Given a string, I need to print out all permutations of the string. How should I do that? I have tried
for(int i = 0; i<word.length();i++)
{
for(int j='a';j<='z';j++){
word = word.charAt(i)+""+(char)j;
System.out.println(word);
}
}
Is there a good way about doing this?
I'm not 100% sure that I understand what you are trying to do. I'm going to go by your original wording of the question and your comment to #ErstwhileIII's answer, which make me think that it's not really "permutations" (i.e. rearrangement of the letters in the word) that you are looking for, but rather possible single-letter modifications (not sure what a better word for this would be either), like this:
Take a word like "hello" and print a list of all "versions" you can get by adding one "typo" to it:
hello -> aello, bello, cello, ..., zello, hallo, hbllo, hcllo, ..., hzllo, healo, heblo, ...
If that's indeed what you're looking for, the following code will do that for you pretty efficiently:
public void process(String word) {
// Convert word to array of letters
char[] letters = word.toCharArray();
// Run through all positions in the word
for (int pos=0; pos<letters.length; pos++) {
// Run through all letters for the current position
for (char letter='a'; letter<='z'; letter++) {
// Replace the letter
letters[pos] = letter;
// Re-create a string and print it out
System.out.println(new String(letters));
}
// Set the current letter back to what it was
letters[pos] = word.charAt(pos);
}
}
OH .. to print out all permutations of a string, consider your algorithm first. What is the definition of "all permutations" .. for example:
String "a" would have answer a only
String "ab" would have answer: ab, ba
String "abc" would have answer: abc acb, bca, bac, cba, cab
Reflect on the algorithm you would use (write it down in english) .. then translate to Java code
While not the most efficient, a recursive solution might be easiest to use (i.e. for a string of length n, go through each of the characters and follow that with the permutations of the string with that character removed).
EDIT: Ok... you changed your request. Permutations is a whole other story. I think this will help: Generating all permutations of a given string
Not sure what you are trying to do... Example 1 is to get the alphabet one letter next to another. Example 2 is to print whatever you gave us there as an example.
//Example 1
String word=""; //empty string
for(int i = 65; i<=90;i++){ //65-90 are the Ascii numbers for capital letters
word+=(char)i; //cast int to char
}
System.out.println(word);
//Example 2
String word="";
for (int i=65;i<=90;i++){
word+=(char)i+"rse";
if(i!=90){ //you don't want this at the end of your sentence i suppose :)
word+=", ";
}
}
System.out.println(word);
Related
This question already has answers here:
Java program to find the character that appears the most number of times in a String?
(8 answers)
Closed 6 years ago.
I got a task from my university today:
Write a program that reads a ( short ) text from the user and prints the so called max letter (most common character in string) , that the letter which the greatest number of occurrences of the given text .
Here it is enough to look at English letters (A- Z) , and not differentiate between uppercase and lowercase letters in the count of the number of occurrences .
For example, if : text = " Ada bada " so should the print show the most common character, this example it would be a.
This is an introductory course, so in this submission we do not need to use the " scanner - class" . We have not gone through this so much.
The program will use the show message input two get the text from user .
Info: The program shall not use while loop ( true / false ) , "return " statement / "break " statement .
I've been struggling with how I can get char values into a table.. am I correct I need to use array to search for most common character? I think I need to use the binarySearch, but that only supports int not char.
I'll be happy for any answers. hint's and solutions. etc.. if you're very kind a full working program, but again please don't use the things I have written down in the "info" section above.
My code:
String text = showInputDialog("Write a short text: ");
//format string to char
String a = text;
char c = a.charAt(4);
/*with this layout it collects number 4 character in the text and print out.
* I could as always go with many char c... but that wouldn't be a clean program * code.. I think I need to make it into a for-loop.. I have only worked with * *for-loops with numbers, not char (letters).. Help? :)
*/
out.print( text + "\n" + c)
//each letter into 1 char, into table
//search for most used letter
Here's the common logic:
split your string into chars
loop over the chars
store the occurrences in a hash, putting the letter as key and occurrences as value
return the highest value in the hash
As how to split string into chars, etc., you can use Google. :)
Here's a similar question.
