Thanks so much for watching this video, everyone! If you liked this video, I would also recommend my Udemy course, "11 Essential Coding Interview Questions": www.udemy.com/11-essential-coding-interview-questions/?couponCode=SOLVE2 (You'll get a discount through the link above :)
Randomly stumbled on this video. Gotta love that he underlines these interviews are NOT about code, puts no snippet in the video and still people write random code thinking they know better while they don't. Amazing.
Holy shit I am studying 8th grade and still got a solution! Code(in python) def multiply_to_20(list): for num in list: if 20/num in list: return num,20/num
Hello Cs Dojo, i am one of your big fan, i love the way you teach these concepts, i really want you to make a series of data structures problems ( if possible with JavaScript, that would be amazing ) Thanks and Regards ❤
Alternate: determine the factors or values involving 20 ... 20 = 20&10&5&4&2&1. Note: Scan if the elements contains negative values, in which the product of two factors of 20 can be applied.
I like your videos :) I've been programming for 7 years now but I like the style you explain things. One more thought on this problem though; why not get all the divisors of the number you want to get and then checking the set of given integers if it contains the required ones. I think your solution is O(nlogn) and this approach would yield O(n). What do you think? greetz from Germany
more&more&more algorithm videoossss pleasee!!!! (The algorithms that facebook microsoft amazon etc asks) I don't know which language do u write but i actually wanna see something like this -> 1- a problem 2- an approach to solve the problem. 3- code writing part 4- and make code simplier. Because it's not easy to write the best code when solving the problem. And if we want to work in big companies we have to write simple&understandable codes. You might help us about writing a simple code/think.
Steps to do this: 1> First sort the array 2> Then have two pointers, the first pointing to the first elem(front) and the second pointing to the last one(back) 3> Multiply the two values that the pointer pointing to: If the multiplication of the two numbers is less than 20, then move the front pointer If the multiplication of the two numbers is more than 20, then move the back pointer If the multiplication is 20, then you a match The algorithm's complexity is O(n logn) + O(n) < O(n^2)
This will not always work, as their can be negative factors. Say you have the string: (-4, -5, 4, 10) The obvious answer here is -4*-5=20 Your answer would do as follows: Sort the list: (-5, -4, 4, 10) -5*10=-50 -50
But his approach is O(n) making the sorting basically inefficent. But if he would not use hashmap and would have a O(n^2) approach, this would be good, yeah.
what's the complexity of your way? Assuming the worst case ex: [2 12 48 24] multiply to 48. If you use a hash table to store the numbers that you have seen you would do something like 2 -> 2, 12 ->12 and 48 - > 48. If you hash a number and nothing is found in the hash table, it means that the desired factor is not on the paper, so the question is what's the complexity of a lookup in a hash table?. I googled this up and the complexity is O(n) in the worst case, so it would be O(n^2) which is the same thing as checking every possible pair.
That surprise me, too. I think he didn't thought about data structures at all. He just throw a few names and that was it. Why would you use a dictionary (or a hash table) when you don't care about key and values? A hash set? maybe... if values were unique, but is hard to tell without context...
we can do this problem more optimally if i dont generalise it.since we already know the factors of 20 i can just apply binary search for each factor.for example first i find if 20 is there,if it is there then i search for 1.else if 20 is not there i check -20, and then -1.then check other factors like this way. time complexity would be o(logn *6) which is equal to o(logn).isnt this better???
There is a prove out there, that you can theoretically find integer in hastable in O(1). Worst case is O(n), yeah, but you could use O(1) for the average Case, which makes it linear instead of quadratic.
I like the manner you teach others, I think you to have a lot of potentials I encourage you to continue and publish more stuff in udemy and here, blessings
Well, if you ever had to implement all the sorting algorithms yourself you would know that sorting alone has an even worse complexity than just: Iterating through each value and if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n]
import math s=set() a=set([2,4,1,6,5,40,-1]) for i in range(1,(int(math.sqrt(20))+1)): if 20%i==0: s.add(i) s.add(int(20/i)) cs=s.intersection(a) for i in cs: if 20%i==0: n=int(20/i) if n in cs: break print(i,n)
You don't need a powerful computer to program because what you'll be coding at first probably won't be that be that resource- intensive. But the most important thing would be the run-time for whatever you're actually writing. So if you write a resource-intensive program, that would usually be due to the program itself, not the environment's. The integrated development environments (IDE's) aren't that heavy but text-editors can run on anything.
