Create a program with the lowest amount of characters to reverse each word in a string while keeping the order of the words, as well as punctuation and capital letters, in their initial place.
By "Order of the words", I mean that each word is split by an empty space (" "), so contractions and such will be treated as one word. The apostrophe in contractions should stay in the same place. ("Don't" => "Tno'd").
(Punctuation means any characters that are not a-z, A-Z or whitespace*).
Numbers were removed from this list due to the fact that you cannot have capital numbers. Numbers are now treated as punctuation.
For example, for the input:
Hello, I am a fish.
it should output:
Olleh, I ma a hsif.
Notice that O, which is the first letter in the first word, is now capital, since H was capital before in the same location.
The comma and the period are also in the same place.
More examples:
This; Is Some Text!
would output
Siht; Si Emos Txet!
I've tried this:
public static String reverseWord(String input)
{
String words[]=input.split(" ");
StringBuilder result=new StringBuilder();
for (String string : words) {
String revStr = new StringBuilder(string).reverse().toString();
result.append(revStr).append(" ");
}
return result.toString().trim();
}
I have tried to solve your problem. It's working fine for the examples I have checked :) Please look and let me know :)
public static void main(String[] args) {
System.out.println(reverseWord("This; Is Some Text!"));
}
public static boolean isAlphaNumeric(String s) {
return s != null && s.matches("^[a-zA-Z0-9]*$");
}
public static String reverseWord(String input)
{
String words[]=input.split(" ");
StringBuilder result=new StringBuilder();
int startIndex = 0;
int endIndex = 0;
for(int i = 0 ; i < input.length(); i++) {
if (isAlphaNumeric(Character.toString(input.charAt(i)))) {
endIndex++;
} else {
String string = input.substring(startIndex, endIndex);
startIndex = ++endIndex;
StringBuilder revStr = new StringBuilder("");
for (int j = 0; j < string.length(); j++) {
char charToAdd = string.charAt(string.length() - j - 1);
if (Character.isUpperCase(string.charAt(j))) {
revStr.append(Character.toUpperCase(charToAdd));
} else {
revStr.append(Character.toLowerCase(charToAdd));
}
}
result.append(revStr);
result.append(input.charAt(i));
}
}
if(endIndex>startIndex) // endIndex != startIndex
{
String string = input.substring(startIndex, endIndex);
result.append(string);
}
return result.toString().trim();
}
Call the reverseWord with your test string.
Hope it helps. Don't forget to mark it as right answer, if it is :)
Here is a proposal that follows your requirements. It may seem very long but its just comments and aerated code; and everybody loves comments.
public static String smartReverseWords(String input) {
StringBuilder finalString = new StringBuilder();
// Word accumulator, resetted after each "punctuation" (or anything different than a letter)
StringBuilder wordAcc = new StringBuilder();
int processedChars = 0;
for(char c : input.toCharArray()) {
// If not a whitespace nor the last character
if(!Character.isWhitespace(c)) {
// Accumulate letters
wordAcc.append(c);
// Have I reached the last character? Then finalize now:
if(processedChars == input.length()-1) {
reverseWordAndAppend(wordAcc, finalString);
}
}
else {
// Was a word accumulated?
