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Er werd een SQL Server Developer gevraagd...22 maart 2015

### If I knew what the cross apply operator was

4 antwoorden

If they want the word its join, a description could be "the Cartesian product of multiplying the tables" and an example would be "like merging two matrices except every column may be of incongruent variables creating a result with polynomial rows, (which is likely no longer normalized to the form level of the database)" Minder

An Apply operator is a join type that the query engine evaluates in sequential order because the right-hand table object can reference data in the left-hand table object. The Cross Apply operator is the Apply operator that acts like an Inner Join, eliminating any records in the left-hand table object that have a null match in the right-hand table object. Minder

An Apply operator is a join type that the DB engine evaluates in sequential order because the right-hand table object can reference data in the left-hand table object. The Cross Apply operator is the Apply operator that acts like an Inner Join, eliminating any records in the left-hand table object that have a null match in the right-hand table object. Minder

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### How to discover if the given number is Power of 2 ?

4 antwoorden

return (( x!=0) &amp;&amp; (x &amp; (x -1) == 0)) a number which is power of two will have single 1 in binary representation ex. 16 00010000 x-1 will have pattern 00001111 hence x&amp;(x-1) will always be 0 for power of two numbers. Need to check condition for x!=0 since 0 is not a power of two. Special case to handle. Minder

if you can figure this out inside an hour, congratulations this isnt something that should be in an interview to be honest the level of difficulty of the answer that algo came up with is at the level of an upper division course that has a proof of it written in wikipedia i honestly like the brute force solution Minder

Divide the number by 2, and repeat, until the result is a not an integer number (in this case is not a power of 2) or the result is 1 (in this case the number was a power of 2). Minder

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### write a shell script to detect ddos attack

2 antwoorden

The key in these questions is to cover the fundamentals, and be ready for the back-and-forth with the interviewer. Might be worth doing a mock interview with one of the MobiKwik or ex-MobiKwik Server Engineer experts on Prepfully? They give real-world practice and guidance, which is pretty helpful. prepfully.com/practice-interviews Minder

### Come up a most efficient way to determine if a integer is odd or even.

2 antwoorden

int number = ...; if(number % 2) { odd } else { even }

Compilers are smart enough to understand the result of n%2. Don't over-complicate it. Minder

### why you are looking for change?

2 antwoorden

The job that I have doesn’t pay enough

I answered honestly that i am looking for increase in salary

### Binary tree , left node has smallest value . Print 5 10 20 25 and then “count of nodes=4”…

2 antwoorden

recursion ... defined "current=head" in the method so it doesn't changes ofcourse, even if I sent current.next in recursion... don't be nervous! Minder

Did not get this question? Could you please eleborate it more?

### - Override equals and hashcode - Implement ArrayList - question on wait/ notify - Design a ticket system

2 antwoorden

Was able to answer every question. HR person said that he had got a good feedback from the panel and they will update me as early as possible. But, got rejected for no specific reason. Minder

Also for overriding hahcode and equal what was the input gievn to you?

### What is the end plastic connector on a CAT 5 cable

2 antwoorden

RJ45 connector

Did you undergo face to face interview?

### Implement a queue using two stacks.

2 antwoorden

This one was easy.

#include #include using namespace std; //class MyQueue template class MyQueue { public: MyQueue(){} // pay attention to this ~MyQueue() {} void push(T element) ; void pop() ; bool isEmpty() ; T front() ; T back() ; private: stack stack1 ; stack stack2 ; }; // stack1: newest elements // stack2: oldest elements template void MyQueue::push(T element) { stack1.push(element) ; } // pop the oldest element in the queue template void MyQueue::pop() { if ( stack2.empty()) // change to oldest { while (!stack1.empty()) { stack2.push(stack1.top()); stack1.pop(); } } stack2.pop(); } // oldest template T MyQueue::front() { if( stack2.empty() ) { while (!stack1.empty()) { stack2.push(stack1.top()); stack1.pop(); } } return stack2.top() ; } template T MyQueue::back() { if( stack1.empty() ) { while (!stack2.empty()) { stack1.push(stack2.top()); stack2.pop(); } } return stack1.top(); } template bool MyQueue::isEmpty() { if (stack1.empty() &amp;&amp; stack2.empty()) { return true ; } else return false ; } Minder

### Write a function that compares two strings and returns a third string containing only the letters that appear in both.

1 antwoorden

string getcommstring(string s1,string s2) { string s; int size1 = s1.size() ; int size2 = s2.size() ; int hash[30] ; for (int i = 0 ; i &lt; 30 ; i++) { hash[i] = 0 ; } for ( int i = 0 ; i &lt; size1 ; i++ ) { hash[s1[i] - 'a'] = 1 ; } for (int i = 0 ; i &lt; size2 ; i++) { if ( hash[s2[i] - 'a'] == 1 ) { s.push_back(s2[i]) ; hash[s2[i]-'a']= 0 ; } } return s ; } Minder

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