You are amazing! I just started my Data Structures and Algorithms course, and I almost had a heart attack as we covered both linear and binary search algorithms and now have to do math problem for the quiz. Both (0)logn and theta problems. But the most important thing is to understand the logic and what the code is doing. You're doing an awesome job at explaining this!! Thank you.
I'm cramming for my first face-to-face technical interview in 2 weeks and I have to say that I wish I found your channel sooner. Thank you for your effort in making these videos.
Binary Search Implementation (I tried before seeing your approach): def binary_search(seq, val): print(f"Base Array: {seq}") seq.sort() print(f"Sorted array: {seq}") midpoint = len(seq)//2 while seq[midpoint] != val: if val > seq[midpoint]: midpoint = midpoint + len(seq[midpoint+1:])//2 elif val < seq[midpoint]: midpoint = len(seq[:midpoint]) return f"value is at {midpoint} index in {seq}"
Thank you for your videos. I came across your algorithm series this morning and spent the day watching all 5 videos and learning from your examples. Now I can use them to make my own Python library of tips & tricks to use when I program. Keep up the good work!
When I studying this using the below list : sequence_a = [1,2,4,5,6,7,8,9,12,14,16,17,22] list_a = 1 it will return 0 as sequence, which looks a bit strange to me...so I add +1 to it, haha print(binary_search(sequence_a, list_a)+1) Btw, it's a good video! Thanks for teaching me!
For anyone wondering. This is a recursive solution. a = [1,2,3,4,5,6] def binarySearch(value, low, high, list): if low value: return binarySearch(value, low, middle-1, list)
#here we need to check the upper half as the value is bigger than midpoint elif list[middle] < value: return binarySearch(value,middle+1, high, list) else: return f"Value {value} is not in the list!"
Thanks for the tutorials Derrick they are so easy to understand and well explained. I’m self learning Java and Python for the past three months and your videos are very helpful. Good Teacher.
Hey Derrick , this is the first time I came across your video and it is so articulate and lucid . I am already subscribing and looking forward to learn more from you .
Cannot tell if he is smiling or being held at gun point with those facial expressions lol. At times he seems forced other times he generally seems interested.
I think your solution would work fine too! I haven't tested so don't hold me too it though, haha. In my head I was just adding half of the remaining positions to the current beginning position to get the midpoint is why I wrote it that way
Your algorithm is not quite right. if sequence_a = [2, 4, 5, 6, 9, 9, 9, 10, 12, 13, 14] and item_a = 9 it will return index 5. It is generally accepted that binary search should find the first occurrence, which is 4. Can you correct this problem?
Should this work with a sequence containing duplicate values? It isn't in my testing but I could just be wrong. Make item_a = 22 and perform the binary search on this sequence: sequence_a = [2,4,5,6,7,8,8,9,11,13,14,17,22,22,777]. It should return 12, but returns 13 (the index of the second instance of 22 in the sequence). Is this just a limitation that should be accounted for?
Hey Derrick. Really enjoy your videos and I think is brilliant the animations you are now adding to the beginning of your videos to explain the theory. I'm trying to create a complexity matrix to determine how complex a project is based on user's input. Was thinking on using radio buttons or something similar for the user to select the input and then generate the result based on the selections. Any ideas on how best to accomplish this (it would need to be a web app)? I am new to Python. Do you think Django would work?
Hey Eduardo! Thank you for the kind words! Sorry for the slow reply on this one. Django would definitely be able to handle it for you if you wanted to go down that route. You could collect the input using an HTML form, pass that to your view in django, do your calculations within your view, and then return it back to an HTML page. Django does take a bit of set up in the beginning, but once you get it running, there isn't really a limit to what you can do with it. If you decide to go down that route let me know if you get stuck anywhere - happy to make some tutorial videos around website creation!
