You are very happy because you advanced to the next round of a very important programming contest. You want your best friend to know how happy you are. Therefore, you are going to send him a lot of smile emoticons. You are given an int smiles: the exact number of emoticons you want to send.
You have already typed one emoticon into the chat. Then, you realized that typing is slow. Instead, you will produce the remaining emoticons using copy, paste, and possibly some deleting.
You can only do three different operations:
Copy all the emoticons you currently have into the clipboard.
Paste all emoticons from the clipboard.
Delete one emoticon from the message.
Each operation takes precisely one second. Copying replaces the old content of the clipboard. Pasting does not empty the clipboard. You are not allowed to copy just a part of the emoticons you already have. You are not allowed to delete an emoticon from the clipboard.
Return the smallest number of seconds in which you can turn the one initial emoticon into smiles emoticons.
How to find the recurrence relation for this problem?
Use this code and add seconds according to output of this program. A error of infinite loop may be occur correct it.
/* Input : 95 Output : 64,16,8,4,2,1
Input : 69 Output : 64,4,1
Input : 277 Output : 256,16,4,1
Input : 4 Output : 4
Input : 1 Output : 1
Input : 1042 Output : 1024,16,2
*/
import java.util.Scanner;
class Emotions
{
public static void main(String args[])
{
Scanner a = new Scanner(System.in);
System.out.print("Enter no. of Emotions : ");
int emo = a.nextInt();
value(emo);
}
public static void value(int x)
{
int emo = x;
int n = 1;
do {
n = n * 2;
}while(emo > n);
n = n/2;
System.out.println(n);
while((emo - n) != 1)
{
value(emo-n);
}
}
}
Related
I recently encountered a problem in one of my coding interview tests. The problem is as follows.
Suppose there is a service to send messages to a user. Each message has the length of maximum 30 characters. This service receives a complete message and then breaks it into sub-messages, each of size 30 characters at most. But there is an issue with the service. It doesn't guarantee the order in which the sub-messages are received by the user. Hence, for every sub-message, it appends a suffix (k/n) where k denotes the kth sub-message out of the n sub-messages. This suffix is also considered when counting the number of characters in the sub-message which cannot exceed 30. Find the minimum number of sub-messages required to send.
Eg-1:
message: The quick brown fox jumps over the lazy dog
The first sub-message can be: The quick brown fox jumps (1/2) but
the above is incorrect as it exceeds 30 characters. This has 31 characters.
So,the correct sub-messages are:
The quick brown fox (1/2)
jumps over the lazy dog (2/2)
So, the answer is 2.
Eg-2:
message: The quick brown fox jumps over the lazy tortoise
So,the correct sub-messages are:
The quick brown fox (1/3)
jumps over the lazy (2/3)
tortoise (3/3)
So, the answer is 3.
Eg-3:
message: Hello My name is
sub-message: Hello My name is
Answer = 1.
Note: A word cannot be broken across sub-messages. Assume no word is greater than 30 characters in length. If its a single message, then no need to use the suffix
My approach: If the total character length of string is less than 30 then return 1. If not, then get sub-message till character count is 30, checking per word. But now it gets complicated as I don't know the value of n in the suffix. Is there a simpler way to approach the problem?
Thanks for posting this, I do enjoy these sorts of problems.
As domen mentioned above, there is a bit of a challenge here in that you do not know how many lines are required. Thus you do not know whether to allow for 2 (or more) digits for the message number / total message count. Also, could you use Hexadecimal (16 messages require a single digit, or even a base 62 format number (0-9, then A-Z followed by a-z)?
You could of course use a guess and say, if the input is more than, say, 200 characters, then you might use a two digit message number, but then if the message was a single letter followed by a single space repeated 100 times, then you could probably get away with single digit message numbers.
So, you might find that you need to run the algorithm a couple of times. I will assume that a single digit message number is acceptable for this problem, you can enhance my solution to use base 52 message numbers if you like.
My approach uses 2 classes:
Create a class MessageLine that represents a single line of the message.