There's a common program asked to write in schools to calculate the frequency of a letter in a given String. The only thing you gotta do here is find which letter has the maximum frequency. Here's a code that illustrates it:
String s <--- value entered by user
char max_alpha=' '; int max_freq=0, ct=0;
char c;
for(int i=0;i<s.length();i++){
c=s.charAt(i);
if((c>='a'&&c<='z')||(c>='A'&&c<='Z')){
for(int j=0;j<s.length();j++){
if(s.charAt(j)==c)
ct++;
} //for j
}
if(ct>max_freq){
max_freq=ct;
max_alpha=c;
}
ct=0;
s=s.replace(c,'*');
}
System.out.println("Letter appearing maximum times is "+max_alpha);
System.out.println(max_alpha+" appears "+max_freq+" times");
NOTE: This program presumes that all characters in the string are in the same case, i.e., uppercase or lowercase. You can convert the string to a particular case just after getting the input.
I guess this is not a good assigment, if you are unsure about how to start. I wish you for having better teachers!
So you have a text, as:
String text = showInputDialog("Write a short text: ");
The next thing is to have a loop which goes trough each letter of this text, and gets each char of it:
for (int i=0;i<text.length();i++) {
char c=text.charAt(i);
}
Then comes the calculation. The easiest thing is to use a hashMap. I am unsure if this is a good topic for a beginners course, so I guess a more beginner friendly solution would be a better fit.
Make an array of integers - this is the "table" you are referring to.
Each item in the array will correspond to the occurrance of one letter, e.g. histogram[0] will count how many "A", histogram[1] will count how many "B" you have found.
int[] histogram = new int[26]; // assume English alphabet only
for (int i=0;i<histogram.length;i++) {
histogram[i]=0;
}
for (int i=0;i<text.length();i++) {
char c=Character.toUppercase(text.charAt(i));
if ((c>=65) && (c<=90)) {
// it is a letter, histogram[0] contains occurrences of "A", etc.
histogram[c-65]=histogram[c-65]+1;
}
}
Then finally find the biggest occurrence with a for loop...
int candidate=0;
int max=0;
for (int i=0;i<histogram.length;i++) {
if (histogram[i]>max) {
// this has higher occurrence than our previous candidate
max=histogram[i];
candidate=i; // this is the index of char, i.e. 0 if A has the max occurrence
}
}
And print the result:
System.out.println(Character.toString((char)(candidate+65));
Note how messy this all comes as we use ASCII codes, and only letters... Not to mention that this solution does not work at all for non-English texts.
If you have the power of generics and hashmaps, and know some more string functions, this mess can be simplified as:
String text = showInputDialog("Write a short text: ");
Map<Char,Integer> histogram=new HashMap<Char,Integer>();
for (int i=0;i<text.length();i++) {
char c=text.toUppercase().charAt(i));
if (histogram.containsKey(c)) {
// we know this letter, increment its occurrence
int occurrence=histogram.get(c);
histogram.put(c,occurrence+1);
}
else {
// we dunno this letter yet, it is the first occurrence
histogram.put(c,1);
}
}
char candidate=' ';
int max=0;
for (Char c:histogram.keySet()) {
if (histogram.get(c)>max) {
// this has higher occurrence than our previous candidate
max=histogram.get(c);
candidate=c; // this is the char itself
}
}
System.out.println(c);
small print: i didn't run this code but it shall be ok.
All that I am doing in my project is taking two values(that I am reading from two different excel files) and checking how similar they are.! I tried using the pattern and matcher classes which works perfectly fine when both the words are exactly the same (as in organisation and organisation/s). In my data I have say something like (employee and employment), I just need "employ" as the common string between the two, in which case..pattern and matches fails.! I am stuck with this since a week.I have about 700 rows in the first excel file and about 9000 in the other. Each cell value that I am reading into the program using java, I am storing them in two separate variables. Next, i tried using 4 for loops to match word by word and character by character to find only those characters that match between the two.I have pasted the coded for the for loop implementation. Four for loops are like driving me nuts.! Any help in completing this would be greatly appreciated.
String str1 = "Cover for employees of the company";
String str2 = "Employment Agencies ";
String str,strfinal;
String[] count1 = str1.split("\\s+");
String[] count2 = str2.split("\\s+");
char[] count11 = str1.toCharArray();
char[] count22 = str2.toCharArray();
for(int i=0;i<count1.length;i++)
{
for(int j=0;j<count2.length;j++)
{
for(int m=0;m<count1[i].length();m++)
{
for(int n=0;n<count2[j].length();n++)
{
if(count11[m]==count22[n])
{
// please look at the logic that I am looking for to implement
}
}
}
}
}
Expected output: employ
one more concept that I am trying to implement (in order to make my program more efficient) is..
cover ----(compared with) employment. First character itself does not match.Implies go to the next word in the second string. Once all words in the second string are traversed and checked for, go to the next word in the first string and compare this word with all the words in the second string.
