That was gold!

:) por favor que lo traduzcan a español

I think this solution may break down in the list section if the elements within the list are dicts? Need to test it though to be sure.

Thank you (!!)

Beautiful and simple explaination. I subscribed the channel for its similicity.

How will the search service filter out properties that are not available (that data is in the Booking MySql)?

Thanks for sharing! The recursive implementation is indeed simple and useful.

did we forget an arrow, for feeding information, from the S3 cluster to the Queuing service? How will a new client, that just logged in, will get the actual modified chunk locally on his machine?

are you in the data structures dungeon?

You did a great job explaining it, you got a new subscriber

On what platform do you make your presentations?

You're my saviur 🙏

👍

Data Engineering problems please. This is an ANTI PATTERN in Big Data. If you employed this to find a difference, your batch would take days or weeks. Data Engineering problems please.

Well done! thank you

it will better work if you put in left = -1 and right = array.lenght in the begin))))

Thank you for your detailed explanations and helpful tutorials. I truly appreciate your efforts. The animations in your tutorials are especially beneficial for visual learners like myself. Could you please share: 1. The application you use for writing over VSCode? 2. The software you use to create the illustrations of the linked list? 3. How you split the screen to show both VSCode and the illustrations simultaneously? Thank you once again!

why youTube take too much months to recommend this playlist

You are the best, Liz! Thank you!!!

2:00 I like what she said about imperfectionism. She seems wise, I like her.

Awesome explanations

Thank you for a really thorough explanation of the algorithm, as well as the brute force option. I found it really helpful.

for idx, num in enumerate(set(lst),1): if idx != num: return ind

Great video. Subbed and liked. All my CS classes have me coding C and C++. So far I'm able to understand your coding example. What I'm really after is help writing a recursive function to find unique combinations of K elements from a list of N elements. I have to write it in C and then translate to x86 assembler. Do you know where I can find some good information on how to write a recursive algorithm for this in C. The ChatGBT solution wraps recursion in a for loop. I would love a walk-through of what's happening on the stack.

Here is the code that will work on all data structures def run_on_dict(data): for key, value in data.items(): if isinstance(value, dict): run_on_dict(value) elif isinstance(value, list): for element in value: if isinstance(element, (dict, list)): run_on_dict(element) else: print(f"{key}: {element}") else: print(f"{key}: {value}")

Hello Elizabeth your tutorial on algorithm and data structure is the best, afte watching through your videos I can now understand data structure and have built confidence to go into code challenge

I was able to solve this problem by just taking a copy of the original array and sorting it and then using a for loop to check how many elements have shifted function (arr) { let diffLen = 0; const sortedCopy = […arr].sort((a, b) => a - b); for (let i = 0; i < arr.length; i++) { if (arr[i] !== sortedCopy[i]) { diffLen++ } return diffLen; } Would this not be allowed? Seems simpler. I m starting to recognize patterns a little more and more and was able to see that you can just compare to the sorted version and count the number of diffs thanks to watching videos like yours with the other pointer and sliding window techniques

Thanks a lot! It's really a great idea!)

great video - thanks

you are my virtual teacher.,.. keep uploading content.. ❤❤

for i in range(min(list), max(list)): if i not in list: print(i)

This was immensely useful. Good that you took ideas from Stanford presentation.

That time complexity is wrong. It is exponential not linear

this is very sad it has only less than hundred views. very good materials, thank you

Your dog is in the back like "This again? Great."

Did you end up joining fb?

Thanks for these coding challenge series, do upload more of these...

for those who want to try the problem without dynamic arrays there are a couple other ways here is one i tossed together using masks. no sorting or juggling data, i did the code in JS but easily ported to other languages. it is fun to think of multiple solutions when on the job, and compare costs of each method. note i got lazy with my camel casing. i didn't do strenuous testing so somebody let me know if i messed up // randomly chose longest meeting 10hours with more masks you can do longer meetings const lengthMask = [0,0b1,0b11,0b111,0b1111,0b11111,0b111111,0b1111111,0b11111111,0b111111111,0b1111111111]; // use bits to mask out 1 hour blocks for meetings // if half hour blocks are desired, the max time the business could be open in a day is 16 hours for a 32bit mask // in other languages, we could use 64 or 128 bit integer masks // // input: array of pairs of start, end times of meetings already scheduled // returns: a bitmask where a '1' means a 1 hour block when a meeting is in progress // function createMeetingsMask( meetings ) { const l = meetings.length; let companyMask = 0; for( let i = 0; i < l; i += 2 ) { let startTime = meetings[i]; let endTime = meetings[i+1]; let mask = lengthMask[endTime-startTime] << startTime; companyMask |= mask; } return companyMask; } // if there are multiple open blocks of time of same length, we could return an array of pairs // right now we just return starting time and length of one of the largest free blocks function longestFree( meetingsMask ) { let longest = 0; let longeststart = 0; let curlen = 0; let curstart = 0; for( let i = 0; i < 24; i++ ) { if( meetingsMask & 1 ) { if( curlen > longest ) { longest = curlen; longeststart = curstart; } curlen = 0; } else { if( curlen == 0 ) curstart = i; curlen++; } meetingsMask >>>= 1; } return[longeststart, longest]; } // (0, 6) and (17, 23) are when biz is closed // array of all meetings' start and end times // doesn't matter how array is put together, does not need to be sorted const testMeetingData = [0,6,17,23,9,11,12,14,13,17]; let freeMask = createMeetingsMask( testMeetingData ); let longeststart = -1; let longesttime = -1; [longeststart, longesttime] = longestFree( freeMask ); console.log( longeststart + ' ' + longesttime );

Thank you! It's a good explanation :)

East Or West Liz is the Best.... Thank you for such clearly explained Graph Series.

Thank you for the great explanation! Concise and easy to follow!

Very Well Explained i like your teaching way and Content very Helpful Thank you so much

amazing video! 😇

I didn't initially recognize this problem as a two pointer problem. My first thought was sliding window. Can you expand on how you chose that tool from looking at the problem? Also, I'm surprised that you didn't get array out of bounds issues for [1,2,3] (and [4,3,2,1]). For [1,2,3], when leftPointer increments to 2, the next conditional evaluation should have read arr[3] before checking leftPointer < arr.length - 1.

So helpful!

Awesome

I'm not understanding why at @26:24, the time complexity is `O(logn)`. He mentions it's because of the height of the tree, but there's still more than `O(n)` nodes (recursive calls) in the tree. I was thinking the time complexity (without the n operations at each node), would be `O(2^n)`... For each call, we end up with 2 more calls. Any help here would be awesome!

great video, Liz!

why do we need a block service to chunk file when there is already a chunker in client side?

Thanks for the great explanation