Cognizant Java Coding Interview | Nothing is impossible if you solve code like this

In this video, we covered Cognizant Java Coding Interview Questions. The candidate got selected in the company with the offer of 9 LPA.
We solved below problem
1.) Place all the occurrence of even number in the array before odd numbers using Java
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[@atulyt2129]

Really helpful as a fresher, thank you, subbed ✌️ Please keep posting such videos

[@k283535]

You guys are helping many needy people

[@DriveandThrive]

I liked his approach of course pointers is better but this is a simpler method for a junior developer position. Also, as far as 0(n) you might as well make this even simpler imo. Like: public static ArrayList evenThenOdd(int[] array){ ArrayList even = new ArrayList(); ArrayList odd = new ArrayList(); for(int i = 0; i < array.length; i++) { if(array[i] % 2 == 0) { even.add(array[i] ); } else { odd.add(array[i]); } } even.addAll(odd); System.out.println(even); return even; } If you are going the simple route just go really easy. If you want to do 0(1) ok even better but no need to over complicate things.

[@reeeeel2858]

We can use a queue to store odd elements,nd later copy them in to array without having to run the whole loop once again for odd numbers , nd hence loop runs only for the count of odd numbers . If order should be maintained,else we can use 2 pointer.

[@akashm9974]

We can do it in a single pass (iterating all the element only once) & in-place (not creating an extra array, using the input array) by 2-pointers approach. public static void main(String[] args) { int[] arr = {1,2,5,4,7,8,11,20}; int i = 0; for(int j = 0; j < arr.length; j++){ if(arr[j]%2 == 0){ int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } System.out.println(Arrays.toString(arr)); } !! order will change for odd number !! Hope this is helpful :)

[@vamsisai8924]

Noice

[@sourabhbidikar4143]

As per my understanding if we take auxiliary space it's okay but you should never ever manipulate input data until and unless asked explicitly

[@ravishukla7493]

Solved this question using a single loop, although my first approach was also similar to the solution provided in the video.

[@adityau5823]

Please make more videos for java developer great going 💯💯💯🙌🙌🙌🙌🙌🙌🙌. Pls bring someone 1 year experience people also pure devlopment based like only spring one .

[@Thinkdifferentsagnic]

Here is another solution which might be useful for some- We declare another array having double the length of the intial array and then start putting the odd elements on the new array starting from 0th index till initial array's length also simultaenously, we put the even elements in the new array starting from the index , equal to the length of the inital array till needed. Finally we put a condition which checks till when was odd elements and even elements were pushed into the new array and display till those counter values. Code (Hope it's Helpful)- class oddeven { public static void main(String[] args) { int count=0; int[] arr1={1,2,5,4,7,8,11,20}; int [] b=new int[(arr1.length)*2]; int evcount=arr1.length; for(int i=0;i

[@redstarentertainment2621]

You made it more complex... I have posted a solution using PHP. See if you can find the comment.

[@vaibhavraj8126]

Nice work brother keep it up

[@AnuragTripathy93]

There's a company that still asks such easy questions and allows extra space. Good to know!!!

[@kapilrana4043]

great bro they are asking very easy problem infosys level problem is much tougher thant this for 9.5lpa role

[@shamgavhane6219]

Very helpful sir 👍

[@anandkumar-ir8zj]

You provided real scenario . Some youtubers are faking interview questions to get views , by putting much difficult questions

[@venkatbhaskar3027]

ha ha this is a biggest fake show, stupid

[@ankitbudhewad1964]

Nice, Please create one video on Design pattern And java 8 features Explain therotical part and program please.

[@falaksajjad4910]

It is also constructed using for loop in a single method .

[@Laughing_india_]

this is too easy i think nowadays they don't ask that kind of questions they must be ask about dynamic programming

[@onepercentbetter3313]

Haha

[@souravkumar-hl2ot]

Haha

[@digvijayyamagekar7139]

This is too easy because this is not real interview 😂😂

[@amoghyelasangikar5836]

It depends on company

[@mahidharreddy1213]

Yes it’s true

[@Shorts_n_Laughs]

Wow man you are lucky... They ask very easy question for 9LPA

[@ScientificMyths2706]

Using Stream().filter() we can process the array twice one for even no and other for odd!! Then using string builder concat both the arrays!!

[@musicmasti1938]

Thank you sir

[@monikadesai8251]

Superb 👏👏

[@TiwariAman2912]

During Gen C(4 LPA),They asked me to reverse an array through swaping, Transpose a Matrix,Basic linear search and Reverse a string.