@@CodeWithDerrick thanks for your reply. if you could do some videos related to this idea that would be great. I'm not a developer so it can get tricky to get this up and running. I have some experience with RPGIV but I now work as Business Analyst. My idea is quite simple. have a table with rows and columns and allow users to select the options using radio buttons and show the risk level at the end. for both the radio buttons and the risk level I'm planning on using Bootstrap. in the meantime I'll keep trying and see how far I can get. keep up the good work and thanks for the help.
I can't figure out how to use binary search and user input. I want to use binary search to look at a list of names the user typed in , ask the user for a name to search and display if the name was there or not. I doubt you'll see this but I thought I would put it out there. Can anyone help me?
def search(list,target): for item in list: if list[item]==target: print(i) list=[4,3,6,8,7,9] target=7 search(list,target) 4 #print the index where item belongs
Derrick ,I just completed a python course ,however it is a basic one i am still finding myself rigid while coding how do they I can develop myself to a level where i can write codes smoothly?? Thanks
im confused by the 7th line. Isnt beginning index always going to equal the same so then it would just cancel out? How could beginning index be 2 different values in the same line?
what do you actually mean by complexity? is it the time required for the p;rogram to run? if we use quick sort could you please how much time will the program take
I stumbled on the answer to this in another video. For certain sizes of input arrays, begin+end can create an integer overflow in some programming languages, so adding the extra steps prevents integer overflow errors. I don't understand what that means, but apparently that's the reason for this format.
Thank you so much Derrick, I've been rewatching the playlist almost daily. I have a question: I've tested this on an unsorted list and it did find the index of the element I'm searching for, does this mean that this can work on an unsorted list as well? Many thanks!
It's something I was using by 1981 (When I started with CP/M and PDP11 and VAX sistems). What You are showing it'a the startig procedure of a binary search. What about duplicates? It was already used in 1957 (Peterson, William Wesley (1957). "Addressing for random-access storage". IBM Journal of Research and Development.). Finally, I think U studied well. Pls Read also "Moore School Lectures" from the University of Pennsylvania's Moore School It's an old 1946 Compendium. Here is the link en.wikipedia.org/wiki/Moore_School_Lectures
I don't get what is the point of this? Why not just return the index of the target number? Wouldn't it be the same either way? This just seems like an overly complicated way to do that. Not bashing you or anything but I just don't see the point of all this extra code
#tafelpoten die de winkel kottenphark in de schappen heeft Tafelpoten = [30, 50, 70, 80, 85, 90, 100, 120, 150] Tafelpoten.sort() MaatVanKlant = int(input("Welkom bij meubelwinkel khottenphark, wij verkopen hier verschillende maten tafelpoten, wat is de maat tafelpoot die u graag wilt?")) #zorgt ervoor dat er binair gezocht wordt def binair_zoeken(lijst, maat): links = 0 rechts = len(lijst) - 1 while links 0: dichtstbijzijnde_maten.append(Tafelpoten[index-1]) if index < len(Tafelpoten): dichtstbijzijnde_maten.append(Tafelpoten[index]) if len(dichtstbijzijnde_maten) > 1: print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar. daarentegen zijn de maten {dichtstbijzijnde_maten} wel vergelijkbaar met de maat die u gekozen heeft") elif len(dichtstbijzijnde_maten) == 1: print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar. daarentegen zijn de maten {dichtstbijzijnde_maten[0]} wel vergelijkbaar met de maat die u gekozen heeft") else: print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar.")
Man I really wish you would bring this series back for other algo techniques, you are my fav python instructor on YT
Thanks a lot for this videos man, they are so simple, but they explain everything in perfect detail
You explained this like I'm 5 years old and thank you bc it makes perfect sense now 😭🙌🏾
You are amazing! I just started my Data Structures and Algorithms course, and I almost had a heart attack as we covered both linear and binary search algorithms and now have to do math problem for the quiz. Both (0)logn and theta problems. But the most important thing is to understand the logic and what the code is doing. You're doing an awesome job at explaining this!! Thank you.
I'm cramming for my first face-to-face technical interview in 2 weeks and I have to say that I wish I found your channel sooner. Thank you for your effort in making these videos.
How did your interview go?
