Arun chakravarthy Alagar samy. Technical Skill. Java Code Challenge - Basic Java

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1 Arun chakravarthy Alagar samy DATE OF LAST ACTIVITY TEST Java Code Challenge - Basic LANGUAGES USED Java Technical Skill CANDIDATE RESULT SKILL THRESHOLD Strongly recommended regarding technical potential. Recommended regarding technical potential. More information needed. Not recommended if technical ability is crucial. TASKS IN TEST CORRECT SUBMISSIONS PLAGIARISM LANGUAGE EASY Speed Limit 2 Java EASY A New Alphabet 1 Java MEDIUM Phone List 4 Java

2 EASY Speed Limit SHOWN AS Problem A MEMORY LIMIT 1024 MB CORRECTNESS Accepted PLAGIARISM Not detected CPU TIME LIMIT 1 second LANGUAGE USED Java ATTEMPTS 2 Bill and Ted are taking a road trip. But the odometer in their car is broken, so they don t know how many miles they have driven. Fortunately, Bill has a working stopwatch, so they can record their speed and the total time they have driven. Unfortunately, their record keeping strategy is a little odd, so they need help computing the total distance driven. You are to write a program to do this computation. For example, if their log shows Speed in miles per hour Total elapsed time in hours this means they drove 2 hours at 20 miles per hour, then 6 2 = 4 hours at 30 miles per hour, then 7 6 = 1 hour at 10 miles per hour. The distance driven is then = = 170 miles. Note that the total elapsed time is always since the beginning of the trip, not since the previous entry in their log. Input The input consists of one or more data sets (at most 10). Each set starts with a line containing an integer n, 1 n 10, followed by n pairs of values, one pair per line. The first value in a pair, s, is the speed in miles per hour and the second value, t, is the total elapsed time. Both s and t are integers, 1 s 90 and 1 t 12. The values for t are always in strictly increasing order. A value of

3 1 for n signals the end of the input. Output For each input set, print the distance driven, followed by a space, followed by the word miles. Sample Input 1 Sample Output miles 180 miles 90 miles

4 TotalDistance2.java package com.assessment.question1; import java.util.arraylist; import java.util.list; import java.util.scanner; public class TotalDistance2 { public static void main(string[] args) { Scanner sc = new Scanner(System.in); List<Integer> results = new ArrayList<Integer>(); try { int val = 0; while ((val = sc.nextint())!= -1) { int datasetcount = val; int prevhourentered = 0; int totalmiles = 0; int itr = 0; while ((itr++) < datasetcount) { int speed = sc.nextint(); int currhourentered = sc.nextint(); totalmiles += speed (currhourentered - prevhourentered); prevhourentered = currhourentered; results.add(totalmiles); for (int miles : results) { System.out.println(miles + " miles"); sc.close(); catch (Exception e) { System.out.println("Main class error :" + e.getmessage()); finally { if (sc!= null) sc.close();

5 EASY A New Alphabet SHOWN AS Problem B MEMORY LIMIT 1024 MB CORRECTNESS Accepted PLAGIARISM Not detected CPU TIME LIMIT 1 second LANGUAGE USED Java ATTEMPTS 1 A New Alphabet has been developed for Internet communications. While the glyphs of the new alphabet don t necessarily improve communications in any meaningful way, they certainly make us feel cooler. You are tasked with creating a translation program to speed up the switch to our more elite New Alphabet by automatically translating ASCII plaintext symbols to our new symbol set. The new alphabet is a one-to-many translation (one character of the English alphabet translates to anywhere between 1 and 6 other characters), with each character translation as follows: Photo by r. nial bradshaw ( Original New English Description Original New English Description at symbol n []\[] brackets, backslash, brackets b 8 digit eight o 0 digit zero c ( open parenthesis p D bar, capital D d ) bar, close parenthesis q (,) parenthesis, comma, parenthesis e 3 digit three r Z bar, capital Z f # number sign (hash) s $ dollar sign g 6 digit six t '][' quote, brackets, quote h [-] bracket, hyphen, bracket u _ bar, underscore, bar i bar v \/ backslash, forward slash j _ underscore, bar w \/\/ four slashes k < bar, less than x { curly braces l 1 digit one y `/ backtick, forward slash

