sorry to be offtopic but does any of you know a method to get back into an instagram account?? I somehow lost my account password. I would love any help you can offer me!
@Kyson Bruce I really appreciate your reply. I found the site on google and I'm in the hacking process now. I see it takes quite some time so I will reply here later when my account password hopefully is recovered.
Thanks a ton Kevin - was stuck trying to solve this with a recursive approach. Was getting complicated with only 40% test cases passing. Your explanation and approach helped immensely! :-)
Thank you for everything, Kevin. Also liked that you talked about time and space complexity in the end. Can you please talk about these in every video. That would be really helpful. Thanks again!
Funny thing to share: I fainted twice but expressions in both were opposite - "Looking at the prompt I fainted, looking at the explanation in less than 7 mins fainted again" LOL!!.... thanks for spreading knowledge!!
Small optimization: since we know the range of values (every person has a weight between 0 and limit) we can sort with counting sort -> O(n + limit) time, O(limit) space.
I feel like there are so many different kinds of questions people can ask as leetcode has 1000+ problems. Besides studying the higher frequency ones, how do you remember specific techniques? Like sometimes I forget how to do longest palindromic subsequence. And then I go in a rabbit hole of just being frustrated with not remembering everything I learn even if I understand the code. I don't wanna say to memorize but to some extent do you have a main bank of high value questions that you just need to know front and back?
Hey Frank I know what you mean...there are so many problems and it can def be overwhelming at times. Like you said I don't really suggest memorizing, but I would suggest (similar to what you're saying about understanding a bank of questions) understanding / knowing certain algorithms / techniques that you can apply to a handful of problems. I hope this helps and lmk if you have any other questions and as always thanks for your support!
For 1) It's not a huge consideration in this case. Suppose we have weights [ 2, 4, 6 ] with limit 10. While technically, [ 4, 6 ] is a more optimal boat than [ 2 , 6 ], since there's only space for 2 people, there have to be 2 boats going out anyways. It doesn't matter is the last boat is being carried out by [2], or [4]. However, for 2) the optimization becomes a lot harder. What you would do is "bounce" between the lowest and highest pointers. Imagine weights [2, 4, 6, 10], with limit = 10, and k = 3. First, you'd fill 1 boat with 10, move to index with 6. Now that the boat is full, initialize a second boat, and fill it with the right (6), and the left (2). Move both your pointers. Now, attempt to fill your boat either with the highest value (4), or the lowest value (4). In our case, these values are the same, so we exit with 2 boats.
hi, i tried the same logic but it doesnt work for on leet code for the following test data [44,10,29,12,49,41,23,5,17,26] 50 can you let me know if you checked this test case ? i have been beating my head on this for a few days.
All that he said in Nice and Simple way. let me explain the proof in a very simple way: we sorted the people by their weight......then there are two cases: CASE-I: "minimum weight" man can't sit with "maximum weight" man it implies that "maximum weight" man can't sit with any other man hence "maximum weight" man sits alone so decrement only high-end pointer. CASE-II: "minimum weight" man can sit with "maximum weight" man it implies that "minimum weight" man can sit with any other man since we want to maximize the amount of weight carried by a single boat we sit "minimum weight" man and "maximum weight" man together. so increment low-end pointer and decrement high-end pointer.
Yeah it was probably the sorting but that's what makes this algorithm work. I think another way to do it without sorting is to use a hashmap where through one pass you can go through the unsorted peoples list. You can store the complement that can add up to the i-th's numbers limit as well as the number you're currently on together. This way you only go through the array once
thanks Mazhar! If you like my explanations be sure to check out the interviewing service I created thedailybyte.dev/?ref=kevin I recommend joining the annual plan if you can!
Good point, really depends on that underlying algorithm like you mentioned. Off the top of my head, in Java, I believe it's using quick sort or some hybrid between quick sort and tim sort so I don't believe we'd be using extra memory but correct me if you think I'm wrong :)
@@KevinNaughtonJr the sort you're using defined by java.util.Arrays actually uses two sorts like you mentioned, quicksort for primitives and merge sort for objects using a comparator. In this case, it would use quicksort which uses log(n) memory, but I believe you said constant memory around 6:18. TIL
> Everyone Dies™ ✅ > putting problem intuition in video description ✅ > describing time and space complexities ✅ > mixing it up and using a new outro song ✅ You die in the game, you die in real life. Gotta simulate real life to where if you can't solve leetcode problems in production, people die!!! 😱 Good video though 😂
IT WOULD REALLY HELP IF YOU DIDN'T NAVIGATE AWAY FROM THE CODE AND GET A CLOSE UP ON YOUR FACE EVERYTIME YOU WANT TO EXPLAIN SOMETHING. cos some ppl would like to keep seeing the code while you're explaining .....PS: Like Nick White Does...plss implement this.
If only I came up with that algorithm Need to spend more time analyzing the problem and not always going for brute force. My code became extremely long and complex to a point where I couldnt trace lol.
Considering All Possible Cases class Solution { public int numRescueBoats(int[] p, int limit) { Arrays.sort(p); int l=0; int h=p.length-1; int count=0; while(llimit) { count++; h--; } else { count++; l++; h--; } } return count; } }