Does the role you are doing at google depend on how often you need to solve these types of problems? For example ,would a "web solutions engineer" have to do this much algorithm based work VS a "usual" backend software engineer? I am a web engineer and have never needed to problem solve in this kind of way - the problem solving is more about putting things together (webpack, babel, front end frameworks, testing frameworks, middleware things like cloud functions, UI components etc) and less algorithmic problem solving (which as mentioned I've never had to do in 4 years of experience).
Dude I like ur videos a lot. It's more than enough for python. Btw why did u stop you should have continued. Make a university on ur own. I bet that u will get a lot number of students
1:24 At this point I knew how to approach the problem, however, this is the sad bit, I cannot remember the name of the algorithm that I needed to implement, so I'd Google the answer, but it sad because this is basic CS fundamental concepts and I cannot remember my stuff... I need to practice more lol. Great video.
Mmmm at the point you describe, to implement the naive solution you don't need any particular named algorithm ( as far as I am aware), just two for nested loops. Hope you have made progress in this period of time, it's just a matter of practice.
for me, it's better to take 20 and after that divide it by every single element in array. then, look into array - does it has the result of dividing :)
In this case you check all elements in the list in the second loop, for every 1 iteration in the first loop. As mentioned it would be more efficient to do one loop, and add every number you encounter to a seperate array. If the current number in the first array equals twenty when multiplied by another number in the array you created, you only have to loop over the numbers you already encountered, so there are less steps every iteration.
we can do this problem more optimally if i dont generalise it.since we already know the factors of 20 i can just apply binary search for each factor.for example first i find if 20 is there,if it is there then i search for 1.else if 20 is not there i check -20, and then -1.then check other factors like this way. time complexity would be o(logn *6) which is equal to o(logn).isnt this better???
I might find a solution which is O(n). 1- Initialize 2 dictionary. Saying first,second. 2- Start iterating the array. 3- Check number if it is in the second dictionary as a key. If not add it to first dictionary with zero value. Divide it by 20 and add result to the second dictionary with value the number that we first encountered. 4- Repeat 3. If number found in the second dictionary, voila ! You found a solution. Second dictionary, key-value. time complexity: O(n) Memory: O(2n) (I guess ???)
Of course the following solution not better, however, we can receive solution for the O(n * log n). Okay. We have some target value and we need to find two index of severals numbers in this sequence which multiply is equals to target. We can check each numbers in this sequence and for the each number we can find other number which multiply equals to targer with binary search. Frankly we have the following stuff: a[i] * a[j] = target => a[j] = target / a[i].
Jason E. because with a hash table you can check to see if the element is present in constant time. with an array, you'd have to search the list which would take logarithmic time on average at best. basically with a hash table you can see if a number is present relatively instantly
In this case hashset is good instead of hashtable. because you are going to deal with just a key(or say a single data) in each point and also it avoids duplicates.
ok, is it just me or is the 1st problem is misleading? when i first read it, i thought find multipliers of 20 like 40,80... not find a combination of two numbers, which if you multiplied them the result would be 20.
Hey could we do that take the first number from the array and multiply it with rest elements if it's 20 than save that number else take other number(either the second one?
@CS Dojo hi, whats the reason for using hash table for maintaining already read values given in 2nd explanation because anyways we will have to read those values every time until we get desired output. I'm beginner in programming & after watching the solution, i had a question about space complexity. Thanks in advance
Hi cs dojo and iam from India and here to ask a question on my career.i was just 14 years old boy and have a lot of interest in robotics . So I need your suggestions to start my first advanced coding and what can I do in my future
Hi guys...For this kind of question, is it proper to use this method?(Just for self-learning...write in JAVA) 1. 20 divided by every elements of that array, for example, 20/array[0] = 10, 20/array[1] = 5, etc 2. Then try to match the divided number in that array 3. If it can be found, then print our that pair of 2 numbers (Use nested for loop for the matching just need a few line of codes...is it legal?) THX !!!