6 m []\/[] brackets, slashes, brackets z 2 digit two For instance, translating the string Hello World! would result in: [-]3110 \/\/0 Z1 )! Note that uppercase and lowercase letters are both converted, and any other characters remain the same (the exclamation point and space in this example). Input Input contains one line of text, terminated by a newline. The text may contain any characters in the ASCII range (space through tilde), as well as 9 (tab). Only characters listed in the above table (A Z, a z) should be translated; any non-alphabet characters should be printed (and not modified). Input has at most characters. Output Output the input text with each letter (lowercase and uppercase) translated into its New Alphabet counterpart. Sample Input 1 Sample Output 1 All your base are belong to `/0 _ Z Z3 8310[]\[]6 ']['0 _ $. Sample Input 2 Sample Output 2 What's the Frequency, Kenneth? \/\/[-]@'][''$ ']['[-]3 # Z3(,) _ 3[]\[](`/, <3[]\[][]\[]3']['[-]? Sample Input 3 Sample Output 3 A new D[-]@83']['!

7 SwitchConverter.java package com.assessment.question2; import java.util.scanner; public class SwitchConverter { public static void main(string[] args) { Scanner sc = new Scanner(System.in); try { String input = sc.nextline().tolowercase(); String[] encode = { "@", "8", "(", " )", "3", "#", "6", "[-]", " ", "_ ", " <", "1", "[]\\/[]", "[]\\[]", "0", " D", "(,)", " Z", "$", "']['", " _ ", "\\/", "\\/\\/", "{", "`/", "2" ; char[] letters = input.tochararray(); for (char ch : letters) { int ascii = (int) ch; if (ascii >= 97 && ascii <= 122) System.out.print(encode[ascii - 97]); else System.out.print(ch); sc.close(); catch (Exception e) { finally {

8 MEDIUM Phone List SHOWN AS Problem C MEMORY LIMIT 1024 MB CORRECTNESS Accepted PLAGIARISM Not detected CPU TIME LIMIT 4 seconds LANGUAGE USED Java ATTEMPTS 4 Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let s say the phone catalogue listed these numbers: Emergency 911 Alice Bob In this case, it s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob s phone number. So this list would not be consistent. Input The first line of input gives a single integer, 1 t 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 n Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits. Note that leading zeros in phone numbers are significant, e.g., 0911 is a different phone number than 911. Output For each test case, output YES if the list is consistent, or NO otherwise. Sample Input 1 Sample Output 1

9 NO YES

10 PhoneDirectory2.java package com.assessment.question1; import java.util.arraylist; import java.util.collection; import java.util.collections; import java.util.list; import java.util.scanner; public class PhoneDirectory2 { public static void main(string[] args) { Scanner sc = new Scanner(System.in); ArrayList<String> phoneentries = null; ArrayList<String> results = new ArrayList<String>(); try { int val = sc.nextint(); do { int phoneentry = sc.nextint(); phoneentries = new ArrayList<String>(); while ((phoneentry--) > 0) { phoneentries.add(sc.next()); BinarySearchTree bst = new BinarySearchTree(phoneEntries); if (!bst.invalidtree) results.add("yes"); else results.add("no"); while ((--val) > 0); for (String result : results) { System.out.println(result); catch (Exception e) { System.out.println("Main class error :" + e.getmessage()); e.printstacktrace(); sc.close(); class TreeNode<E> { private E data; private TreeNode<E> left; private TreeNode<E> right; / Constructs a new node with the given data and references to the given left and right nodes. public TreeNode(E data, TreeNode<E> left, TreeNode<E> right) { this.data = data; this.left = left; this.right = right;