[@cloudtech5260]

Will try to solve these problems 👍

[@moyeensyed6637]

Thank you

[@stayvloging]

i have done this question in 3 minute similarly like this without watching your video

[@cnshobbs6748]

Fresher be like - Wts going on here🙄😑

[@tejasnagapure6734]

Bro but fresher's always try to find out the easy solution as compare to experienced

[@riturajghosh937]

If the order can be compromised then two pointer approach can be used for better space complexity. And that will be optimal solution for this problem.

[@sanazbegum1844]

In java pointer is not used

[@yashpawar7881]

@@sanazbegum1844 lol he didn't mean that pointer, two pointer approach is a method where we use two variables to solve problem

[@sanazbegum1844]

@@yashpawar7881 ok bro I am not aware of that and i am a 2 nd year cse student

[@greyhat6599]

We can also use two pointer method . Take Start and end variables . Run a loop which will iterate over array . Swap both the nos . if start is odd and end is even. That's it ! ( Disadvantage, here is , the order of elements get disturbed )

[@cloudtech5260]

GreyHat, You are absolutely right with this approach 👍

[@sanyamkothari4469]

order will be the same bro!

[@sahil5694]

@@sanyamkothari4469 nope dude, order will change for odd numbers as it will be added from the end of the array... so the first occurence odd number will reach at the end the array.

[@sanyamkothari4469]

@@sahil5694 ok , so he's talking about his approach . I thought it is about the approach in the vedio😅😅.

[@priya_world]

GreyHat can you please tell me how you build a logic

[@reeeeel2858]

I got this question in an exam of wipro for 3.5 or 3.6 package, I guess.

[@DriveandThrive]

another approach could be: public static void evenThenOdd(int[] arr) { int i = 0; int j = arr.length - 1; while (i < j) { if (arr[i] % 2 == 0) { i++; } else if (arr[j] % 2 != 0) { j--; } else { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } for (int k = 0; k < arr.length; k++) { System.out.println(arr[k]); }

[@josephpraveen6341]

Ur doing O(n square) .. but it is possible to do with O(n) using pointers

[@lonestoryteller5430]

It was easy. I was hoping for a solution without using extra space. I think that is possible.

[@ranuchakravarti3277]

This is fresher's interview and mentioned 2yrs exp. ,and i was waiting for catch 😂😂

[@akashtawade9264]

Even and Odd program interviewer ask when you are fresher candidate and by the way it's very very very easy problem. Interviewer should ask dynamic programming instead of this.

[@cloudtech5260]

Hi Akash, Thanks for the feedback. We will try to cover dynamic programming in upcoming videos. 👍

[@RohanKumar-yj7sz]

My Solution: List al = new ArrayList(); int e = 0; for (int i : arr) { al.add(i % 2 == 0 ? e++ : al.size(), i); } return al.stream().mapToInt(k -> k).toArray();

[@girirajsinghrathore1749]

This should have been solved with space complexity o(1)

[@rahul_singh_rajput3292]

if order doesn't matter then we can do this Q in O(N) and O(1) space using Two pointer ... otherwise we can do this Q in O(N) space and time if only order is matter not the element should be in sorted order .. if it's matter then we need to sort the arday first.. then the TC will be n*log(n) for sorting

[@predatorgaming895]

Freshers can do this with even more optimization

[@Gametime00789]

Are these the questions they ask for coding round???...such simple ones??

[@Timeless_Trader]

Hi, I think of similar logic just that instead is creating new array, can't we use same array. Logic is like this. (I don't know how to code this). Start from left --> if number is odd > put it at n-1 place: Else if even don't change it's place. Count number of even or odd while traversing. If anybody can help me out on this, it would be much appreaciated.

[@TheReddKite]

This is so easy question,I don't think this is an actual interview fake

[@vedantkulkarni5437]

This kind of questions i used to do in first year 🙆🏻‍♂️😂

[@MG-Anandhu]

Can i just print the elements instead of storing it in another array .

[@vineethyadav4221]

For freshers what is the package for springboot if we are good at spring boot? and if we done a project on springboot is it a plus point ?

[@cloudtech5260]

Yes, having springboot in your tech stack id plus point, you may get better package.

[@iiTzAkash]

optimization itna easy interview

[@rahulg415]

Can't we give odd number condition in else block instead of another for loop ?

[@cloudtech5260]

Hi Rahul, No. As we don't know the number of even number. If you add else then it will be even odd combination in between.

