Software development engineer in test Interview Questions

Possible implementation in C++ -------------------------------------------------- int CounterStep(int counter,int N); int LocatePosition(int *cards,int N, int startPointer,int value); using namespace std; void main() { int cards[20]; int N=20; int POS = 0,TOP = 0; // Initializing everything to zero for(int i=0;i

Written in Python although it can be re-written in C/C++ if required. def generateSpecialSeries(numberOfCards): specialSeries = [] if numberOfCards > 0: for i in range(numberOfCards, 0, -1): specialSeries.append(i) currentSpecialElement = 1 currentIndex = 1 while currentIndex < numberOfCards: indexOfCurrentSpecialElement = numberOfCards - currentSpecialElement specialSeries[currentIndex], specialSeries[indexOfCurrentSpecialElement] = specialSeries[indexOfCurrentSpecialElement], specialSeries[currentIndex] currentSpecialElement += 1 currentIndex += currentSpecialElement + 1 return specialSeries specialSeries = generateSpecialSeries(50) print specialSeries specialElement = 1 specialElementSum = 1 currentIndex = 1 while currentIndex < 50: print specialSeries[currentIndex] == specialElement specialElement += 1 specialElementSum += specialElement currentIndex += specialElement + 1

// Short and simple C++ algorithm #define N 10 void compute(int stack[]) // Make sure N elements are allocated { int j=0; for(int i=0; i

I misunderstood the question, and wrote the program for printing number of a's or b's etc., present in the sequence when I showed him he asked me to code according to his requirement, the program should produce output without any errors,during this time he was writing something in the laptop. I wrote program correctly. He said ok.

void PrintAlphaNum( string str ) { if ( (str == null) || str.Length == 0 ) { return; } char lastChar = str[0]; int count = 0; foreach( char ch in str ) { if ( ch == lastChar ) { count++; } else { Console.Write("{0}{1}, ", count, lastChar ); count = 1; lastChar = ch; } } if ( count > 0 ) { Console.WriteLine("{0}{1}", count, lastChar ); } }

#Iinclude #include main() { char str[ ]; printf("Enter the input\n"); scanf("%c",&str); printf("input:%c",str); int len= strlen(str); int alp[26]; if(str!='NULL' && len!=0) { for(int i=0;i

#include #include using namespace std; int main() { int list[10] = {5, 4, 3, 6, 7, 1, 2, 9, 8, 10}; for (int i=0; i= 0 && next list[next]) { int temp = list[i]; list[i] = list[next]; list[next] = temp; next--; } else next++; } } for (int i=0; i<10; i++) { cout << list[i] << " "; } return 0; }

you still have two loops #include #include using namespace std; int main() { int list[10] = {5, 4, 3, 6, 7, 1, 2, 9, 8, 10}; for (int i=0;j=i; i<10; i++;j++) { I can't remember, but you use two indices within one loop to sort using bubble sort } for (int i=0; i<10; i++) { cout << list[i] << " "; } return 0; }

These were the questions for written round? Written test was pen paper test or an online test?

yes those questions were for written technical round and it was -pen and paper based test

Interviews for the same were on the same day?

there is a solution in O(n) time

@ Interview Candidate: can you give some details about the logic ?

import java.util.HashMap; public class MSQuestion2 { public static void main(String[] args) { // TODO Auto-generated method stub CustomLinkedList lst = new CustomLinkedList(); lst.head=new Node1(1); lst.incrementsize(); Node1 node2 = new Node1(2); lst.head.next=node2; Node1 node3 = new Node1(3); node2.next=node3; lst.printList(lst); // lst.reverseList(lst); lst.printList(lst); CustomLinkedList clonelst= lst.clone(lst); lst.printList(clonelst); } } class Node1 { int data; Node1 next; Node1 random; public Node1(int data) { this.data=data; this.next=this.random=null; } } class CustomLinkedList { Node1 head; static int size; public CustomLinkedList() { this.head=null; this.size=1; } public CustomLinkedList(Node1 head) { this.head=head; this.size=1; } public CustomLinkedList clone(CustomLinkedList lst) { // TODO Auto-generated method stub CustomLinkedList lst2 = new CustomLinkedList(); HashMap map = new HashMap(); Node1 n = lst.head; Node1 clonednode = null; while(n!=null) { clonednode = new Node1(n.data); map.put(n, clonednode); n=n.next; } n = lst.head; while(n!=null) { clonednode=map.get(n); clonednode.next=map.get(n.next); clonednode.random=map.get(n.random); n=n.next; } lst2.head=map.get(lst.head); return lst2; } public void printList(CustomLinkedList lst) { Node1 n = lst.head; while(n!=null) { System.out.println(n.data); n=n.next; } } public void reverseList(CustomLinkedList lst) { Node1 n=lst.head; Node1 prev =null; Node1 next = null; while(n!=null) { next=n.next; n.next=prev; prev=n; head=n; n=next; } } public void add(CustomLinkedList lst,int data) { Node1 node= new Node1(data); node.next=null; size++; } public void incrementsize() { this.size++; } }

you can modify the lists to form circular lists. that both are merged to each other without temp.

In case link lists are sorted, then per-form merge sort tech. otherwise above seems fine

Suck my dick to know the correct answer..!

I have written a program using Hashmap and StringBuilder.

An efficient solution to the problem can done using simple pattern matching. A sample solution goes as follows : public class SampleMain { /** * @param args */ public static void main(String[] args) { String charStream = "ABABCBABABABABABATGBHacbfoabcbahasacbJKOPLMNBGDESRABADACAJALAMNBADDHEMSZPQ"; String noise = "ABCPQR"; System.out.println("CharStream : " + charStream + "\n" + "Noise : " + noise + "\n" + "FliteredCharStream : " + func(charStream, noise)); } private static String func(String charStream, String noise) { String regex = "[" + noise.toLowerCase() + noise.toUpperCase() + "]"; Pattern p = Pattern.compile(regex); Matcher m = p.matcher(charStream); return m.replaceAll(""); } }

import java.util.Arrays; public class Solution{ public static void main(String[] args){ System.out.println(remove("as","asasfsg")); } public static String remove(String s1, String s2){ boolean[] flags = new boolean[255]; char[] a1 = s1.toCharArray(); char[] a2 = s2.toCharArray(); String str = ""; Arrays.fill(flags,false); for(int i=0; i=a1.length && i