Software Engineer In Test Interview Questions

NC

#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; }

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..!

Answered technical question correctly but, all the puzzles were not correct.

Get a call from a3logics for interview for the position on Qa-Test Engineer. can u suggest me what type of questions and rounds of interview i can expect

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++; } }

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

int maxSum(int arr[], int size) { int maxSum= 0; int currMaxSum = 0; for (int i = 0; i < size ; i++) { currMaxSum += arr[i]; if (currMaxSum < 0) { currMaxSum = 0; } if (maxSum < currMaxSum ) { maxSum = currMaxSum ; } } returb naxSum; } }

private static int findMaxSubSeqSum(ArrayList allRecords) { int maxSum = 0; int prevSum = 0; int currentSum = 0; for (int i = 0; i maxSum) { maxSum = currentSum; } if (currentSum < prevSum) { currentSum = 0; } } return maxSum; }

could not answer

array1 [5] = {1,3,7,10,11} array2 [4 ]= {2,4,5,6} position of median will be=(5+4)/2=4(floor value) now,simply check whether which of the two,array1[4/2] or array2[4/2] is smaller. time complexity: O(1) Considering that size of the 2 arrays was known prior..!