I'm trying to display one character at a time in processing. If I type the letter "Q" for example, how can I make it so that when I type another character, it displays in the same place as the previous one but deleting the previous one in the process?
ultimately I'm trying to make a program that would show any pressed key on different fonts; but when I attempted this, the characters just keep adding in succession.
As of now, this gives me characters in succession:
String letters = "";
void setup() {
size(100, 100);
stroke(255);
fill(0);
textSize(16);
}
void draw() {
background(204);
text(letters, 0, 50);
}
void keyPressed() {
if (key == BACKSPACE) {
if (letters.length() > 0) {
letters = letters.substring(0, letters.length()-1);
}
} else if (textWidth(letters+key) < width) {
letters = letters + key;
}
}
How is it possible to have only one character showed at all times?
Look at this line:
letters = letters + key;
Here you're adding key to the letters variable. So if letters is "ABC" and key is 'X', then letters will be "ABCX" after this line runs.
If all you want to do is display the most recent key pressed, why don't you just use the key value directly? That's exactly what it holds:
void draw() {
background(0);
text(key, 25, 25);
}
Related
i am trying to set maximum length in string value and put '..' instead of removed chrs like following
String myValue = 'Welcome'
now i need the maximum length is 4 so output like following
'welc..'
how can i handle this ? thanks
The short and incorrect version is:
String abbrevBad(String input, int maxlength) {
if (input.length <= maxLength) return input;
return input.substring(0, maxLength - 2) + "..";
}
(Using .. is not the typographical way to mark an elision. That takes ..., the "ellipsis" symbol.)
A more internationally aware version would count grapheme clusters instead of code units, so it handles complex characters and emojis as a single character, and doesn't break in the middle of one. Might also use the proper ellipsis character.
String abbreviate(String input, int maxLength) {
var it = input.characters.iterator;
for (var i = 0; i <= maxLength; i++) {
if (!it.expandNext()) return input;
}
it.dropLast(2);
return "${it.current}\u2026";
}
That also works for characters which are not single code units:
void main() {
print(abbreviate("argelbargle", 7)); // argelb…
print(abbreviate("🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰", 4)); // 🇩🇰🇩🇰🇩🇰…
}
(If you want to use ... instead of …, just change .dropLast(2) to .dropLast(4) and "…" to "...".)
You need to use RichText and you need to specify the overflow type, just like this:
Flexible(
child: RichText("Very, very, very looong text",
overflow: TextOverflow.ellipsis,
),
);
If the Text widget overflows, some points (...) will appears.
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 would like to send a list of elements inside a structure via serial port but the output produced by Arduino is abnormal.
A little help? What is the reason for this abnormal output?
const int menu_max_item = 20;
int menu_num_item = 0;
typedef struct item_menu{
String text;
void (*func)(void);
} t_item_menu;
t_item_menu arr_menu[menu_max_item];
void menu_add_item(String txt, void (*f)(void)){
arr_menu[menu_num_item].text = txt;
arr_menu[menu_num_item].func = f;
menu_num_item++;
}
void fn_nd_function(){
Serial.println('test');
}
void print_menu_lcd(){
for(int x = 0; x < 4 && x < menu_num_item; x++){
lcd.setCursor(0,x);
lcd.print(arr_menu[x].text);
}
}
void setup(){
Serial.begin(9600);
for(int i = 0; i < 2; i++) menu_add_item("item " + i, fn_nd_function);
}
void loop() {
print_menu_lcd();
delay(1000);
}
Real output
item
tem
em
Desired output
item 1
item 2
item 3
You have a couple of errors...
This code:
void fn_nd_function(){
Serial.println('test');
}
test is NOT a single character is it? So why do you have it in single quotes?
But more importantly this which is the cause of your bad output:
menu_add_item("item " + i, fn_nd_function);
"item" + i is NOT how you concatenate a number to the end of the character string "item". This is C++ not Java or Python. You'll have to build that string separately. Please don't be tempted to use the String class as that can cause other issues.
What is happening now is that you are passing "item" which is a pointer to the character array stored somewhere in memory holding the characters 'i', 't', 'e' and 'm'. When you add 1 to that pointer you end up with a pointer pointing to the 't' and when you add 2 you end up with a pointer pointing to the 'e'. So when you print from those pointers you only get the part after what that pointer points to.
You need to have a line ahead of that to build the string first. Something along the lines of:
char str[7] = "item "; // Note the two spaces to leave room for the digit
str[5] = i + '0'; // Add '0' to convert single digit to ascii
menu_add_item(str, fn_nd_function);
I have an Array of strings:
private var phrase:Array = ["You will be given a series of questions like this:\n2 + 2 =\n(click or press ENTER to continue)","You can use the Keyboard or Mouse\nto deliver the answer\n\"ENTER\" locks it in.\n(click or press ENTER to continue)","\nClick Here\n to start."];
I have a conditional later in the script to see if the phrase[0] is equal to the instructText.text, so I put a "test" directly after the assignment as below:
instructText.text = phrase[0];
if (instructText.text == phrase[0]) {
trace("phrase zero");
}
else {
trace("nottttttttt");
}
//OUTPUT: nottttttttt
I've tried various combinations of phrase[0] as String and String(phrase[0]), but haven't had any luck.
What am I missing?
Turns out that the text property of the TextField class converts the "Line Feed" characters (the "\n", ASCII code of 1010=A16) to the character of "Carriage Return" (the ASCII code of 1310=D16).
So, you need a LF to CR conversion (or vise-versa) to make a homogeneous comparison of what is stored in the property against what you have in the array element:
function replaceLFwithCR(s:String):String {
return s.replace(/\n/g, String.fromCharCode(13));
}
if (instructText.text == replaceCRwithLF(phrase[0])) {
trace("They are equal :)");
}
else {
trace("They are NOT equal :(");
}
// Output: They are equal :)
P.S. To get the code of a character, you may utilize the charCodeAt() method of the String class:
trace("\n".charCodeAt(0)); // 10
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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;
}
}
}