@@swarooprajpurohit110 bro did not get hired
Binary Search Implementation (I tried before seeing your approach):
def binary_search(seq, val):
print(f"Base Array: {seq}")
seq.sort()
print(f"Sorted array: {seq}")
midpoint = len(seq)//2
while seq[midpoint] != val:
if val > seq[midpoint]:
midpoint = midpoint + len(seq[midpoint+1:])//2
elif val < seq[midpoint]:
midpoint = len(seq[:midpoint])
return f"value is at {midpoint} index in {seq}"
Thank you for your videos. I came across your algorithm series this morning and spent the day watching all 5 videos and learning from your examples. Now I can use them to make my own Python library of tips & tricks to use when I program. Keep up the good work!
You are sooooo young and smart. I feel so inadequate in regard to thinking capacity 😭. But your videos are amazing, please don’t stop!
When I studying this using the below list :
sequence_a = [1,2,4,5,6,7,8,9,12,14,16,17,22]
list_a = 1
it will return 0 as sequence, which looks a bit strange to me...so I add +1 to it, haha
print(binary_search(sequence_a, list_a)+1)
Btw, it's a good video! Thanks for teaching me!
yes, the list is zero refrenced in code, thus one needs to add 1 again when printing out to natural world.
For anyone wondering.
This is a recursive solution.
a = [1,2,3,4,5,6]
def binarySearch(value, low, high, list):
if low value:
return binarySearch(value, low, middle-1, list)
#here we need to check the upper half as the value is bigger than midpoint
elif list[middle] < value:
return binarySearch(value,middle+1, high, list)
else:
return f"Value {value} is not in the list!"
print(binarySearch(6, 0, len(a),a))
Thanks for the tutorials Derrick they are so easy to understand and well explained. I’m self learning Java and Python for the past three months and your videos are very helpful. Good Teacher.
Thanks alot man for the tutorial very well explained you deserve more subs
Thank you very much for this amazing video. I've been searching everywhere for this but there are too complex. But this one is...Schway!!😀
Excellent tutorial. Love the algorithm explanation at the end to re-iterate our understanding.
You are doing a great job man. The videos are short, well explained and very easy to understand. Please keep posting.
For beginners like me, make sure that the while loop has a
and also for midpint can be (start + end )//2
Hey Derrick , this is the first time I came across your video and it is so articulate and lucid . I am already subscribing and looking forward to learn more from you .
Really loved this. Simple and precise. Thanks
Absolute clutch, as a comp sci student, you explained better than the teacher (no offence to teacher lol)
Cannot tell if he is smiling or being held at gun point with those facial expressions lol. At times he seems forced other times he generally seems interested.
😂😂
I think this code is a little simpler:
key=a #the_number_you_are_searching
first=0
last=len(L)-1
pos=-1 #the_position_of_the_number
while first
keep uploading ...Nicely,Thoroughly explained...
Why we should use the formula midpoint= (start+end-start)/2 instead of (start +end)/2 ?
I think your solution would work fine too! I haven't tested so don't hold me too it though, haha. In my head I was just adding half of the remaining positions to the current beginning position to get the midpoint is why I wrote it that way
@@CodeWithDerrick ☺️ thank you
It does work indeed since your begin_index and end_index get new values on every iteration :)
Thanks amazing it was!
Please upload video on linear search also!
Beautiful videos you angel, so clear and simple.Thank you!!!
This solution is better than lead code selection
def binary_search(arr, target):
low, high = 0, len(arr) - 1
while low
Thanks a lot for the series!! I watch it all
Your algorithm is not quite right. if sequence_a = [2, 4, 5, 6, 9, 9, 9, 10, 12, 13, 14] and item_a = 9 it will return index 5. It is generally accepted that binary search should find the first occurrence, which is 4. Can you correct this problem?
Thanks a lot for this videos man
i do node.js and php but u made it clear for as weel thanx
Who else feels embarrassed? 🤣🤣
Note: Really thank you for this content. It helped me a lot.
Super clear explanation! Thank you so much.