Yep you are already better off by NOT including much more complexity by using a sorting algorithm, like I see in other comments. And I had to implement several sorting algorithms including calculation of complexity. Yep you do what I would do: Iterating through each value and if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n]
Allopat true, overlook that one. Now to come to think of it, I don't think is necessary to sort the array. It would only increase time complexity. So look for + pair lcm & - pair lcm
does it matter if you have negative numbers? they are still integers right? sorry, i just notice you still having more attention dealing with negative numbers than irrational numbers. I just don't see your point.
Am i dumb or could you just set a loop with the condition of 2 numbers multiplied together to make 20 and then just write the algorithm were it will just multiply every number together until the condition is satisfied?
The KyloRylo multiplying every combination of numbers would be expensive. it's much better just to take the original number, know what you'd need to multiply to 20 and compare to that number
But how if you getting stuck in the middle of programming? I mean, you've already thinking about the solution on paper but it's not implemented on code yet
why not just calculate 20%n where n is number in array and if answer is zero, push it into seperate array, then iterate through array and if another number gives zero, multiply it by all values in seperate array and see if result is20
hello , i would like to ask if these companies like amazon and google recruite only candidates who studied computer science and have the certification or if you are not a computer science student but you do have expérience in coding and using pyhton for example
Hello, CS DOJO can u please tell me if this is a good way to solve it? i dont know if it's inefficient or something because im not using hash tables. public string EncuentraMultiplos(int [] arreglo,int numFind,int start) { while (start < arreglo.Length) { for (int i = 0; i < arreglo.Length; i++) { if (arreglo[start] * arreglo[i] == numFind) return "encontrado"; } return EncuentraMultiplos(arreglo,numFind,start+1); } return "no encontrado"; }
Well how about just iterating through each value and only if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n]. Even hashing is more complex than necessary.
Thanks so much for watching this video, everyone!
If you liked this video, I would also recommend my Udemy course, "11 Essential Coding Interview Questions": www.udemy.com/11-essential-coding-interview-questions/?couponCode=SOLVE2
(You'll get a discount through the link above :)
CS Dojo I think you make enough money at Google to afford to give that course for free though ;)
He left his job at google to do this.
Hey have you tried his course? is it good?
Google is gonna be phone screening me in a couple of weeks and im wondering if its any good.
Its not that I don't wanna buy the course. it's just that I'm wondering if its worth spending time on it. So I don't waist valuable studying time.
Randomly stumbled on this video. Gotta love that he underlines these interviews are NOT about code, puts no snippet in the video and still people write random code thinking they know better while they don't.
Amazing.
He wasted a paper
rip the amazon rain-forest 2017
All rathes are one and the same they are shitts may be....Dhruv rathe is also shitty like you
Pranav Shastri that paper was already wasted
ikr? #TreeLivesMatter
@@pranupranav5563 I think he was joking, dude. Don't hate. (From a rathee)
You should be a professor at a college bro. Really making a change.
Madhu, you've just given him a great advice for an additional career path :) and your comment should be awarded with a heart
Holy shit I am studying 8th grade and still got a solution!
Code(in python)
def multiply_to_20(list):
for num in list:
if 20/num in list:
return num,20/num
you have good approach for explaining things
Sort the array and use two pointer technique
Hello Cs Dojo, i am one of your big fan, i love the way you teach these concepts, i really want you to make a series of data structures problems ( if possible with JavaScript, that would be amazing )
Thanks and Regards ❤
Alternate: determine the factors or values involving 20 ... 20 = 20&10&5&4&2&1. Note: Scan if the elements contains negative values, in which the product of two factors of 20 can be applied.
explaing each algorithm with time complexity O(n),O(n2),log(n),nlogn
I like your videos :)
I've been programming for 7 years now but I like the style you explain things.
One more thought on this problem though;
why not get all the divisors of the number you want to get and then checking the set of given integers if it contains the required ones.