if(wordAcc.length() > 0) {
reverseWordAndAppend(wordAcc, finalString);
}
// Append non-letter char to final string:
finalString.append(c);
}
processedChars++;
}
return finalString.toString();
}
private static void reverseWordAndAppend(StringBuilder wordAcc, StringBuilder finalString) {
// Then reverse it:
smartReverse(wordAcc); // a simple wordAcc.reverse() is not possible
// Append word to final string:
finalString.append(wordAcc.toString());
// Reset accumulator
wordAcc.setLength(0);
}
private static class Marker {
Integer position;
String character;
}
private static void smartReverse(StringBuilder wordAcc) {
char[] arr = wordAcc.toString().toCharArray();
wordAcc.setLength(0); // clean it for now
// Memorize positions of 'punctuation' + build array free of 'punctuation' in the same time:
List<Marker> mappedPosOfNonLetters = new ArrayList<>(); // order matters
List<Integer> mappedPosOfCapitals = new ArrayList<>(); // order matters
for (int i = 0; i < arr.length; i++) {
char c = arr[i];
if(!Character.isLetter(c)) {
Marker mark = new Marker();
mark.position = i;
mark.character = c+"";
mappedPosOfNonLetters.add(mark);
}
else {
if(Character.isUpperCase(c)) {
mappedPosOfCapitals.add(i);
}
wordAcc.append(Character.toLowerCase(c));
}
}
// Reverse cleansed word:
wordAcc.reverse();
// Reintroduce 'punctuation' at right place(s)
for (Marker mark : mappedPosOfNonLetters) {
wordAcc.insert(mark.position, mark.character);
}
// Restore capitals at right place(s)
for (Integer idx : mappedPosOfCapitals) {
wordAcc.setCharAt(idx,Character.toUpperCase(wordAcc.charAt(idx)));
}
}
EDIT
I've updated the code to take all your requirements into account. Indeed we have to make sure that "punctuation' stay in place (and capitals also) but also within a word, like a contraction.
Therefore given the following input string:
"Hello, I am on StackOverflow. Don't tell anyone."
The code produces this output:
"Olleh, I ma no WolfrEvokcats. Tno'd llet enoyna."
Related
I am trying to check if a sentence is the same forwards and backwards or a "sentence palindrome." The sentence "You can cage a swallow, can't you, but you can't swallow a cage, can you?" should return (True) as a palindrome. Ignore everything that is not a letter.
My problem: Not sure how to compare words specifically. This currently works for words checking if they are palindromes, but I need to figure out what to change to compare each word.
public static boolean isWordPalindrome(String input) {
Deque<Character> q = new LinkedList<>( );
Deque<Character> q2 = new LinkedList<>( );
Character letter; // One character from the input string
int mismatches = 0; // Number of spots that mismatched
int i; // Index for the input string
int x;
for (i = 0; i < input.length( ); i++)
{
letter = input.charAt(i); // read next character in the string
if (letter.toString().equals(',') || letter.toString().equals('"') || letter.toString().equals('?') || letter.toString().equals('!') || letter.toString().equals('.') || letter.toString().equals(' ')) {
//throwaway.add(letter); //ignore above chars and put in throwaway stack
}
if (Character.isLetter(letter)) // if letter put into q's
{
q.add(letter);
q2.addFirst(letter);
}
} // end of for loop
System.out.println("q: " + q);
System.out.println("q2:" + q2);
while (!q.isEmpty( ))
{
if (!Objects.equals(q.remove(), q2.remove()))
mismatches++;
}
I'd remove all the special characters, split the string by whitespaces and check the list is "symmetrical":
private static boolean isSentancePalindrom(String sentence) {
String[] words = sentence.replaceAll("[^a-zA-Z ]", "").split("\\s+");
for (int i = 0; i < words.length / 2; ++i) {
if (!words[i].equalsIgnoreCase(words[words.length - i - 1])) {
return false;
}
}
return true;
}
I came up with this:
public static boolean checkString(String str) {
str = str.replaceAll("[,\\?\\.!]+", "").toUpperCase();
String[] split = str.split(" ");
String[] reverse = new String[split.length];
System.arraycopy(split, 0, reverse, 0, split.length);
List<String> listOfSring = Arrays.asList(split);
List<String> reversListOfSring = Arrays.asList(reverse);
Collections.reverse(reversListOfSring);
return reversListOfSring.equals(listOfSring);
}
I hope it would help!
I've been really struggling with a programming assignment. Basically, we have to write a program that translates a sentence in English into one in Pig Latin. The first method we need is one to tokenize the string, and we are not allowed to use the Split method usually used in Java. I've been trying to do this for the past 2 days with no luck, here is what I have so far:
public class PigLatin
{
public static void main(String[] args)
{
String s = "Hello there my name is John";
Tokenize(s);
}
public static String[] Tokenize(String english)
{
String[] tokenized = new String[english.length()];
for (int i = 0; i < english.length(); i++)
{
int j= 0;
while (english.charAt(i) != ' ')
{
String m = "";
m = m + english.charAt(i);
if (english.charAt(i) == ' ')
{
j++;
}
else
{
break;
}
}
for (int l = 0; l < tokenized.length; l++) {
System.out.print(tokenized[l] + ", ");
}
}
return tokenized;
}
}
All this does is print an enormously long array of "null"s. If anyone can offer any input at all, I would reallllyyyy appreciate it!
