazing explanation! Thank you so much for the help on this <3

So smart to add background music while you're unable to articulate well the n/m complexity. I've broken my ears

Bit confusing 😂

static long repeatedString(String s, long n) { int strLength = s.length(); long q = 0, r = 0; q = n / strLength; r = n % strLength; long partialStrLen = (r == 0) ? 0 : r; long aCount = q * getLetterCount(s, s.length()) + getLetterCount(s, partialStrLen); return aCount; } public static long getLetterCount(String s, long strLength) { long count = 0; for (int i = 0; i < strLength; i++) { if (s.charAt(i) == 'a') count++; } return count; } public static void main(String[] args) { Scanner sc = new Scanner(System.in); String s = sc.next(); long n = sc.nextLong(); long aCount = repeatedString(s, n); System.out.println(aCount); sc.close(); }

what is helper function used for?

At max, minimum of size of both arrays can be duplicates, so instead of doing length1 * length2, you can do min(length1, length2)

solved using similar concept , but use recussion , although liked your solution more

I'm trying to do a level order traversal stores the node in list<list<node>> and then for each level, in an interation, i updates their right to next node's address. -- does not work tho.

it's enough to find one unbalanced subTree to say that the whole tree is unbalanced class Solution { boolean isBalanced = true; public boolean isBalanced(TreeNode root) { if (root == null) return true; dfs(root); return isBalanced; } private int dfs(TreeNode root) { if (root == null) return -1; int left = dfs(root.left); int right = dfs(root.right); if (Math.abs(left - right) > 1) { isBalanced = false; } return Integer.max(left + 1, right + 1); } }

Nice solution, made one with a lot of apparently useless if else statements.

bruh with one sentence, you answer the question that I've been struggling for days. I really don't understand the question XD

the game[i] = 1 part is very smart

your code had some errors: this soltuion passed the testcases kindly verify: class Solution { public Node connect(Node root) { if (root == null) { return null; } Node levelStart = root; while (levelStart != null) { Node current = levelStart; Node dummy = new Node(); // Dummy node to maintain the structure Node tail = dummy; while (current != null) { if (current.left != null) { tail.next = current.left; tail = tail.next; } if (current.right != null) { tail.next = current.right; tail = tail.next; } current = current.next; } levelStart = dummy.next; // Move to the next level } return root; } }

thank you!

So you're basically looping from the back of the array and testing each element as if it was bribed. Would it be possible to loop from the front of the array and test each element as if it was the briber?

"I don't know how you come up with this unless your really smart" Ya thats the vibe every time I cant do a problem in O(n) time =[

From a more mathematical perspective: Let x = distance from starting node to node where cycle begins Let y = distance from node where cycle begins to where fast and slow pointer meets Let c = length of the cycle Let n = number of cycles travelled When slow and fast meet: slow pointer travelled x + y distance fast pointer travelled 2(x + y) distance -> extra distance travelled is a multiple of cycle length -> 2(x + y) - (x + y) = nc -> x + y = nc (also means slow travelled distance of nc) -> x = nc - y We can infer than if slow pointer continues travelling from (x + y), after x more steps it will reach the cycle starting point. Which is why if we move head at the same rate as slow in the second part, their meeting point equals to start of cycle.

I had this problem interviewed at Google back in 2021, but not binary tree, it was N-ary Tree. I was using topological sorting to resolve it.(make every node pointing to its parent node) Not at good as yours though.

Thank you. I have always wondered why there was a pre-increment and a post increment, and this is a perfect use case for post-increment. It's the small things sometimes. I've never seen that before.

good solution, thank you

Where can we learn more about dynamic programming for problems related?

4:41

Thanks I was breaking my head over this

Thanks for mang this helpful vid, just wondering. What does the int columns A[0].length do?>

This one🔥

hi i am sorry if i am bothering you... in this problem i tried to make like a duplicate and make one go to the right and the other to the left and little by little replace them one of them would be you used as a sort of pointer to guide which value goes next and the other one to change the values. something like this: class Solution { public TreeNode invertTree(TreeNode root) { return mirrorTree(root,root); } public TreeNode mirrorTree(TreeNode normalTree,TreeNode invertedTree) { if(normalTree != null) { mirrorTree (normalTree.right, invertedTree.left); invertedTree.val = normalTree.val; mirrorTree (normalTree.left, invertedTree.right); }return invertedTree; } } but it did not work it only works with half, i assume it gets normalTree and invertedTree as same tree when it should be different but i am not sure. Just started programming about a month ago and i am still new at this.