I think your solution is O(nlogn) and this approach would yield O(n).
What do you think?
greetz from Germany
how would you find all the divisors of a number?. That's a tough task.
given the number n you can for-i-loop until n/2 and check if n%i == 0 => those are all possible divisors and the time it takes is n/2
that's a different 'n' to the size of array 'n'
oh yours would be O(K) right? if K goes to 1000 and theres only 3 elements in the array n = 3, the complexity would take longer
He is suggesting use of hash table... O(n), O(n)
Great video to see how to approach the problems...
Thank you! Please let me know if there are any other types of videos you'd like to see in the future :)
more&more&more algorithm videoossss pleasee!!!! (The algorithms that facebook microsoft amazon etc asks)
I don't know which language do u write but i actually wanna see something like this -> 1- a problem 2- an approach to solve the problem. 3- code writing part 4- and make code simplier. Because it's not easy to write the best code when solving the problem. And if we want to work in big companies we have to write simple&understandable codes. You might help us about writing a simple code/think.
CS Dojo More and more algorithm based questions.
Steps to do this:
1> First sort the array
2> Then have two pointers, the first pointing to the first elem(front) and the second pointing to the last one(back)
3> Multiply the two values that the pointer pointing to:
If the multiplication of the two numbers is less than 20, then move the front pointer
If the multiplication of the two numbers is more than 20, then move the back pointer
If the multiplication is 20, then you a match
The algorithm's complexity is O(n logn) + O(n) < O(n^2)
sunny grewal you can do it in O(n) :P
This will not always work, as their can be negative factors.
Say you have the string:
(-4, -5, 4, 10)
The obvious answer here is -4*-5=20
Your answer would do as follows:
Sort the list:
(-5, -4, 4, 10)
-5*10=-50
-50
But his approach is O(n) making the sorting basically inefficent. But if he would not use hashmap and would have a O(n^2) approach, this would be good, yeah.
what o and n refers to.
you can use set with O(n)
Take first Number and search for another in the remaining array using binary search, repeat this until we find a pair. Hence only nlogn complexity.
He found a solution O(n) thought^^
BugRushMedia yes thanks for pointing out. But that’s using extra space!
Raj Chaudhary yeah sure, but mostly we are interested in time complexity
Yes you are right, I agree with you!
The array may be not sorted before, so doing the sorting adds some complexity to your algorithm too
For x,y in enumerate z
If x*y double equals to 20:
Print(x)
Print(y)
#z is the list that contains the numbers
He speaks english so well.Amazing content.
what's the complexity of your way? Assuming the worst case ex: [2 12 48 24] multiply to 48. If you use a hash table to store the numbers that you have seen you would do something like 2 -> 2, 12 ->12 and 48 - > 48. If you hash a number and nothing is found in the hash table, it means that the desired factor is not on the paper, so the question is what's the complexity of a lookup in a hash table?. I googled this up and the complexity is O(n) in the worst case, so it would be O(n^2) which is the same thing as checking every possible pair.
That surprise me, too. I think he didn't thought about data structures at all. He just throw a few names and that was it.
Why would you use a dictionary (or a hash table) when you don't care about key and values?
A hash set? maybe... if values were unique, but is hard to tell without context...
hmm yes worst case but i believe it is O(1) average case assuming the hash function does its job
Sure, but even taking Big O out of the table, is still confusing (from the code perspective) seeing key values when plain values would do it.
we can do this problem more optimally if i dont generalise it.since we already know the factors of 20 i can just apply binary search for each factor.for example first i find if 20 is there,if it is there then i search for 1.else if 20 is not there i check -20, and then -1.then check other factors like this way. time complexity would be o(logn *6) which is equal to o(logn).isnt this better???
There is a prove out there, that you can theoretically find integer in hastable in O(1). Worst case is O(n), yeah, but you could use O(1) for the average Case, which makes it linear instead of quadratic.
I like the manner you teach others, I think you to have a lot of potentials I encourage you to continue and publish more stuff in udemy and here, blessings
Sort the array. Divide the number by the ith element of the array. Binary search the factor. Complexity would be O(nlogn). Correct?