Thank you in advance
Update: We are supposed to assume that there will be no punctuation or extra spaces, so basically whenever there is a space, it's a new word
If I understand your question, and what your Tokenize was intended to do; then I would start by writing a function to split the String
static String[] splitOnWhiteSpace(String str) {
List<String> al = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (char ch : str.toCharArray()) {
if (Character.isWhitespace(ch)) {
if (sb.length() > 0) {
al.add(sb.toString());
sb.setLength(0);
}
} else {
sb.append(ch);
}
}
if (sb.length() > 0) {
al.add(sb.toString());
}
String[] ret = new String[al.size()];
return al.toArray(ret);
}
and then print using Arrays.toString(Object[]) like
public static void main(String[] args) {
String s = "Hello there my name is John";
String[] words = splitOnWhiteSpace(s);
System.out.println(Arrays.toString(words));
}
If you're allowed to use the StringTokenizer Object (which I think is what the assignment is asking, it would look something like this:
StringTokenizer st = new StringTokenizer("this is a test");
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
which will produce the output:
this
is
a
test
Taken from here.
The string is split into tokens and stored in a stack. The while loop loops through the tokens, which is where you can apply the pig latin logic.
Some hints for you to do the "manual splitting" work.
There is a method String#indexOf(int ch, int fromIndex) to help you to find next occurrence of a character
There is a method String#substring(int beginIndex, int endIndex) to extract certain part of a string.
Here is some pseudo-code that show you how to split it (there are more safety handling that you need, I will leave that to you)
List<String> results = ...;
int startIndex = 0;
int endIndex = 0;
while (startIndex < inputString.length) {
endIndex = get next index of space after startIndex
if no space found {
endIndex = inputString.length
}
String result = get substring of inputString from startIndex to endIndex-1
results.add(result)
startIndex = endIndex + 1 // move startIndex to next position after space
}
// here, results contains all splitted words
String english = "hello my fellow friend"
ArrayList tokenized = new ArrayList<String>();
String m = "";
int j = 0; //index for tokenised array list.
for (int i = 0; i < english.length(); i++)
{
//the condition's position do matter here, if you
//change them, english.charAt(i) will give index
//out of bounds exception
while( i < english.length() && english.charAt(i) != ' ')
{
m = m + english.charAt(i);
i++;
}
//add to array list if there is some string
//if its only ' ', array will be empty so we are OK.
if(m.length() > 0 )
{
tokenized.add(m);
j++;
m = "";
}
}
//print the array list
for (int l = 0; l < tokenized.size(); l++) {
System.out.print(tokenized.get(l) + ", ");
}
This prints, "hello,my,fellow,friend,"
I used an array list since at the first sight the length of the array is not clear.
I've been looking and I can't find anywhere how to write a word count using 3 methods. Here is what the code looks like so far. I'm lost on how to use the methods. I can do this without using different methods and just using one. Please help!!!
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.print("Enter a string: ");
String s = in.nextLine();
if (s.length() > 0)
{
getInputString(s);
}
else
{
System.out.println("ERROR - string must not be empty.");
System.out.print("Enter a string: ");
s = in.nextLine();
}
// Fill in the body with your code
}
// Given a Scanner, prompt the user for a String. If the user enters an empty
// String, report an error message and ask for a non-empty String. Return the
// String to the calling program.
private static String getInputString(String s) {
int count = getWordCount();
while (int i = 0; i < s.length(); i++)
{
if (s.charAt(i) == " ")
{
count ++;
}
}
getWordCount(count);
// Fill in the body
// NOTE: Do not declare a Scanner in the body of this method.
}
// Given a String return the number of words in the String. A word is a sequence of
// characters with no spaces. Write this method so that the function call:
// int count = getWordCount("The quick brown fox jumped");
// results in count having a value of 5. You will call this method from the main method.
