THIS is a valid solution to this problem. I was looking at the solutions with the most upvotes, and their solution was to mathematically solve it by adding up all the longest increasing subarray and decreasing subarray, and then subtracting the lowest of the two from the totalCandy. I hate that leetcode solutions have become mathematically solutions rather than programming. Solutions for the CarFleet problem is a hashmap with the key being the product of a formula. Then returning all the unique keys. Like what is that, the solution should go through a process that imitates the problem. Now we’re seeing problems that are purely solved by inventing a formula for the specific issue. Rather than preparing you for problems you’ll face regularly
I came up with the O(n^2) solution with a while loop of while(candyIncreasedThisLoop){ do a pass and add 1 if the rules arent followed for each } and I was proud of myself, but damn this code is not only neat is damn near beautiful 😢 Thank you for the great explanation as always you are a huge help in my career journey as a new grad at the moment. I can't thank you enough for your years of backlogs of coding solutions I am learning from right now. Keep up the amazing work!
An alternate to the last step: if ratings[i] > ratings[i+1]: arr[i] = max(arr[i], arr[i+1]+1) That was more intuitive for me: if ratings[i] > ratings[i+1] and arr[i]
Thanks a lot bro, I actually just solved it had to pause the video at the mid of the intuition because that was the missing part that I needed, went and solved it then got back.
I derived my solution by going deep into the diffs array and trying to reduce it to other problems, my soln ended up going to 30+ lines so i was quite disappointed to go into the solutions and see that ppl had done it in like 10 lines
You mentioned that no extra memory is needed, yet your solution clearly allocated a new array to track the number of candies given. Doesn't that make the space complexity O(n)?
Hi, thanks for providing this brilliant solution. I understand that after left->right, every kid with higher ranking compared with their left neighbor will have more candies, and after right->left, every kid with higher ranking compared with their right neighbor will have more candies. However, how to make sure after second round from right to left, every kid with higher ranking compared with their left neighbor still have more candies?
Seems like this problem can be solved in one pass too, but I have a hard time understanding that lol. Would've been nice if you could explain the onepass method because I can understand your explanations only. Others take way too much time drawing, or do a poor job of explaining.
@@Stay_away_from_my_swamp_waterno, apparently, you can use three variables up, down and peak and then do something which I don't understand to complete this in one pass and O(1) space complexity. It's the top solution on solutions page (by vanAmsen). I don't think I can post links here, the spam filter will probably go off.
And this is how your boss think of giving salary to you
Oh I was waiting for you :) yes the problem is poorly worded I feel
Thanks man you're so good that I never had to wait for you to code.
THIS is a valid solution to this problem.
I was looking at the solutions with the most upvotes, and their solution was to mathematically solve it by adding up all the longest increasing subarray and decreasing subarray, and then subtracting the lowest of the two from the totalCandy. I hate that leetcode solutions have become mathematically solutions rather than programming. Solutions for the CarFleet problem is a hashmap with the key being the product of a formula. Then returning all the unique keys. Like what is that, the solution should go through a process that imitates the problem. Now we’re seeing problems that are purely solved by inventing a formula for the specific issue. Rather than preparing you for problems you’ll face regularly
I came up with the O(n^2) solution with a while loop of while(candyIncreasedThisLoop){ do a pass and add 1 if the rules arent followed for each } and I was proud of myself, but damn this code is not only neat is damn near beautiful 😢 Thank you for the great explanation as always you are a huge help in my career journey as a new grad at the moment. I can't thank you enough for your years of backlogs of coding solutions I am learning from right now. Keep up the amazing work!
Same dude
same
An alternate to the last step:
if ratings[i] > ratings[i+1]:
arr[i] = max(arr[i], arr[i+1]+1)
That was more intuitive for me:
if ratings[i] > ratings[i+1] and arr[i]
very well explained!
just like interview style!
Thanks for uploading this. I had requested for this in the comments in one of your videos.
You are totally more than awesome
You explained this concept in such an awesome manner I have no words for it
Thanks a lot bro, I actually just solved it had to pause the video at the mid of the intuition because that was the missing part that I needed, went and solved it then got back.
Amazing explanation !!!!. Greedy Questions are just mind blowing.
Yea, greedy hits you up as if you never know what has happened.
I only thought about iterating through left -> right..always amazing
same 🤧
literally the best explaination
I derived my solution by going deep into the diffs array and trying to reduce it to other problems, my soln ended up going to 30+ lines so i was quite disappointed to go into the solutions and see that ppl had done it in like 10 lines
isn't it O(n) space solution as we are using extra array to store candies ?
You mentioned that no extra memory is needed, yet your solution clearly allocated a new array to track the number of candies given. Doesn't that make the space complexity O(n)?
Yep, that's my bad. Space is O(n)
@@NeetCodeIO Thanks for clarifying! appreciate your efforts and explanations 👍🏼
Best Explanation 👏🏻 Nice work great intution
how you get so good at logic, damn nice and nice , best approach
Thank You So Much for this wonderful video...🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Quite much easier problem than lots of "easy" ones.
Best explanation ever
Great explaination ❤
Hi, thanks for providing this brilliant solution. I understand that after left->right, every kid with higher ranking compared with their left neighbor will have more candies, and after right->left, every kid with higher ranking compared with their right neighbor will have more candies. However, how to make sure after second round from right to left, every kid with higher ranking compared with their left neighbor still have more candies?
I have seen a 0(1) space solution in lc solution section. Could u explain it in a future vid
good job man. very helpful series......
You're a genius
Well explained 👍
Genius explanation... Wow
First LC hard I solved on my own! :)
Cool buddy
Wait is over guys 😅
Finally solved the problem. Thank you so much!
solve on your own,give your everything to solve the problem
@@RishabhSingh-xn3xu 🤣
11:52 precisely. Did you make that same mistake there?
Seems like this problem can be solved in one pass too, but I have a hard time understanding that lol. Would've been nice if you could explain the onepass method because I can understand your explanations only. Others take way too much time drawing, or do a poor job of explaining.
@@Stay_away_from_my_swamp_waterno, apparently, you can use three variables up, down and peak and then do something which I don't understand to complete this in one pass and O(1) space complexity. It's the top solution on solutions page (by vanAmsen). I don't think I can post links here, the spam filter will probably go off.
@@aisteelmemesforaliving I'm also looking for a better explanation of that algorithm. If I figure it out, I'll make a video.
@@stefanopalmieri9201 please don't forget to reply to this if you actually do. Highly appreciated.
@@aisteelmemesforaliving It's uploaded now: ua-cam.com/video/_l9N_LcplLs/v-deo.html
any luck?@@aisteelmemesforaliving
the problem is not enough described in leetcode
12:08 stuck 2 days because i did not take that into account
Thanks for the daily
Can you explain one pass approach?
was waiting for this
this one looks easy after seeing solution ,but how to endup with this intuition !!!
This was asked to me in an interview and of course, I messed it up😭
I missed that "welocme back" this morning...
very clever
Was waiting for this :)
How to get referrals brother, I worked really hard , solved 500+ problems on leetcode but still I m suffering to get interviews
Please help me I m sm out of hope right now
I feel you man, me too.
How i can get big brain like your?
It's O(n) additional memory, isn't it?
Now do it using O(1) space
This is an O(N) space solution but a O(1) space solution does exist for this problem.
The best
I have no idea how I could figure this out on my own tbh
Corporate in a nutshell
im too dumb for this shit chief
not pertaining to the question or this video, but what's the difference between this channel and the original channel? @neetcode