SDE - 1

Approach : 
1) Make a count array and initialize all characters by -1, i.e., all considered as absent.
2) Traverse the given string, and the value of count[x] will be the index of ‘x’ if ‘x’ appears only once. Else, the value is going to be either -1 or -2.
3) Now, traverse the count array and check if the ith character occurs only once (count[i] has a positive value) and appears before the current result, then update the result.
4) Add the character present at the result index in the arraylist.

Approach :
1) I solved it in O(N^2) approach .
2) Sort the array
3) Initially map all the elements of the array to their index in a Hash Map or a Hash Set
3) Now , run 2 for loops and for every x^2 + y^2 ,check if there exists a z^2 s.t x^2+y^2=z^2 and the index of z^2 is different than both the indices of x and y

This was a very standard DP problem and I had already solved it on platforms like LeetCode and CodeStudio so I was able to come up with the logic and code it preety fast.

Steps :
1) Create a two-dimensional array, ‘dp’, where ‘dp[i][j]’ will denote the total number of ways to make j value by using i coins.
2) Run 2 loops ,1st one from 1 to value and second one throught the array denominations.
3) Fill dp array by the recurrence relation as follows:
ways[i][j] = ways[i-1][j] + ways[i][j-denominations[i]] (if j-denominations[i]>=0)
Where first term represents that we have excluded that coin and,
Second term represents that we have included that coin.
4) Base Case : Fill dp[0]=0 (as we can make 0 amount by giving no coins at all) and the remaining indices with INT_MAX value
5) Finally , if dp[value]==INT_MAX , return -1 as it is not possible to make "value" using the given coins else return dp[value] itself


TC : O(N*V) where N=number of coins and V=Value to be made
SC : O(N*V)

Answer :
1) Views in SQL are kind of virtual tables.
2) A view also has rows and columns as they are in a real table in the database. 
3) We can create a view by selecting fields from one or more tables present in the database. 
4) A View can either have all the rows of a table or specific rows based on certain condition.

Answer :

Method 1 – Taking a Backup of a MySQL Table Using INSERT INTO

CREATE TABLE table_backup; 
INSERT INTO table_backup SELECT * FROM table;

This method creates the structure of the table including indexes and then loading the data in via one statement.


Method 2 – Taking a Backup of a MySQL Table Using mysqldump

mysqldump -u{backup_user} -p{backup_password} from_db_name table_to_backup > backup_file.sql

The backup is taken using mysqldump and can be directed to a location of choice. Disk space is therefore not a consideration for the data drive, rather it is necessary just for the location being backed up to.

Answer : 

Constructors :
1) A Constructor is a block of code that initializes a newly created object.
2) A Constructor can be used to initialize an object.
3) A Constructor is invoked implicitly by the system
4) A Constructor is invoked when a object is created using the keyword new.
5) A Constructor doesn’t have a return type.
6) A Constructor’s name must be same as the name of the class.
7) A Constructor cannot be inherited by subclasses.

Method :
1) A Method is a collection of statements which returns a value upon its execution.
2) A Method consists of Java/C++ code to be executed.
3) A Method is invoked by the programmer.
4) A Method is invoked through method calls.
5) A Method must have a return type.
6) A Method’s name can be anything.
7) A Method can be inherited by subclasses.

Approach : 

I) Using Hashing :

Basically, we need to find a common node of two linked lists. So we hash all nodes of the first list and then check the
second list.

Note : Hash the nodes not the node value

1) Create an empty hash set.
2) Traverse the first linked list and insert all nodes' addresses in the hash set.
3) Traverse the second list. For every node check if it is present in the hash set. If we find a node in the hash set,
return the node

TC : O(m+n)
SC : O(min(m,n))

II) Using Two pointers :

1) Initialize two pointers ptr1 and ptr2 at the head1 and head2.
2) Traverse through the lists,one node at a time.
3) When ptr1 reaches the end of a list, then redirect it to the head2.
4) Similarly when ptr2 reaches the end of a list, redirect it the head1.
5) Once both of them go through reassigning, they will be equidistant from the collision point
6) If at any node ptr1 meets ptr2, then it is the intersection node.
7) After second iteration if there is no intersection node it returns NULL.

TC : O(m+n) where m=Length of LL-1 and n=Length of LL-2
SC : O(1)

Approach 1 :
1) Initialise a string named rev with the given string .
2) Reverse the string rev.
3) Now, if s is a palindrome then we would have rev==s , return true if rev==s and false otherwise

TC : O(N) where N=length of the string
SC : O(1)

Approach 2(Using Two Pointers) : 
1) Keep pointer i=0 and j=n-1 where n=length of the string
2) Run a loop till i<=j and check if s[i]==s[j] for the condition of palindrome
3) If at any point s[i]!=s[j] then simply return false from the loop
4) Else return true after the end of the loop.

TC : O(N)
SC : O(1)

Steps : 
Let's call 5L jar as X and 3L jar as Y. 

1) We'll take the first jar, i.e, X. Fill X completely. 
2) Then empty the X content in Y. We are left with 2 L in the X. 
3) Now we will empty the Y jar and then Fill 2L from X into Y. We are left with 0L in X and 2L in Y.
4) Now again we will fill X with 5L and then empty till Y gets full. 
5) As Y is already full with 2L from previous operation so it can only accomodate 1L now from X. So when 
Y jar is full automatically we will have 4L left in X. This is how we will calculate 4L

Answer : 

1) Garbage Collection in Java is a process by which the programs perform memory management automatically. 
2) The Garbage Collector(GC) finds the unused objects and deletes them to reclaim the memory. I
3) In Java, dynamic memory allocation of objects is achieved using the new operator that uses some memory and the 
memory remains allocated until there are references for the use of the object.
4) When there are no references to an object, it is assumed to be no longer needed, and the memory, occupied by the 
object can be reclaimed. 
5) There is no explicit need to destroy an object as Java handles the de-allocation automatically.

The technique that accomplishes this is known as Garbage Collection. Programs that do not de-allocate memory can eventually crash when there is no memory left in the system to allocate. These programs are said to have memory leaks.

Garbage collection in Java happens automatically during the lifetime of the program, eliminating the need to de-allocate memory and thereby avoiding memory leaks.

In C language, it is the programmer’s responsibility to de-allocate memory allocated dynamically using free() function. This is where Java memory management leads.

Answer :

a) It is an IT company that caters to the telecommunication domain. The business involved in the telecommunication domain is interesting and can widen your chances of switching into various fields be it in software, hardware or networking profiles. Also, Amdocs carries a good brand name in its domain.

b) Amdocs employees get to enjoy a positive and happy work environment.

c) It is truly an employee friendly organisation.I say this because Amdocs policies are employee oriented above anything else.

Tip 1 : The cross questioning can go intense some time, think before you speak.
Tip 2 : Be open minded and answer whatever you are thinking, in these rounds I feel it is important to have opinion.
Tip 3 : Context of questions can be switched, pay attention to the details. It is okay to ask questions in these round, like what are the projects currently the company is investing, which team you are mentoring. How all is the work environment etc.
Tip 4 : Since everybody in the interview panel is from tech background, here too you can expect some technical questions. No coding in most of the cases but some discussions over the design can surely happen.