[@rahulg415]

@@cloudtech5260 Thanks for the response!

[@AkashGupta-yo1wl]

impressive how he build a logic ...i wonder how much it will take me to build logic like that.

[@cyber_dbs]

what!

[@AkashGupta-yo1wl]

@@cyber_dbs the way that guy is coding in video.

[@cyber_dbs]

don't mind me, but I was just surprised to see someone getting impressed by watching how to solve one of the easiest coding problem to solve.

[@AkashGupta-yo1wl]

@@cyber_dbs well i am just began my coding journey and for me it is really something ahead from what i am doing right now...

[@chandanpatra3850]

2 pointer approch you can easily solved this question. Because this unsorted array

[@pranoydas9654]

Three for loop. What will be the complexity?

[@sumanmodak1846]

This is very easy level question and I can't believe it that it is for 9Lpa.. I can do hard level questions and have buit full stack projects in Mern stack and in php frameworks also. But still unemployed. Sach me yaar..naseeb apna apna

[@cloudtech5260]

Hi Suman, Keep trying, you will soon get the job. Best of luck 👍

[@sourabhprajapat6637]

can we connect suman

[@GowthamiKaranam]

How to prepare for logical code??

[@Shorts_n_Laughs]

Are they still recruiting for the same post?

[@xolo2617]

Very easy problem ***

[@ashaymishra7082]

Too easy🙂

[@prashanthreddy-sd9ju]

Miku 6 to 7 years experience unna kuda firstly they will ask core Java question only its true

[@raviravi-gg5ck]

Alternative approach List list1 = Stream.array(arr1).filter(e -> e%2==0).collect(Collectors.toList); List list2= Stream.array(arr1).filter(e -> e%2 !=0).collect(Collectors.toCollection(()->(list1)); System.out.println(list2);

[@sudarshanpatil2227]

hey, can you solve this without using extra array?

[@amitmahato6404]

Two pointers. First iterate over the array and count even numbers. Take two pointers l=0, r=count+1, as odd numbers will be placed after even numbers. Then again iterate over the array and check if it's even make a[l]=a[i], then l++. Vice versa for odd numbers

[@zehansayed3631]

9 lpa is very less 20 lpa is good

[@themoonlight1922]

Try to do it in place

[@saurabhgupta6681]

Lol... I thought, I need much more preparation and leetcode practice 😂

[@Amit6316-w6e]

Sir i want to ask something to you, I m a commerce student but I m very interested in coading is this better for me.?

[@cloudtech5260]

Yes. If you get in IT industry definitely it is good for you 👍

[@abhishek7713]

😅 ye or exp walo ko puchenge 😅😅

[@1313durgesh]

we can also use Streams

[@cloudtech5260]

Yes

[@PatelIndra]

This program done by using only 1 for loop ..Guys ..We don't want take more than 1 for loop

[@surajchougala]

I am having 2.8 years experience in java developer and what package should I need to ask

[@cloudtech5260]

Hi Suraj, it depends on the company, but the general rule is Average package = years of experience* 2 Good package = years of experience* 3 Best package = years of experience* 4 These values are general guidelines, some people can get even higher or lower packages. Best of luck 👍

[@mriduljain8092]

Sama banda interviewer accenture wali video me bhi tha..... How is this possible ?

[@cloudtech5260]

Hi Mridul, Mock interview 👍

[@AdBoss12]

But in which app they are coding please can anyone tell i want to download that

[@cloudtech5260]

You can use eclipse for coding.

[@AdBoss12]

@@cloudtech5260 thank you sir i can learn a lot from your videos

[@zamanali713]

Mazaak tha kya ye😊... It can easily be solved with time complexity of O(N).....with just one for loop....in javascript....Why to use so many for loops.....

[@rathneshpolavarapu9628]

Even in 5lpa jobs asking to implement linkedlists how cognizant giving 9 lakhs for this question ❓ This is something looks fishy

[@cloudtech5260]

You can compare the packages offered before and after covid 19. Getting 9-10LPA in 2-3 years of experience is very easy in MNCs.

[@prudhvi_reddy]

I think it is not a 9lpa interview 🙄...some fresher's hiring interview

[@ranganath4110]

This is too easy I am fresher I am looking for a job I didn't get any chance

[@hussainrizvi7316]

this can be done in contant space and O(N ) time, this video tittle is only a clickbait ....

