Video

simply amazing man!

java.lang.NullPointerException: Cannot invoke "java.util.TreeMap.floorKey(Object)" because "<local3>" is null at line 19, TimeMap.get at line 71, __Driver__.__helperSelectMethod__ at line 91, __Driver__.__helper__ at line 112, __Driver__.main Can you help me resolve this ?

glad to have found your channel

Very clear explanation

perfect

you explain thing very nicely.

very well explained man all my doubts are clear now regarding this problem

Nicee explanation! Thank you

0 sec ..

Is there a way to solve this at linear time if we use a hashmap?

Excellent explanation, thanks.

int maxDepth(TreeNode* root) { int maxDepth = 0; // Initialize the maximum depth int count = 0; // Initialize the current depth counter dfs(root, count, maxDepth); return maxDepth; } private: void dfs(TreeNode* node, int count, int &maxDepth) { if (node == NULL) return; count++; // Increment counter to reflect current depth if (count > maxDepth) { maxDepth = count; // Update maximum depth } dfs(node->left, count, maxDepth); dfs(node->right, count, maxDepth); } }; brother this code works but i cant understand how every recursion call maintain its own count variable

this is linear search approach, try binary search

Hey Fisher, thanks for the video, quick one .. why do keep assigning the list in line "19 and 20".. can't you simply run it without assigning the list?

Great approach sir !

I loved the explanation. The thing is I can understand from he code, but I cannot understand and then code. How to think in a way to write code not read the code? Any tips especially for recursion?

Excellent logic. Super easy. Didn't think this way. Thanks a lot!

Thank you for sharing and explaining the video. If possible, please also share time and space complexity at the end. I think based on above solution it should be as below but please correct it if it is incorrect. Time complexity: For creating the map, if the number of users are m and maximum number of language spoken by a user is n. The time complexity is O(mn) If total Friendships are k , then creating the boolean array is O(kn) If there are l languages, we are iterating l times * number of friendships k times. O(lk) Total time complexity is O(mn)+O(kn)+O(lk) Space complexity: For creating the map, if the number of users are m and maximum number of language spoken by a user is n. The space complexity is O(mn) If total Friendships are k , then space complexity of boolean array is O(k) The size of teach set at the end cannot be more than that of number of users m thus space complexity is O(m) Total space complexity is O(mn)+O(k)+O(m) which will be equal to O(m*n) as we can drop the other smaller k and m.

thanks!!!!

Amazing!!!!!

great concept

You are a god among men.

This is by far the best explanation I've seen on this problem.

You made it look so easy. Though It took a few minutes to understand the find method but the overall approach is pretty simple and I have easily replicate that

I don't think you've added enough #fishercode tags in your video.

beautiful! thank you!

I've implemented even a faster solution pastebin ref -> 1yPw0SSG

Why make a video, if you only have the Brute Force Approach to Discuss

lol rly same code for all problems

Thank you sir.....it took me a while to understand your explanation but now it's finally clear!!

Any one also getting exception?

Nice man, other youtube videos just write the recursive answer without really explaining it. But you can explain it in a simple way

🧠

Why in the code you increment j until it reaches i and only then increment i, but in the presentation you increment i first and then at some unclear point you increment j?

Best visualization out there. Thanks

Guy, great explanation with code and graph!!! thank you, keep going.

In line 8, can't we just create the treemap as a new TreeMap, why do we equal it to "map.get(username[i])"? We are already mapping it with username, the newly created tree map at line 10 while putting it into the map.

Sir will u come back to CZcams again

Amazing content, thank you!

Thanks but your code not passed successsfully if you not add last line like this complete Query is: # Write your MySQL query statement below SELECT s.student_id, s.student_name, sub.subject_name, COUNT(e.subject_name) AS attended_exams FROM Students s JOIN Subjects AS sub LEFT JOIN Examinations e ON s.student_id=e.student_id AND sub.subject_name = e.subject_name GROUP BY s.student_id, sub.subject_name ORDER BY s.student_id, sub.subject_name;

i don't understand your explanation, why are you incrementing i before j? j is nested inside the i loop but you started by looping through i before j??

recently got asked with this problem i used a similar approach (almost identical)

Nice explanation, thanks👍

one of the most optimized approch thanks bro

Give important topics leetcode bro give question sheet

SELECT a.student_id, a.student_name, b.subject_name, COUNT(c.subject_name) AS attended_exams FROM Students as a JOIN Subjects as b LEFT JOIN Examinations as c ON a.student_id = c.student_id AND b.subject_name = c.subject_name GROUP BY a.student_id, b.subject_name ORDER BY a.student_id, b.subject_name

Can someone explain why there are two lines of stack.pop() 😭 I understood first one but not second

you way of explanation is good

QUESTION: using set instead of map may has some issues. what if there is two duplicate in the sliding window and one is just about to be outside the window and needed to be removed. we cannot set.remove(e), since another one may be still in the window

Goated video. The helper function makes it similar to the iterative solution and now this way sticks in my mind 😃