// For this assignment you may assume that
// words will be separated by exactly one space.
private static int getWordCount(String input) {
// Fill in the body
}
}
EDIT:
I have changed the code to
private static String getInputString(String s) {
String words = getWordCount(s);
return words.length();
}
private static int getWordCount(String s) {
return s.split(" ");
}
But I can't get the string convert to integer.
You have read the name of the method, and look at the comments to decide what should be implemented inside the method, and the values it should return.
The getInputString method signature should be:
private static String getInputString(Scanner s) {
String inputString = "";
// read the input string from system in
// ....
return inputString;
}
The getWordCount method signature should be:
private static int getWordCount(String input) {
int wordCount = 0;
// count the number of words in the input String
// ...
return wordCount;
}
The main method should look something like this:
public static void main(String[] args) {
// instantiate the Scanner variable
// call the getInputString method to ... you guessed it ... get the input string
// call the getWordCount method to get the word count
// Display the word count
}
count=1 //last word must be counted
for(int i=0;i<s.length();i++)
{
ch=s.charAt(i);
if(ch==' ')
{
count++;
}
}
Use trim() and split() on 1-n whitespace chars:
private static int getWordCount(String s) {
return s.trim().split("\\s+").length;
}
The call to trim() is necessary, otherwise you'll get one extra "word" if there is leading spaces in the string.
The parameter "\\s+" is necessary to count multiple spaces as a single word separator. \s is the regex for "whitespace". + is regex for "1 or more".
What you need to do is, count the number of spaces in the string. That is the number of words in the string.
You will see your count will be off by 1, but after some pondering and bug hunting you will figure out why.
Happy learning!
You can do this by :
private static int getWordCount(String input) {
return input.split("\\s+").length;
}
Use String.split() method like :
String[] words = s.split("\\s+");
int wordCount = words.length;
I'm not sure what trouble you're having with methods but I dont think you need more than one, try this: it uses split to split up the words in a string, and you can chose the delimeters
String sentence = "This is a sentence.";
String[] words = sentence.split(" ");
for (String word : words) {
System.out.println(word);
}
then you can do:
numberOfWords = words.length();
if you want to use 3 methods, you can call a method from your main() method that does this for you, for example:
public String getInputString() {
Scanner in = new Scanner (System.in);
System.out.print("Enter a string: ");
String s = in.nextLine();
if (s.length() > 0) {
return s;
} else {
System.out.println("ERROR - string must not be empty.");
System.out.print("Enter a string: ");
return getInputString();
}
}
public int wordCount(String s) {
words = splitString(s)
return words.length();
}
public String[] splitString(String s) {
return s.split(" ");
}
Based on your code i think this is what you're trying to do:
private static int getWordCount(String input) {
int count = 0;
for (int i = 0; i < input.length(); i++) {
if (input.charAt(i) == ' ') {
count++;
}
}
return count;
}
Here's what I've done:
I've moved the code you were 'playing' with into the right method (getWordCount).
Corrected the loop you were trying to use (I think you have for and while loops confused)
Fixed your check for the space character (' ' not " ")
There is a bug in this code which you'll need to work out how to fix:
getWordCount("How are you"); will return 2 when it should be 3
getWordCount(""); will return 0
getWordCount("Hello"); will return 0 when it should be 1
Good luck!
Better use simple function of spilt() with arguments as space
int n= str.split(" ").length;
public static int Repeat_Words(String arg1,String arg2)
{
//It find number of words can be formed from a given string
if(arg1.length() < 1 || arg2.length() < 1)
return 0;
int no_words = 99999;
char[] str1 = arg1.toCharArray();
char[] str2 = arg2.toCharArray();
for(int x = 0; x < str1.length; x++)
{
int temp = 0;
for(int y = 0; y < str2.length; y++)
{
if(str1[x] == str2[y])
temp++;
}
if(temp == 0)
return 0;
if(no_words > temp)
no_words = temp;
temp = 0;
}
return no_words;
}
I was given this problem to solve. I have only the slightest idea on how it should be implemented, and I'm all too new with programming and stuffs, and would love to hear your comments on this.
Say given a string in the form "abc1234defgh567jk89", and I must create a new string "a1b2c3d5e6f7j8k9".