The program as is returns None every time unless you purposely choose a midpoint equal to the value you're searching for
Thanks man! excellent explanation, greetings from Argentina
Should this work with a sequence containing duplicate values? It isn't in my testing but I could just be wrong. Make item_a = 22 and perform the binary search on this sequence: sequence_a = [2,4,5,6,7,8,8,9,11,13,14,17,22,22,777]. It should return 12, but returns 13 (the index of the second instance of 22 in the sequence). Is this just a limitation that should be accounted for?
2:52 hahaha bruh that glitch in your face, sorry but I can't help but laugh but laugh it off. Btw really good video, it helped me a lot ;)
While coders where too busy focusing on the code, you was busy focusing on him yeh :)
@@rebornpixelsproduction u should focus on english yeh :)
Great video and explanation!
Thank you so much. Huge help.
Subscribed
Hey Derrick. Really enjoy your videos and I think is brilliant the animations you are now adding to the beginning of your videos to explain the theory. I'm trying to create a complexity matrix to determine how complex a project is based on user's input. Was thinking on using radio buttons or something similar for the user to select the input and then generate the result based on the selections. Any ideas on how best to accomplish this (it would need to be a web app)? I am new to Python. Do you think Django would work?
Hey Eduardo! Thank you for the kind words! Sorry for the slow reply on this one. Django would definitely be able to handle it for you if you wanted to go down that route. You could collect the input using an HTML form, pass that to your view in django, do your calculations within your view, and then return it back to an HTML page. Django does take a bit of set up in the beginning, but once you get it running, there isn't really a limit to what you can do with it.
If you decide to go down that route let me know if you get stuck anywhere - happy to make some tutorial videos around website creation!
@@CodeWithDerrick thanks for your reply. if you could do some videos related to this idea that would be great. I'm not a developer so it can get tricky to get this up and running. I have some experience with RPGIV but I now work as Business Analyst. My idea is quite simple. have a table with rows and columns and allow users to select the options using radio buttons and show the risk level at the end. for both the radio buttons and the risk level I'm planning on using Bootstrap. in the meantime I'll keep trying and see how far I can get. keep up the good work and thanks for the help.
why the midpoint is begin_index + ? why just second part ( end -begin)//2 only?
you should cast as jack frost. ♥
Is there going to be any more videos on algorithms? Great work!
I can't figure out how to use binary search and user input. I want to use binary search to look at a list of names the user typed in , ask the user for a name to search and display if the name was there or not. I doubt you'll see this but I thought I would put it out there. Can anyone help me?
@Mister Blo0 as he did, user could input the target as an argument in function
Bro pls post more videos on this datas and the algorithms pls ........................
🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
def search(list,target):
for item in list:
if list[item]==target:
print(i)
list=[4,3,6,8,7,9]
target=7
search(list,target)
4 #print the index where item belongs
I think it should be if item == target... And this is not binary search so your comment is not valid.
Great vid! tysm!
Thank you !
Thanks a lot. Very great explanation.
gracefully done!
hello Derrick, I have a question. What if in the sorted list there is a duplicated number/ numbers? Thanks a lot! :)
it will pick the first index of the located value
Derrick ,I just completed a python course ,however it is a basic one i am still finding myself rigid while coding how do they I can develop myself to a level where i can write codes smoothly?? Thanks
im confused by the 7th line. Isnt beginning index always going to equal the same so then it would just cancel out? How could beginning index be 2 different values in the same line?
very useful brother. thanks a lot.
great videos man!
Awesome, Thank you!
thanks man this video helps alot
thanks, really help me out!
what do you actually mean by complexity?
is it the time required for the p;rogram to run?
if we use quick sort could you please how much time will the program take
thank you
please do a video on data structures and linked list too
Is it wrong if I find the mid_point like:
midpoint = (begin_index + end_index) // 2
Thank you :)
Thanks, this is helpful.
Please upload the vedios for Merge sort, shell sort & heap sort as well.🥺❤️
i noticed a issue if you pass [1, 2, 3, 4, 5] as the sequence the while loop goes infinity
Awesome
Great video
Thanks very good Explanation. Why not put
middle = (begin + end)//2 ?