Okay.. so this is what I am looking for right now.. Any help will be greatly appreciated.
Thanks!
So I am writing a scrabble word suggestion program that I decided to do because I wanted to learn sets (don't worry, I at least got that part) and referencing info/data not created within the program. Im pretty new to Java (and programming in general), but I was wondering how to pull words from a word list .FIC file in order to check them against words generated from the letters inputted.
To clarify, I have written a program which takes a series of letters and returns a set of every possible word created from those letters. for example:
input:
abc
would give a set containing the "words":
a, ab, ac, abc, acb, b, ba, bc, bac, bca, c, ca, cb, cab, cba
What I am asking, really, is how to check those to find the ones contained in the .FIC file.
The file is the "official crosswords" file from the Moby project word list and I am still (very) shaky on parsing and other file dealing-with methods. I am continuing to research so I dont have any prototype code for that.
Sorry if the question isn't entirely clear.
edit: here is the method that makes the "words" to make it easier to understand the idea. The part I don't understand is specifically how to pull a word(as a string) from the .FIC file.
private static Set<String> Words(String s)
{
Set<String> tempwords = new TreeSet<String>();
if (s.length() == 1)
{ // base case, last letter
tempwords.add(s);
// System.out.println(s); uncomment when debugging
}
else
{
//set up to add each letter in s
for (int i = 0; i < s.length(); i++)
{ //cut the i letter out of the string
String remaining = s.substring(0, i) + s.substring(i+1);
//recursion to add all combinations of letters onto the current letter/"word"
for (String permutation : Words(remaining))
{
// System.out.println(s.substring(i, i+1) + permutation); uncomment when debugging
//add the full length words
tempwords.add(s.substring(i, i+1) + permutation);
// System.out.println(permutation); uncomment when debugging
//add the not-full-length words
tempwords.add(permutation);
}
}
}
// System.out.println(tempwords); uncomment when debugging
return tempwords;
}
I dont know if it is the best solution, but i figured it out (hobbs the line thing helped a lot, thank you). I found that this works:
public static void main(String[] args) throws FileNotFoundException
{
Scanner s = new Scanner(new FileReader("C:/Users/Sean/workspace/Imbored/bin/113809of.fic"));
while(true)
{
words.clear();
String letters = enterLetters();
words.addAll(Words(letters));
while(s.hasNextLine()) {
String line = s.nextLine();
String finalword = checkWords(line, words);
if (finalword != null) finalwordset.add(finalword);
}
s.reset();
System.out.println(finalwordset);
System.out.println();
System.out.println("_________________________________________________________________________");
}
}
A few things:
The checkWords method checks if the current word from the file is in the generated list of "words"
The enterletters method takes user inputted letters and returns them in a string
The Words method returns a set of strings of all of the possible combinations of the characters in the given string, with each character used up to as many times as it appears in the string and no repeated "words" in the returned set.
finalwordset and words are arraylists of strings defined as instance variables(i would put them in the main method but I'm lazy and it doesn't matter for this case)
I am very sure there is a better/more efficient way to do this, but this at least works.
Finally: I decided to answer rather than delete because I didn't see this answered anywhere else, so if it is feel free to delete the question or link to the other answer or whatever, at this point it is to help other people.
I have a serious problem with extracting terms from each string line. To be more specific, I have one csv formatted file which is actually not csv format (it saves all terms into line[0] only)
So, here's just example string line among thousands of string lines:
(split() doesn't work.!!! )
test.csv
"31451 CID005319044 15939353 C8H14O3S2 beta-lipoic acid C1C[S#](=O)S[C##H]1CCCCC(=O)O "
"12232 COD05374044 23439353 C924O3S2 saponin CCCC(=O)O "
"9048 CTD042032 23241 C3HO4O3S2 Berberine [C##H]1CCCCC(=O)O "
I want to extract "beta-lipoic acid" ,"saponin" and "Berberine" only which is located in 5th position.
You can see there are big spaces between terms, so that's why I said 5th position.
In this case, how can I extract terms located in 5th position for each line?
One more thing: the length of whitespace between each of the six terms is not always equal. the length could be one, two, three, four, or five, or something like that.
Because the length of whitespace is random, I can not use the .split() function.