Note that there are corresponding [digits] & [characters] group and since there may be more of one type over the other, the output has only matching sequence and ignore extra digits or characters in this case '4' & 'g' & 'h'.
I know I will have to use 2 sets of queues to store both types of elements, but I do not know how else to proceed from here.
Would appreciate if you could share a pseudocode or a Java(prefably) version, since I am learning thru this language now.
Thank you.
Pseudocode:
Queue letterQueue;
Queue numberQueue;
for (every character in the string) {
if (it's a letter) {
if (numberQueue is not empty) {
add the letters alternating into the buffer (stringbuilder), and purge buffers
}
add newest letter to letterqueue
}
if (it's a number) {
add newest letter to numberqueue
}
}
add any remaining unprocessed letters to the queue (this will happen most of the time)
return contents of string buffer
You will need:
Queue, probably a LinkedList
StringBuilder
String.toCharArray
Character
Code:
import java.util.LinkedList;
import java.util.Queue;
public class StringTest {
private static String str ="abc1234defgh567jk89";
private static String reorganize(String str) {
Queue<Character> letterQueue = new LinkedList<>();
Queue<Character> numberQueue = new LinkedList<>();
StringBuilder s = new StringBuilder();
for (char c : str.toCharArray()) {
if(Character.isLetter(c)) {
if (!numberQueue.isEmpty()) processQueues(letterQueue, numberQueue, s);
letterQueue.offer(c);
} else if(Character.isDigit(c)) {
numberQueue.offer(c);
}
}
processQueues(letterQueue, numberQueue, s);
return s.toString();
}
private static void processQueues(Queue<Character> letterQueue, Queue<Character> numberQueue, StringBuilder s) {
while(!letterQueue.isEmpty() && !numberQueue.isEmpty()) {
s.append(letterQueue.poll());
s.append(numberQueue.poll());
}
letterQueue.clear();
numberQueue.clear();
}
public static void main(String... args) {
System.out.println(reorganize(str));
}
}
See this hint:
String str = "abc1234defgh567jk89";
String c = str.replaceAll("\\d", ""); // to store characters
String d = str.replaceAll("\\D", ""); // to store digits
Try this:
public static void main(String[] args) {
String str = "abc1234defgh567jk89";
String c = str.replaceAll("\\d", "");
String d = str.replaceAll("\\D", "");
String result = "";
int j = 0, k = 0;
int max = Math.max(c.length(), d.length());
for (int i = 0; i < max; i++) {
if (j++ < c.length())
result = result + c.charAt(i);
if (k++ < d.length())
result = result + d.charAt(i);
}
System.out.println(result);
}
Output:
a1b2c3d4e5f6g7h8j9k
This code is inside the main function:
Scanner input = new Scanner(System.in);
System.out.println("Type a sentence");
String sentence = input.next();
Stack<Character> stk = new Stack<Character>();
int i = 0;
while (i < sentence.length())
{
while (sentence.charAt(i) != ' ' && i < sentence.length() - 1)
{
stk.push(sentence.charAt(i));
i++;
}
stk.empty();
i++;
}
And this is the empty() function:
public void empty()
{
while (this.first != null)
System.out.print(this.pop());
}
It doesn't work properly, as by typing example sentence I am getting this output: lpmaxe. The first letter is missing and the loop stops instead of counting past the space to the next part of the sentence.
I am trying to achieve this:
This is a sentence ---> sihT si a ecnetnes
Per modifications to the original post, where the OP is now indicating that his goal is to reverse the letter order of the words within a sentence, but to leave the words in their initial positions.
The simplest way to do this, I think, is to make use of the String split function, iterate through the words, and reverse their orders.
String[] words = sentence.split(" "); // splits on the space between words
for (int i = 0; i < words.length; i++) {
String word = words[i];
System.out.print(reverseWord(word));
if (i < words.length-1) {
System.out.print(" "); // space after all words but the last
}
}
Where the method reverseWord is defined as:
public String reverseWord(String word) {
for( int i = 0; i < word.length(); i++) {
stk.push(word.charAt(i));
}
return stk.empty();
}
And where the empty method has been changed to:
public String empty() {
String stackWord = "";
while (this.first != null)
stackWord += this.pop();
return stackWord;
}
Original response
The original question indicated that the OP wanted to completely reverse the sentence.