Well, if you ever had to implement all the sorting algorithms yourself you would know that sorting alone has an even worse complexity than just: Iterating through each value and if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n]
It can be done by
for loop
for(int i=0 :i< arraysize ; i++)
{ for (int j=0; i< arraysize ; i++)
{ if((arr[i] * arr[j])==20)
cout
Like your approach, I acutely enjoyed it.
Thanks.. It's really awesome the a non-CS guy teaching CS :D
Really kind of you to share those stuff. Thank you very much, Dojo❤
CS DOJO : Add a coding problem of combination of dynamic programming and binary tree or trees... That would be great !
import math
s=set()
a=set([2,4,1,6,5,40,-1])
for i in range(1,(int(math.sqrt(20))+1)):
if 20%i==0:
s.add(i)
s.add(int(20/i))
cs=s.intersection(a)
for i in cs:
if 20%i==0:
n=int(20/i)
if n in cs:
break
print(i,n)
Hi YK... your way of explaining is really simple and easy to understand. Can you start Datastructures and algorithm course. Would be of great help
20/X = Y where X is one of the numbers in the list then check if Y is also in the list
You cant skip the coding if thats required by the interviewer. You can explain as you did but then you have to implement the code for it as well.
should i need a laptop for learning coding or i can continue with my pc one and what kind of specs would you recomend me please reply
You don't need a powerful computer to program because what you'll be coding at first probably won't be that be that resource- intensive. But the most important thing would be the run-time for whatever you're actually writing. So if you write a resource-intensive program, that would usually be due to the program itself, not the environment's. The integrated development environments (IDE's) aren't that heavy but text-editors can run on anything.
Does the role you are doing at google depend on how often you need to solve these types of problems? For example ,would a "web solutions engineer" have to do this much algorithm based work VS a "usual" backend software engineer?
I am a web engineer and have never needed to problem solve in this kind of way - the problem solving is more about putting things together (webpack, babel, front end frameworks, testing frameworks, middleware things like cloud functions, UI components etc) and less algorithmic problem solving (which as mentioned I've never had to do in 4 years of experience).
Dude I like ur videos a lot. It's more than enough for python. Btw why did u stop you should have continued. Make a university on ur own. I bet that u will get a lot number of students
1:24 At this point I knew how to approach the problem, however, this is the sad bit, I cannot remember the name of the algorithm that I needed to implement, so I'd Google the answer, but it sad because this is basic CS fundamental concepts and I cannot remember my stuff... I need to practice more lol.
Great video.
Mmmm at the point you describe, to implement the naive solution you don't need any particular named algorithm ( as far as I am aware), just two for nested loops. Hope you have made progress in this period of time, it's just a matter of practice.
Tutorial channels don't grow this fast, kudos bro!! approaching 1 M already
Please make a video on what topics and how should I prepare to crack all the rounds of google interview....
The paper he mentioned is similar to cache
for me, it's better to take 20 and after that divide it by every single element in array. then, look into array - does it has the result of dividing :)
MiM MeLeRy sounds like a good possible way. Divide by 20 and check for remainder
So very important. I need more vedios... Thank you.
Love from India.
that is so good... you explained very well
You can also solve that problem with a linear complexity by using only an array
We are thinking in the same direction
Thanks for the video , thanks for sharing your experience, we need more videos about Algorithms & logic ..
For x in lists:
for y in lists:
If(x*y==20):
print(x,y)
I got the output.Is this optimal one??
In this case you check all elements in the list in the second loop, for every 1 iteration in the first loop. As mentioned it would be more efficient to do one loop, and add every number you encounter to a seperate array. If the current number in the first array equals twenty when multiplied by another number in the array you created, you only have to loop over the numbers you already encountered, so there are less steps every iteration.
Wouldn't it be faster to sort and binary search for tuples? E.g. (2 10) (5 4) ... You could even filter out non divisors while you sort
i learn programing in highschool and your videos and tips really help me. thank you.
we can do this problem more optimally if i dont generalise it.since we already know the factors of 20 i can just apply binary search for each factor.for example first i find if 20 is there,if it is there then i search for 1.else if 20 is not there i check -20, and then -1.then check other factors like this way. time complexity would be o(logn *6) which is equal to o(logn).isnt this better???
