Capgemini Java Coding Interview | Watch this if you want offer

In this video, we covered Capgemini Coding Interview Questions. The candidate got selected in the company with the offer of 10LPA.
We solved two problems.
1.) Finding the missing number from the array
2.) Convert first half of the String in lower case and second half in upper case

[@ashishsrivastav11111]

Great ..now we are aware of character class and its methods, toUpperCase(), toLowerCase()

[@roshanpatro5777]

Really good questions. Thank you!! Please, make videos on questions related to Collection also.

[@Tusharsharma028]

tysm for teaching begneers

[@TheCorporateCricket-y4j]

One more way to do this, will just go on checking with the reference of index value I will check with arr[0] == I+1 if not the missing element is the I+1

[@parthjethwani1837]

array need to be sorted

[@TheCorporateCricket-y4j]

@@parthjethwani1837 yes

[@sanketpatil6084]

i was asked missing and repeating number program for 3.5 lpa package

[@nikhileshyoutube4924]

Excellent

[@radhikagadve]

If they really ask these questions I'm gonna go into interviews right now

[@deepakraj-bu3ez]

public static void main(String[] args) {
int [] arr = new int []{1,2,3,5, 6,7,8,9} ;
int n= arr. length+1;
int sum = n* (n+1) /2;
for (int i : arr ){
sum -= i;
}
System.out .printIn("Missing number is "+sum);

[@somublackstar]

please don't do DS and Algo, we can get this easily.
Do class related videos like Employee, list of employees , increase their salary, group by etc

[@OS-bt7il]

if only one no is missing then only answer will work.what if 2 no are missing .

[@pca33teja93]

Don't you think for missing number code the formula will result in overflow if n is very very large

[@cloudtech5260]

Hi Teja,
Very good question.
This logic can handle upto n = 46300 elements easily. The calculation n*(n +1) will still fit in int data type. If you want even more elements then we can go for long data type.

[@Aman_Gupta54]

So you don't need any actual DSA problem solving skills for Capgemini??
Is it that easy or am i missing something

[@cloudtech5260]

For Capgemini Java 8 is enough. 👍

[@rahulkharapkar9962]

Cloudtech ,not anymore... I am working on java 11 and it's not this easy for sure they do ask array and some advanced DSA concepts as well.. this looks like fresher interview

[@omprakashnaik693]

@@rahulkharapkar9962 hii bro

[@java_minds1755]

String s = "India is My Country India is My Country";
String s1 = s.substring(0,s.length()/2);
String s2 = s.substring(s.length()/2);
String result = "\0";

result = s1.toLowerCase();
result += s2.toUpperCase();
System.out.println(result);

[@kiranbhopale1249]

What is meaning of "\0" ?

[@surabhisahu1231]

@@kiranbhopale1249 means null
Like 0 in integers

[@ramakrishnakoonapareddy3251]

Ur logic so simple sir

[@bhabanisankar1770]

String s1=s.substring(0,(s.length()/2).toLowerCase();
String s2=s.substring(s.length()).toUpperCase();

[@harshwardhanshetty4683]

@@ramakrishnakoonapareddy3251 i think it's readability is simple

[@shivarajhiremath5963]

Here is the simplest way
public class StringQuestion {
public static void main(String[] args) {
String str="Capgemini Coding Interview Questions Capgemini Coding Interview Questions";
String upper=str.substring(str.length()/2,str.length()).toUpperCase();
String lower=str.substring(0,str.length()/2).toLowerCase();
System.out.println("The final output is "+lower +" " +upper);
}
}

[@kvnagendra5354]

Plz post real time questions

[@cloudtech5260]

Yes will do. 👍

[@gauravkumar6124]

I must have the worst luck, for a 5LAP job, they asked me codewars morse code advance, as a fresher. 🤣🤣

[@cloudtech5260]

Hi Gaurav,
Which company 😁

[@gauravkumar6124]

@@cloudtech5260 giyzer systems.

[@cloudtech5260]

Great! Have a wonderful career ahead 👍

[@gauravkumar6124]

@@cloudtech5260 na, i got rosted in that interview, i didn't know what i am supposed to do😅😅

[@gauravkumar6124]

@@cloudtech5260 official feedback was that, candidate is arrogant, rude and does not know coding.

[@mohammedajazquadri7869]

Birlliant

[@swapnilbagul7502]

This is not an actual capgemini interview....

[@cloudtech5260]

This is mock interview with actual questions.

[@anitakaruturi]

Sound is too low!

[@cloudtech5260]

We will fix in upcoming videos!! 👍

[@lonestoryteller5430]

He could have taken an variable with initial value of 1 and compare it with every element. If equal, increase value of that variable. Where the variable and element value differ, break the loop and cur value of that variable is the output.

[@AshishKumar-cy7dx]

For that there will be two for loop.time complexity will be O(n^2).here it is O(n).

[@lonestoryteller5430]

@@AshishKumar-cy7dx Nope. If the array is sorted and increasing or decreasing by 1, as given in the problem, it will take only linear TC.

[@AshishKumar-cy7dx]

@@lonestoryteller5430 array is not sorted.for sorting again you need to run two for loop

[@lonestoryteller5430]

@@AshishKumar-cy7dx No brother. Listen to the problem statement carefully. It says there n numbers are given then the interviewer says suppose 1 to 10 and one number is missing. Also, we don't need nested loops to sort the array. Arrays.sort () function will do it in logN TC

[@AshishKumar-cy7dx]

@@lonestoryteller5430 you listen and understand problem statement .u can't use inbuilt function in any interview.its is just one of case in sorted order,you have to write logic for all case.

[@ashoksaho]

If I provide the number in put how u find the last element in this case 10

[@cloudtech5260]

Hi Ashok,
We already know the sum of first 10 numbers using formula sum1 = n*(n+1)/2. We will then take the sum of input array.
Missing number = sum1 - sum of input array

[@alim241081]

capgi bht chindi company hai package ke mamle me

[@Natkhat_reels_n_Vlogss]

With 2 year bond😂🤣

[@85PREMKUMAR]

String lowerUpperString = new StringBuilder().append(str.substring(0, mid).toLowerCase()).append(str.substring(mid).toUpperCase()).toString();