You've got a double-looping construct where you don't really need it.
Consider this logic:
Read each character from the input string and push that character to the stack
When the input string is empty, pop each character from the stack and print it to screen.
So:
for( int i = 0; i < sentence.length(); i++) {
stk.push(sentence.charAt(i));
}
stk.empty();
I assume that what you want your code to do is to reverse each word in turn, not the entire string. So, given the input example sentence you want it to output elpmaxe ecnetnes not ecnetnes elpmaxe.
The reason that you see lpmaxe instead of elpmaxe is because your inner while-loop doesn't process the last character of the string since you have i < sentence.length() - 1 instead of i < sentence.length(). The reason that you only see a single word is because your sentence variable consists only of the first token of the input. This is what the method Scanner.next() does; it reads the next (by default) space-delimited token.
If you want to input a whole sentence, wrap up System.in as follows:
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
and call reader.readLine().
Hope this helps.
Assuming you've already got your input in sentence and the Stack object is called stk, here's an idea:
char[] tokens = sentence.toCharArray();
for (char c : tokens) {
if (c == ' ') {
stk.empty();
System.out.print(c);
} else {
stk.add(c);
}
}
Thus, it will scan through one character at a time. If we hit a space character, we'll assume we've hit the end of a word, spit out that word in reverse, print that space character, then continue. Otherwise, we'll add the character to the stack and continue building the current word. (If you want to also allow punctuation like periods, commas, and the like, change if (c == ' ') { to something like if (c == ' ' || c == '.' || c == ',') { and so on.)
As for why you're only getting one word, darrenp already pointed it out. (Personally, I'd use a Scanner instead of a BufferedReader unless speed is an issue, but that's just my opinion.)
import java.util.StringTokenizer;
public class stringWork {
public static void main(String[] args) {
String s1 = "Hello World";
s1 = reverseSentence(s1);
System.out.println(s1);
s1 = reverseWord(s1);
System.out.println(s1);
}
private static String reverseSentence(String s1){
String s2 = "";
for(int i=s1.length()-1;i>=0;i--){
s2 += s1.charAt(i);
}
return s2;
}
private static String reverseWord(String s1){
String s2 = "";
StringTokenizer st = new StringTokenizer(s1);
while (st.hasMoreTokens()) {
s2 += reverseSentence(st.nextToken());
s2 += " ";
}
return s2;
}
}
public class ReverseofeachWordinaSentance {
/**
* #param args
*/
public static void main(String[] args) {
String source = "Welcome to the word reversing program";
for (String str : source.split(" ")) {
System.out.print(new StringBuilder(str).reverse().toString());
System.out.print(" ");
}
System.out.println("");
System.out.println("------------------------------------ ");
String original = "Welcome to the word reversing program";
wordReverse(original);
System.out.println("Orginal Sentence :::: "+original);
System.out.println("Reverse Sentence :::: "+wordReverse(original));
}
public static String wordReverse(String original){
StringTokenizer string = new StringTokenizer(original);
Stack<Character> charStack = new Stack<Character>();
while (string.hasMoreTokens()){
String temp = string.nextToken();
for (int i = 0; i < temp.length(); i ++){
charStack.push(temp.charAt(i));
}
charStack.push(' ');
}
StringBuilder result = new StringBuilder();
while(!charStack.empty()){
result.append(charStack.pop());
}
return result.toString();
}
}
public class reverseStr {
public static void main(String[] args) {
String testsa[] = { "", " ", " ", "a ", " a", " aa bd cs " };
for (String tests : testsa) {
System.out.println(tests + "|" + reverseWords2(tests) + "|");
}
}
public static String reverseWords2(String s) {
String[] sa;
String out = "";
sa = s.split(" ");
for (int i = 0; i < sa.length; i++) {
String word = sa[sa.length - 1 - i];
// exclude "" in splited array
if (!word.equals("")) {
//add space between two words
out += word + " ";
}
}
//exclude the last space and return when string is void
int n = out.length();
if (n > 0) {
return out.substring(0, out.length() - 1);
} else {
return "";
}
}
}
This can pass in leetcode