Looks simpler. :)
I stumbled on the answer to this in another video. For certain sizes of input arrays, begin+end can create an integer overflow in some programming languages, so adding the extra steps prevents integer overflow errors. I don't understand what that means, but apparently that's the reason for this format.
paperfish113 really? Thats interesting. I think overflow is when a number is too big for computer memory to handle.
@@agesilausii7759
Yes and while (begin + end) may be overflow l, (end-begin) won’t be overflow
TY BRO
Amazing!!!
Perfect!
wowww, super!!
can you do merge sort video
Hey for line 7 why can't it just be begin+end//2
Thumbs up!
nrb approves
Thank you so much Derrick, I've been rewatching the playlist almost daily.
I have a question: I've tested this on an unsorted list and it did find the index of the element I'm searching for,
does this mean that this can work on an unsorted list as well?
Many thanks!
Binary assumes you already have a sorted list. If you know the list to start with is unsorted, run it through a sorting algorithm first.
program code not working properly.IS the code wrong guys ?
It's something I was using by 1981 (When I started with CP/M and PDP11 and VAX sistems). What You are showing it'a the startig procedure of a binary search. What about duplicates? It was already used in 1957 (Peterson, William Wesley (1957). "Addressing for random-access storage". IBM Journal of Research and Development.). Finally, I think U studied well. Pls Read also "Moore School Lectures" from the University of Pennsylvania's Moore School It's an old 1946 Compendium. Here is the link en.wikipedia.org/wiki/Moore_School_Lectures
my number is off by one position what should i do?
midpoint = (begin_index + end_index)//2
neat!
which ide is used in this video
How do you perform 1 through 300
Hey Derrick Sherrill, your video is amazing. But I would request you to please don't speak so fast during the animation or during the explanation.
excuse me, why should "begin" and "end" be assigned to "mid +1" and "mid-1" ?
istead of just "mid"
Beacause of index in position
Hence +1 means after the mid and -1 is before the middle point
I don't get what is the point of this? Why not just return the index of the target number? Wouldn't it be the same either way? This just seems like an overly complicated way to do that. Not bashing you or anything but I just don't see the point of all this extra code
Umm, why am I getting TLE's?
Time limit expired errors. :)
I love you
wut if you have an even amount of items
👍
do you play Minecraft?
Make video on merge sort...... Really need it.....
Is the code same for visual code for python on windows
Abdul Rameez yes
ofc. Visual Code is just an IDE. there is no difference in codes
not working in pycharm. Help me!!
check the indentations
What does it feel like to be born with plastic surgery?
I've tested all of the sites for free points and only GameCrook works.
If I told you that I am lying to you right now, would I be saying the truth?
yes
#tafelpoten die de winkel kottenphark in de schappen heeft
Tafelpoten = [30, 50, 70, 80, 85, 90, 100, 120, 150]
Tafelpoten.sort()
MaatVanKlant = int(input("Welkom bij meubelwinkel khottenphark, wij verkopen hier verschillende maten tafelpoten, wat is de maat tafelpoot die u graag wilt?"))
#zorgt ervoor dat er binair gezocht wordt
def binair_zoeken(lijst, maat):
links = 0
rechts = len(lijst) - 1
while links 0:
dichtstbijzijnde_maten.append(Tafelpoten[index-1])
if index < len(Tafelpoten):
dichtstbijzijnde_maten.append(Tafelpoten[index])
if len(dichtstbijzijnde_maten) > 1:
print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar. daarentegen zijn de maten {dichtstbijzijnde_maten} wel vergelijkbaar met de maat die u gekozen heeft")
elif len(dichtstbijzijnde_maten) == 1:
print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar. daarentegen zijn de maten {dichtstbijzijnde_maten[0]} wel vergelijkbaar met de maat die u gekozen heeft")
else:
print(f"helaas is de maat {MaatVanKlant}cm niet beschikbaar.")