Check if a String Contains all Binary Codes of Size K - Leetcode 1461 - Python

[@Victor-hg6hk]

Man, thank you so much. I just got my dream Google offer and I would've never ever gotten it without your videos. Thank you so much and I'll be joining you in the Seattle office this fall!

[@ChanChan-pg4wu]

It indeed has been a while! Glad to see you back! Always great explanation! This reminds me to revisit permutation & backtracking.

[@AMAZINGGUY1113]

Thought you'd stop after your hard work paid off. Good to see that you're still at it! Keep it up man, it's greatly appreciated :)

[@Mutual_Information]

Interesting to think that there is a subset of the183K subscribers who passed an interview and accepted the job because of this channel. Very strong impact

[@fatihorhan9355]

I'd suggest allocating a boolean array space of size 2^k and changing each corresponding place of the array with the given substring. This allows us to avoid computing the hash function and saves a lot of space since every bool is only 1 bit. (I implemented my solution in c++ and am unsure how it'd go in python 3.)

[@fatihorhan9355]

Beats 85% of runtime and 94% of space in c++.

[@allandogreat]

Yes, good one.

[@pallav7041]

nice approach

[@damirshakenov2198]

Actually, bool is 1 byte (not bit), because the CPU can't address anything smaller than a byte.

[@fatihorhan9355]

@@damirshakenov2198 You're totally right. I forgot that almost all instruction architectures are designed so that smaller size CPUs can interact with is 1 byte. However, STL's boolean vector is designed to optimize for space.

[@trollbaron1329]

Hi NeetCode, would you consider making a video on more advanced topics like string hashing or segment trees?

[@numberonep5404]

Short and to the point as usual! nice pb

[@udayptp]

This was as simple as tricky, loved your explanation

[@VivekSingh-in6rq]

damn, i just got this challenge today and found ur video was released today too 😂😂

[@jasonswift7468]

We can add if ((s.length() - k + 1) * k < (1

[@shantanukumar4081]

Great Explanation !!!

[@halahmilksheikh]

I thought he went up to Google heaven and never returned

[@NeetCode]

Lol good one

[@halahmilksheikh]

@@NeetCode For real though, I thought you retired this channel. I hope you continue it even if it's at a slower cadence (like 1 per week?)

[@aishwaryaranghar3385]

great approach

[@Malmidas]

Keep up the good work, really appreciated!

[@shouryannikam3949]

Is the neetcode 150 list definitive or do you plan on adding more problems to it?

[@hilsCYBER]

Thanks man, great content as usual.

[@nguyenquangduy910]

hey Neet, your solution is brilliant but what if (s) dont have enough sample to return the number of solution we are seeking. E.g s="0110" k=2

[@minarefaat1573]

Hey man We miss you

[@farzinhaddadpour7192]

This is clever!

[@pekarna]

Hi, hashcode is not too efficient here. Since we have a continuous sequence of integers to search for, why not have an array, 0 to 2^k-1? This leads to a quite efficient code :) Runtime: 555 ms, faster than 100.00% of Kotlin online submissions for Check If a String Contains All Binary Codes of Size K. Memory Usage: 68.4 MB, less than 100.00% of Kotlin online submissions for Check If a String Contains All Binary Codes of Size K.

[@mohithadiyal6083]

Awesome 😁😁

[@bh4541]

why O(n*k) can someone explain for me ?: thank u

[@osamaayman3765]

Because the for-loop loops over all the input string of length n and in each iteration we are creating a substring of length k (which takes linear time because it copies all chars to a new string). So, the overall time complexity is O(n*k).

[@beebread._.m8]

saos