11 / Constructs a new node containing the given data. Its left and right references will be set to null. public TreeNode(E data) { this(data, null, null); / Returns the item currently stored in this node. public E getdata() { return data; / Overwrites the item stored in this Node with the given data item. public void setdata(e data) { this.data = data; / Returns this Node's left child. If there is no left left, returns null. public TreeNode<E> getleft() { return left; / Causes this Node to point to the given left child Node. public void setleft(treenode<e> left) { this.left = left; / Returns this nodes right child. If there is no right child, returns null. public TreeNode<E> getright() { return right; / Causes this Node to point to the given right child Node. public void setright(treenode<e> right) { this.right = right; class BinarySearchTree<E extends Comparable<E>> { private TreeNode<E> root = null; private int size = 0; public boolean InvalidTree = false; / Creates an empty tree. public BinarySearchTree() { public BinarySearchTree(Collection<E> col) { List<E> list = new ArrayList<E>(col); Collections.shuffle(list); for (int i = 0; i < list.size(); i++) { add(list.get(i)); / Adds the given item to this BST.

12 public void add(e item) { this.size++; if (this.root == null) { // tree is empty, so just add item this.root = new TreeNode<E>(item); else { // find where to insert, with pointer to parent node TreeNode<E> parent = null; TreeNode<E> curr = this.root; boolean wentleft = true; // InvalidTree = true; while (curr!= null &&!InvalidTree) { // will execute at least once parent = curr; int flen = item.tostring().length(); int slen = curr.getdata().tostring().length(); String fstring = null; String sstring = null; if (flen > slen) { fstring = item.tostring(); sstring = curr.getdata().tostring(); else { sstring = item.tostring(); fstring = curr.getdata().tostring(); if (fstring.startswith(sstring)) { InvalidTree = true; else { if (item.compareto(curr.getdata()) <= 0) { curr = curr.getleft(); wentleft = true; else { curr = curr.getright(); wentleft = false; if (!InvalidTree) { // now add new node on appropriate side of parent curr = new TreeNode<E>(item); if (wentleft) { parent.setleft(curr); else { parent.setright(curr); / Returns item from tree that is equivalent (according to compareto) to the given item. If item is not in tree, returns null. public E get(e item) { return get(item, this.root); / Finds it in the subtree rooted at the given node. private E get(e item, TreeNode<E> node) { if (node == null) { return null; else if (item.compareto(node.getdata()) < 0) { return get(item, node.getleft());

13 else if (item.compareto(node.getdata()) > 0) { return get(item, node.getright()); else { // found it! return node.getdata(); / Returns the number of elements currently in this BST. public int size() { return this.size; / Returns a single-line representation of this BST's contents. Specifically, this is a comma-separated list of all elements in their natural Comparable ordering. The list is surrounded by [] characters. public String tostring() { return "[" + tostring(this.root) + "]"; private String tostring(treenode<e> n) { // would have been simpler to use Iterator... but not implemented yet. if (n == null) { return ""; else { String str = ""; str += tostring(n.getleft()); if (!str.isempty()) { str += ", "; str += n.getdata(); if (n.getright()!= null) { str += ", "; str += tostring(n.getright()); return str; public String tofullstring() { StringBuilder sb = new StringBuilder(); tofullstring(root, sb); return sb.tostring(); / Preorder traversal of the tree that builds a string representation in the given n root of subtree to be sb StringBuilder in which to create a string representation / private void tofullstring(treenode<e> n, StringBuilder sb) { if (n == null) { return; sb.append(n.getdata().tostring()); sb.append("\n"); if (n.getleft()!= null) { sb.append("<"); else if (n.getright()!= null) { sb.append(">"); if (n.getleft()!= null n.getright()!= null) { tofullstring(n.getleft(), sb); tofullstring(n.getright(), sb);