MessageSender a class that collects the MessageLine(s). It has a helper method that and processes the message and returns a list of MessageLines.
Here is the main MessageSender class. If you run it, you can pass a message on the command line for it to process.
package com.gtajb.stackoverflow;
import java.util.LinkedList;
public class MessageSender {
public static void main(String[] args) {
if (args.length == 0) {
System.out.println("Please supply a message to send");
System.exit(1);
}
// Collect the command line parameters into a single string.
StringBuilder sb = new StringBuilder();
boolean firstWord = true;
for (String s: args) {
if (!firstWord) {
sb.append(" ");
}
firstWord = false;
sb.append(s);
}
// Process the input String and create the MessageSender object.
MessageSender ms = new MessageSender(sb.toString());
System.out.println("Input message: " + sb.toString());
// Retrieve the blocked message and output it.
LinkedList<MessageLine> msg = ms.getBlockedMessage();
int lineNo = 0;
for (MessageLine ml : msg) {
lineNo += 1;
System.out.printf("%2d: %s\n", lineNo, ml.getFormattedLine(msg.size()));
}
}
private String msg;
public MessageSender(String msg) {
this.msg = msg;
processMessage();
}
private LinkedList<MessageLine> blockedMessage = new LinkedList<MessageLine> ();
public LinkedList<MessageLine> getBlockedMessage() {
return blockedMessage;
}
private static final int LINE_MAX_SIZE = 30;
/**
* A private helper method that processes the supplied message when
* the object is constructed.
*/
private void processMessage() {
// Split the message into words and work out how long the message is.
String [] words = msg.split("\\s+");
int messageLength = 0;
for (String w: words) {
messageLength += w.length();
}
messageLength += words.length - 1; // Add in the number of words minus one to allow for the single spaces.
// Can we get away with a single MessageLine?
if (messageLength < LINE_MAX_SIZE) {
// A single message line is good enough.
MessageLine ml = new MessageLine(1);
blockedMessage.add(ml);
for (String w: words) {
ml.add(w);
}
} else {
// Multiple MessageLines will be required.
int lineNo = 1;
MessageLine ml = new MessageLine(lineNo);
blockedMessage.add(ml);
for (String w: words) {
// check if this word will blow the max line length.
// The maximum number of lines is 2. It can be anything that is > 1.
if (ml.getFormattedLineLength(2) + w.length() + 1 > LINE_MAX_SIZE) {
// The word will blow the line length, so create a new line.
lineNo += 1;
ml = new MessageLine(lineNo);
blockedMessage.add(ml);
}
ml.add(w);
}
}
}
}
and here is the Message Line class:
package com.gtajb.stackoverflow;
import java.util.LinkedList;
public class MessageLine extends LinkedList<String> {
private int lineNo;
public MessageLine(int lineNo) {
this.lineNo = lineNo;
}
/**
* Add a new word to this message line.
* #param word the word to add
* #return true if the collection is modified.
*/
public boolean add(String word) {
if (word == null || word.trim().length() == 0) {
return false;
}
return super.add(word.trim());
}
/**
* Return the formatted message length.
* #param totalNumLines the total number of lines in the message.
* #return the length of this line when formatted.
*/
public int getFormattedLineLength(int totalNumLines) {
return getFormattedLine(totalNumLines).length();
}
/**
* Return the formatted line optionally with the line count information.
* #param totalNumLines the total number of lines in the message.
* #return the formatted line.
*/
public String getFormattedLine(int totalNumLines) {
boolean firstWord = true;
StringBuilder sb = new StringBuilder();
for (String w : this) {
if (! firstWord) {
sb.append (" ");
}
firstWord = false;
sb.append(w);
}
if (totalNumLines > 1) {
sb.append (String.format(" (%d/%d)", lineNo, totalNumLines));
}
return sb.toString();
}
}
I tested your scenarios and it seems to produce the correct result.
