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Same problem as LeetCode 31: Next Permutation

good video gang, here in 2024

if you don't like math do this: new_i = (i + 1) if j == 8 else i new_j = 0 if j == 8 else (j + 1)

I could not understand the part where for each value of k (guess number), why we took minimum of it? If we are looking for worst case, shouldn't we take a worst guess or all the guesses we made?

Worst explanation ever.

Great solution!

Thank you

Thank you, very nice explanation!

I dont like python syntax its confusing and too much things going on in a single line

As start end are only positive ints. Just have an array of frequencies and binary search the new booking in so its n*log(n)

this was a good explanation my first thought was to do dfs for each cell but I wasn't sure how to check the min width and make sure they are a valid recatangle

best explanation sir. thanks you so much

Thank you

thank you for the clear explanation!

I think an important thing to sort of internalize is that when you insert a parity into the hashmap, you are guaranteed that this is the earliest index at which this parity is found. So everytime you encounter this parity again, you are comparing it with the earliest index you encountered this parity, which gives you the longest length as i keeps increasing. Thanks for the great video, took a little while to digest but seems to be very helpful in solving problems with a similar recurring pattern!

nice one , smart and clear, it takes me a little time to figure out why BFS miss some result, and I tried to use BFS + DFS backtrack, but actually just use your idea will work

Its marked hard now

I am convinced all these hard problems are just one observation away from becoming a medium problem. Like how you made : numMoves > len(graph)*2 return 0

I was getting Stack Overflow in DP but BFS worked fine

gee, improve your English

your explanation helped a lot!! Thanks mate !💌

nice explanation. I made one improvement: if we create seen as a map instead of a set, with the keys being (nx, ny) and the values being cur_k or cur_k - 1, we can ignore any k that is less than what we already have in the map (since a lower k will never result in a better path). This decreased the run time for me a substantial amount.

Thank you, this was very well explained. Now sure how I am supposed to solve this during an interview though..

the worst explanation of all times!

Explanation is not clear at all, you keep saying it's obvious, but it's not.

26:14 This part where you explained the intuition was brilliant ❤

just add a case sumCount[0] = 0 for solving it

surprisingly this guy beat Neetcode in this problem's explanation. Good work !!

What do you mean by TRE at 05:50? out of memory limit?

You nailed it man!

Damn, whats the time complexity?

Thank you! I was struggling with this one, you made it look easy lol

For this input "abcdabcdabdeac" and k = 4, We can print "adcbadcbadbeac" as well "abcdabcdabcdae" ... it is nowhere mentioned in the question that we need to print lexicographically smallest string.

Thanks for sharing!

technically without the rebalance of the tree its still O(n) in the worst case

Rabin Karp is technically O(n + m) where m is the length of the pattern we hash and search for. Here L can be as big as N, so we can simplify it to O(N). But still worth it to know the details. BTW it is O(n + m) is because of the calculating powers which takes time proportional to the L. In the textbook implementations, calculations are done using a while loop taking mod at each step so as to now overflow

I dont know if you'll see this coz I guess you're not currently active in this channel.. but just wanted to let you know that all of your videos are easy to follow and super helpful ❤

very hard to understand

Great explanation

Thank you so much!

why cant i do like these // Option 2: Place the current book on a new shelf int newShelfHeight = solve(books, idx + 1, shelfWidth - currWidth, currHeight + height, shelfWidth, dp);

The problem was very boring but your explanation made it easy. Thank you!!!

Hey Cheng! I recently discovered your channel and it's very helpful. No other channel has covered a lot of hard problems as yours does.. please continue this and your channel will blow up soon as people are moving towards solving more harder problems :)

Adding a dummy node is a really nice trick. Thankyou!!

great explaination dude it really helped a lot...

man u deserve a lot of veiws.. ur problem solving way is very unique and very good.. this the best channel for sloving leetcode problems..

Nice explanation! You make it seem very simple

not clear about ans = [ ]

Simple and robust solution.

Good explanation