Would a HashMap give you guaranteed O(n) complexity, or "on average" O(n) complexity?
O denotes worst time complexity.
Fantastic explanation.
Which data structure would you prefer in this example if the array size is small like 5 maybe?
no need for HashMap, use HashSet
How sir.. Can u explain me
Dude if there are same numbers then we have to increase the count of repeating numbers. So hash map is used instead of hash set
In Java, a HashSet uses a HashMap as a backing type anyway.
I might find a solution which is O(n).
1- Initialize 2 dictionary. Saying first,second.
2- Start iterating the array.
3- Check number if it is in the second dictionary as a key. If not add it to first dictionary with zero value. Divide it by 20 and add result to the second dictionary with value the number that we first encountered.
4- Repeat 3. If number found in the second dictionary, voila ! You found a solution. Second dictionary, key-value.
time complexity: O(n)
Memory: O(2n) (I guess ???)
Wrong
What if the result is at the end of the array, your approach won’t work as expected.
It is Big O(n) always. They are special cases. And there is no way to know if solution is going to be at the end of the array.
Of course the following solution not better, however, we can receive solution for the O(n * log n). Okay. We have some target value and we need to find two index of severals numbers in this sequence which multiply is equals to target. We can check each numbers in this sequence and for the each number we can find other number which multiply equals to targer with binary search. Frankly we have the following stuff: a[i] * a[j] = target => a[j] = target / a[i].
thank you, would you pls advise some website for practicing and developing problem solving skills?
Why use a hash table?
Jason E. because with a hash table you can check to see if the element is present in constant time. with an array, you'd have to search the list which would take logarithmic time on average at best. basically with a hash table you can see if a number is present relatively instantly
Xellos976 cool , I am going to start using them more often now!
Actually, Java I use hashset... never heard hash table before
Ning hash table was there in Java like HashMap. But it was Synchronised
In this case hashset is good instead of hashtable. because you are going to deal with just a key(or say a single data) in each point and also it avoids duplicates.
Really helped the way of approach!!!
We can use 2 pointer technique.
Cool very nice... use array.sort() to order array before..
ok, is it just me or is the 1st problem is misleading? when i first read it, i thought find multipliers of 20 like 40,80... not find a combination of two numbers, which if you multiplied them the result would be 20.
Hey could we do that take the first number from the array and multiply it with rest elements if it's 20 than save that number else take other number(either the second one?
Hello, i am new at coding but i have theory knowledge so can i practice coding on LeetCode?
Your advise would be helpful for me.
do you heard about the 2 pointers technique ?
it will be much better and O(N)
@CS Dojo
hi, whats the reason for using hash table for maintaining already read values given in 2nd explanation because anyways we will have to read those values every time until we get desired output.
I'm beginner in programming & after watching the solution, i had a question about space complexity.
Thanks in advance
Hash table with a good hash function has the retrieval time of O(1). So, you can easily check if the value exists in the table in constant time.
Hi cs dojo and iam from India and here to ask a question on my career.i was just 14 years old boy and have a lot of interest in robotics . So I need your suggestions to start my first advanced coding and what can I do in my future
Hi guys...For this kind of question, is it proper to use this method?(Just for self-learning...write in JAVA)
1. 20 divided by every elements of that array, for example, 20/array[0] = 10, 20/array[1] = 5, etc
2. Then try to match the divided number in that array
3. If it can be found, then print our that pair of 2 numbers
(Use nested for loop for the matching just need a few line of codes...is it legal?)
THX !!!
Yep you are already better off by NOT including much more complexity by using a sorting algorithm, like I see in other comments. And I had to implement several sorting algorithms including calculation of complexity.
Yep you do what I would do: Iterating through each value and if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n]
Well, wouldn't i just be able to ex. in C++ sort the array and do a Binary Search?
Awesome bro...
Keep continuing....
