Software Development 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; }

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..!

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..!

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

algorithm for C- Compiler 1.start . 2.include header files from the library (required for future compiling code) 3.In main create and allocate memory asked for. 4.Go through the conditions and use appropriate queries to have ALU working. 5.go for the end,if not generate error. 6.Stop.

How buckets are foound, how elements are saved