[@pranavdambalkar5365]

int[] myArr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int[] processedArray = new int[myArr.length]; int left = 0; int right = myArr.length - 1; for (int i = 0; i < myArr.length; i++) { if (myArr[i] % 2 == 0) { processedArray[left] = myArr[i]; left++; } else { processedArray[right] = myArr[i]; right--; } } System.out.println(Arrays.toString(processedArray)); hope this helps :D

[@sanketpoojari6057]

How much scripted do you want it to be? CloudTech: Yes!!

[@bharanRK]

Dude if you don't use time/space complexity.. as a factor in your coding.. you will never be able to crack top companies programming rounds

[@KaustavOP]

Is is on campus interview?

[@cloudtech5260]

This is a mock interview 👍

[@pratika-prakhar]

solution is not optimized? you are using two for loops and also using extra space. This can be done by O(n) time complexity and O(1) space complexity

[@cloudtech5260]

Yes. Correct. Will optimize the upcoming solutions. 👍

[@greyhat6599]

Hey chill ! Although, optimizing the problem is always nice practice , but the main motive here is to solve the problem and not to optimize it.....☺️ And He is really doing it well ! I appreciate it !!💝❣️

[@namanpatel7366]

I had the same idea similar to what @GreyHat was emphasising. It improves our time complexity as well as space complexity but the only trade off will be maintaining the order of the numbers. //arrange the array in even numbers first, than odd numbers import java.util.*; class HelloWorld { public static void main(String[] args) { int[] numbers = {1,2,5,4,7,8,11,20}; // if the order of the elements in the array doesn't matter // we already have an idea about our expected result which is even numbers on the right and odd numbers on the left //lets take two pointer approach int i = 0; int j = numbers.length-1; while (i

[@namanpatel7366]

Some Similar problems to practice will be 1) Segregate 0s on the left and 1s on the right side from a given array. While traversing the array only once. int[] numbers = {1,0,1,1,0,0,1,0} 2) Segregate 0s on the left side, 1 in the middle and 2s on the right from a given array. int[] numbers = {1,2,1,1,0,0,2,2,0}

[@sateesh9525]

It's fake call I think😂 for views, simple one and now a days every one using streams if u write like this they think u r not aware about latest coding skills

[@ravikumark2023]

Idhu interview? Adha na nambanum🌚

[@Sayan_Shankhari]

when will you stop using mod to get odd even, just check if last digit is in [0, 2, 4, 6, 8] or not 😂

[@rakipalivela64]

Bye bye, thank you 😂😂😂

[@MrBkkrishna]

if he IIT or IIM and they clear they give 10 laks not for everyone simply fooling around telling of money

[@an_other_world]

And why was i recommended this KZbin?

[@veereshj9421]

Lol no one gives you 9lpa for such questions

[@rohankrishnani]

I don't think these are true interviews, the guy is simply faking it .

[@cloudtech5260]

Hi Rohan, Mock interview 👍

[@avocadorable3695]

same person taking interview in every company,,, lol

[@cloudtech5260]

This is mock interview with real-time questions.

[@kartikrajput9880]

Kafi Faila hua business hai Control Majnu... Control

[@susheelkumar291]

Noob level max++

[@ritwikbhar21]

Very bad solution. Why do you even need an extra array.

[@SagarSingh-il7ui]

It can be solved in single pass without using auxiliary space as well. Let me know if anyone wants the code 😉

[@cloudtech5260]

Hi Sagar, Good day. Please paste the code in the comments section, it will be really helpful. Thanks. 👍

[@SagarSingh-il7ui]

public class YtSolution { // {1,3,2,4,5,10,7,11,9,8} public static void main(String[] args) { int[] a = {1,3,2,4,5,10,7,11,9,8}; int i = 0; for(int j = 0;j

[@xolo2617]

@@SagarSingh-il7ui How will you maintain order then? question says you have to solve this in ascending order ? you need sorting here don't focus only one testcase try to think for all test cases

[@extremeweirdness1528]

Lol I wasn't asked a single coding question in the interview for genCnext and in my opinion don't join these companies also. You will spoil your career.

[@abhijitzende7292]

import java.util.*; // Compiler version JDK 11.0.2 class EvenFirst { public static void main(String args[]) { Scanner sc = new Scanner(System.in); int len = sc.nextInt(); int[] arr = new int[len]; for(int i=0;i

[@cloudtech5260]

Thanks Abhijit 👍

[@easywaytostudy8061]

Java code chalange kzbin.infoltgMz_nyZBE?si=jxae4Zp21gArbDwB

[@gopalkannan4934]

Thank you