Iam a bio chemistry student i dont know why Iam watching but I will say that this small guy is smart
Please!!! Make a video on statistics
Filter out neg, and num over 20. Perform sort alg on array, get lcm
What if you've got -4 and -5?
Allopat true, overlook that one. Now to come to think of it, I don't think is necessary to sort the array. It would only increase time complexity. So look for + pair lcm & - pair lcm
Yeah I did something similar. It reminded me of another interview problem that had a similar solution - not sure where I found it, though.
does it matter if you have negative numbers? they are still integers right? sorry, i just notice you still having more attention dealing with negative numbers than irrational numbers. I just don't see your point.
Am i dumb or could you just set a loop with the condition of 2 numbers multiplied together to make 20 and then just write the algorithm were it will just multiply every number together until the condition is satisfied?
The KyloRylo multiplying every combination of numbers would be expensive. it's much better just to take the original number, know what you'd need to multiply to 20 and compare to that number
he is so young.
he's just a midget. =P
javier mila ?? He looks like he is in his mid twenties
Life is all about realizing that there is a younger Asian kid whose way better than you :P
But how if you getting stuck in the middle of programming? I mean, you've already thinking about the solution on paper but it's not implemented on code yet
To become software engineer which kind degree need.CSE or IT
I suck at math and I wanted to ask if Math isss Like Everywhere like needed all the time in programming
why not just calculate 20%n where n is number in array and if answer is zero, push it into seperate array, then iterate through array and if another number gives zero, multiply it by all values in seperate array and see if result is20
Your paper solution is hash table drills written on paper :)
You are amazing!
Thank man I need that! :)
hello , i would like to ask if these companies like amazon and google recruite only candidates who studied computer science and have the certification or if you are not a computer science student but you do have expérience in coding and using pyhton for example
it helped me a lot thanks
Tip #1 - Think on paper, consider multiple solutions and feel confident about the solution before writing cod
Thank you so much sir
Can you please make some teaching video on software engineer
Thanks a lot! Nice Video!
replace paper with your editor and it maybe is a good thing to actually do literally
dude but you can't use hash table with negative numbers, can you?
You can
Nice explanation bro
The Evolution of a Programmer
School:
10 PRINT "HELLO WORLD"
20 END
College:
(defun hello
(print
(cons 'Hello (list 'World))))
Professional (CS Dojo):
#include
void main(void)
{ char *message[] = {"Hello ", "World"};
int i;
for(i = 0; i < 2; ++i)
printf("%s", message[i]);
printf("
"); }
Guru:
10 PRINT "HELLO WORLD"
20 END
How long have you been coding . I am 14 and I only know JavaScript, CSS and HTML :(
Good u r good
How to apply for Google jobs ?
Can someone please show me how to do this problem on c++? Thank you
Great video.
Hello, CS DOJO can u please tell me if this is a good way to solve it? i dont know if it's inefficient or something because im not using hash tables.
public string EncuentraMultiplos(int [] arreglo,int numFind,int start)
{
while (start < arreglo.Length)
{
for (int i = 0; i < arreglo.Length; i++)
{
if (arreglo[start] * arreglo[i] == numFind)
return "encontrado";
}
return EncuentraMultiplos(arreglo,numFind,start+1);
}
return "no encontrado";
}
Your algorithm takes O(n^2) which is too slow for this case.
With hash table, you can achieve O(n).
thx
Well how about just iterating through each value and only if 20%array[i] == 0 then look up (20/array[i]) within array[i+1..n].
Even hashing is more complex than necessary.
@@FeroChau Well, then you have to consider the internal complexity of copying and hashing the data from the array into a hashtable as well.
I like when he say problem
Will you upload more videos on this series ??????????? PLEASE LET ME KNOW
Yes, more videos coming up. Any particular topic you want to see?
Data Structure basics, please.
I am a commerce students and I want to join Apple as a IOS app developer
How can I do that
Can anyone help me ???
Where did you finish your school work genius?
This is a good video.
What I want is a video in which you won't mentioned you worked for google
what is your boy looks tooo much intelligent
You just wasted a perfectly fine piece of paper 😂