Let me know if we get the job. :-)
You can binary-search on the total number of submessages. That is, start with two numbers L and H, such that you know that L submessages are not enough, and that H submessages are enough, and see whether their average (L+H)/2 is enough by trying to construct a solution under the assumption that that many submessages are involved: If it is, make that the new H, otherwise make it the new L. Stop as soon as H = L+1: H is then the smallest number of submessages that works, so construct an actual solution using that many submessages. This will require O(n log n) time.
To get initial values for L and H, you could start at 1 and keep doubling until you get a high enough number. The first value that is large enough to work becomes your H, and the previous one your L.
BTW, the constraints you give are not enough to ensure a solution exists: For example, an input consisting of two 29-letter words separated by a space has no solution.
I want to split a large string file into large number of very small sub string and then store it so that i can use all sub string further in my program. what will be best way to do so.my program is in C# so Please provide solution only in C#.
Here is input File : https://he-s3.s3.amazonaws.com/media/hackathon/code-monk-hashing/problems/monk-in-the-land-of-pokemons/4326354c-2-input-4326316.txt?Signature=mtbhaQ2%2F6VRAH8A9OfZTSAX6oTQ%3D&Expires=1457955770&AWSAccessKeyId=AKIAJLE6MUHDYS3HN6YQ
all i want to get all number separated by white space in a collection.
is it possible that i can read all number separated by white space without saving them in collection?
Question:
Monk has to visit a land where strange creatures, known as Pokemons, roam around in the wild. Each Pokemon in the land will attack any visitor. They can only be pacified by feeding them their favorite food.
The Pokemon of type X eats one food item of type X.
Monk knows that he will encounter N ponds on the way. At each pond, he will find a food item and then encounter a Pokemon. The i'th pond has the food item of type Ai and the Pokemon of type Bi.
The monk can feed the item at the i'th pond to the Pokemon at the pond if the type matches. Monk may have to carry some food items with him before leaving so as to feed all the Pokemons. Help him find the number of items he must carry, to be to able to pass through the land safely.
Input:
The first line contains T, denoting the number of test cases. Then, T test cases follow.
The first line of each test case contains an integer N. Then, N lines follow.
Each line consists of 2 space-separated integers Ai and Bi.
Output: For each test case, print the answer in a new line.
Constraints:
1 ≤ T ≤ 10
1 ≤ N ≤ 105
1 ≤ Ai, Bi ≤ 106
Code:
using System;
using System.Numerics;
using System.Collections.Generic;
class MyClass
{
static void Main(string[] args)
{
int TestCaseNo = Int32.Parse(System.Console.ReadLine().Trim());
int TestCase;
for (int j = 0; j < TestCaseNo; j++)
{
LinkedList<int> numbers = new LinkedList<int>();
numbers.Clear();
int count = 0;
TestCase = Int32.Parse(System.Console.ReadLine().Trim());
for (int i = 0; i < TestCase; i++)
{
string line = System.Console.ReadLine().Trim();
string[] temp = line.Split(' ');
if (Int32.Parse(temp[0]) == Int32.Parse(temp[1]))
{ }
else
{
if (numbers.Contains(Int32.Parse(temp[1])))
{
numbers.Remove(Int32.Parse(temp[1]));
numbers.AddFirst(Int32.Parse(temp[0]));
}
else
{
numbers.AddFirst(Int32.Parse(temp[0]));
count++;
}
}
}
System.Console.WriteLine(count);
}
}
}
I'm making an air dictionary and I have a(nother) problem. The main app is ready to go and works perfectly but when I tested it I noticed that it could be better. A bit of context: the language (ancient egyptian) I'm translating from does not use punctuation so a phrase canlooklikethis. Add to that the sheer complexity of the glyph system (6000+ glyphs).
Right know my app works like this :
user choose the glyphs composing his/r word.
app transforms those glyphs to alphanumerical values (A1 - D36 - X1A, etc).
the code compares the code (say : A5AD36) to a list of xml values.
if the word is found (A5AD36 = priestess of Bast), the user gets the translation. if not, s/he gets all the possible words corresponding to the two glyphs (A5A & D36).
If the user knows the string is a word, no problem. But if s/he enters a few words, s/he'll have a few more choices than hoped (exemple : query = A1A5AD36 gets A1 - A5A - D36 - A5AD36).
What I would like to do is this:
query = A1A5AD36 //word/phrase to be translated;
varArray = [A1, A5A, D36] //variables containing the value of the glyphs.
Corresponding possible words from the xml : A1, A5A, D36, A5AD36.
Possible phrases: A1 A5A D36 / A1 A5AD36 / A1A5A D36 / A1A5AD36.
Possible phrases with only legal words: A1 A5A D36 / A1 A5AD36.
I'm not I really clear but to things simple, I'd like to get all the possible phrases containing only legal words and filter out the other ones.
(example with english : TOBREAKFAST. Legal = to break fast / to breakfast. Illegal = tobreak fast.
I've managed to get all the possible words, but not the rest. Right now, when I run my app, I have an array containing A1 - A5A - D36 - A5AD36. But I'm stuck going forward.
Does anyone have an idea ? Thank you :)
function fnSearch(e: Event): void {
var val: int = sp.length; //sp is an array filled with variables containing the code for each used glyph.
for (var i: int = 0; i < val; i++) { //repeat for every glyph use.
var X: String = ""; //variable created to compare with xml dictionary
for (var i2: int = 0; i2 < val; i2++) { // if it's the first time, use the first glyph-code, else the one after last used.
if (X == "") {
X = sp[i];
} else {
X = X + sp[i2 + i];
}
xmlresult = myXML.mot.cd; //xmlresult = alphanumerical codes corresponding to words from XMLList already imported
trad = myXML.mot.td; //same with traductions.
for (var i3: int = 0; i3 < xmlresult.length(); i3++) { //check if element X is in dictionary
var codeElement: XML = xmlresult[i3]; //variable to compare with X
var tradElement: XML = trad[i3]; //variable corresponding to codeElement
if (X == codeElement.toString()) { //if codeElement[i3] is legal, add it to array of legal words.
checkArray.push(codeElement); //checkArray is an array filled with legal words.
}
}
}
}
var iT2: int = 500 //iT2 set to unreachable value for next lines.
for (var iT: int = 0; iT < checkArray.length; iT++) { //check if the word searched by user is in the results.
if (checkArray[iT] == query) {
iT2 = iT
}
}
if (iT2 != 500) { //if complete query is found, put it on top of the array so it appears on top of the results.
var oldFirst: String = checkArray[0];
checkArray[0] = checkArray[iT2];
checkArray[iT2] = oldFirst;
}
results.visible = true; //make result list visible
loadingResults.visible = false; //loading screen
fnPossibleResults(null); //update result list.
}
I end up with an array of variables containing the glyph-codes (sp) and another with all the possible legal words (checkArray). What I don't know how to do is mix those two to make legal phrases that way :
If there was only three glyphs, I could probably find a way, but user can enter 60 glyphs max.
I want to limit the no. of character that can be put on JTextField because on my database I have this column that has Sex,Status (which the no. of char. allowed is 1 only).
and Middle Initial (which the no. of char. allowed is 2 only).
This what I have in my mind :
(for Sex,Status column)
String text = jTextField2.getText();
int count = text.();
if (count>1) {
(delete the next character that will be input)
}
(for M.I. column)
String text = jTextField1.getText();
int count = text.();
if (count>2) {
(delete the next character that will be input)
}
is this possible? is there a command that will delete the next character, so the no. of char. is acceptable for my database?
Sure. Just use String#substring.
String middleInitial = "JKL";
middleInitial.substring(0, 2);
System.out.println(middleInitial); // => JK
Similarly, you can use substring(0, 1) for sex.
It might be better if sex is an enum, though.
public enum Sex {
MALE("m"), FEMALE("f");
final String symbol;
private Sex(String symbol) {
this.symbol = symbol;
}
}
Now you can use it like this:
String sex = "male";
Sex.valueOf(sex.toUpperCase());
Or directly
Sex.MALE;
Instead of a text field for sex, you might use a JComboBox so the user can only choose one of the two options. This way you're sure to have valid input.
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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}