For example, in the first line I would get "beta-lipoic" instead "beta-lipoic acid.**
Here is a solution for your problem using the string split and index of,
import java.util.ArrayList;
public class StringSplit {
public static void main(String[] args) {
String[] seperatedStr = null;
int fourthStrIndex = 0;
String modifiedStr = null, finalStr = null;
ArrayList<String> strList = new ArrayList<String>();
strList.add("31451 CID005319044 15939353 C8H14O3S2 beta-lipoic acid C1C[S#](=O)S[C##H]1CCCCC(=O)O ");
strList.add("12232 COD05374044 23439353 C924O3S2 saponin CCCC(=O)O ");
strList.add("9048 CTD042032 23241 C3HO4O3S2 Berberine [C##H]1CCCCC(=O)O ");
for (String item: strList) {
seperatedStr = item.split("\\s+");
fourthStrIndex = item.indexOf(seperatedStr[3]) + seperatedStr[3].length();
modifiedStr = item.substring(fourthStrIndex, item.length());
finalStr = modifiedStr.substring(0, modifiedStr.indexOf(seperatedStr[seperatedStr.length - 1]));
System.out.println(finalStr.trim());
}
}
}
Output:
beta-lipoic acid
saponin
Berberine
Option 1 : Use spring.split and check for multiple consecutive spaces. Like the code below:
String s[] = str.split("\\s\\s+");
for (String string : s) {
System.out.println(string);
}
Option 2 : Implement your own string split logic by browsing through all the characters. Sample code below (This code is just to give an idea. I didnot test this code.)
public static List<String> getData(String str) {
List<String> list = new ArrayList<>();
String s="";
int count=0;
for(char c : str.toCharArray()){
System.out.println(c);
if (c==' '){
count++;
}else {
s = s+c;
}
if(count>1&&!s.equalsIgnoreCase("")){
list.add(s);
count=0;
s="";
}
}
return list;
}
This would be a relatively easy fix if it weren't for beta-lipoic acid...
Assuming that only spaces/tabs/other whitespace separate terms, you could split on whitespace.
Pattern whitespace = Pattern.compile("\\s+");
String[] terms = whitespace.split(line); // Not 100% sure of syntax here...
// Your desired term should be index 4 of the terms array
While this would work for the majority of your terms, this would also result in you losing the "acid" in "beta-lipoic acid"...
Another hacky solution would be to add in a check for the 6th spot in the array produced by the above code and see if it matches English letters. If so, you can be reasonably confident that the 6th spot is actually part of the same term as the 5th spot, so you can then concatenate those together. This falls apart pretty quickly though if you have terms with >= 3 words. So something like
Pattern possibleEnglishWord = Pattern.compile([[a-zA-Z]*); // Can add dashes and such as needed
if (possibleEnglishWord.matches(line[5])) {
// return line[4].append(line[5]) or something like that
}
Another thing you can try is to replace all groups of spaces with a single space, and then remove everything that isn't made up of just english letters/dashes
line = whitespace.matcher(line).replaceAll("");
Pattern notEnglishWord = Pattern.compile("^[a-zA-Z]*"); // The syntax on this is almost certainly wrong
notEnglishWord.matcher(line).replaceAll("");
Then hopefully the only thing that is left would be the term you're looking for.
Hopefully this helps, but I do admit it's rather convoluted. One of the issues is that it appears that non-term words may have only one space between them, which would fool Option 1 as presented by Hirak... If that weren't the case that option should work.
Oh by the way, if you do end up doing this, put the Pattern declarations outside of any loops. They only need to be created once.
So, I'm in need of help on my homework assignment. Here's the question:
Write a static method, getBigWords, that gets a String parameter and returns an array whose elements are the words in the parameter that contain more than 5 letters. (A word is defined as a contiguous sequence of letters.) So, given a String like "There are 87,000,000 people in Canada", getBigWords would return an array of two elements, "people" and "Canada".
What I have so far:
public static getBigWords(String sentence)
{
String[] a = new String;
String[] split = sentence.split("\\s");
for(int i = 0; i < split.length; i++)
{
if(split[i].length => 5)
{
a.add(split[i]);
}
}
return a;
}
I don't want an answer, just a means to guide me in the right direction. I'm a novice at programming, so it's difficult for me to figure out what exactly I'm doing wrong.
EDIT:
I've now modified my method to:
public static String[] getBigWords(String sentence)
{
ArrayList<String> result = new ArrayList<String>();
String[] split = sentence.split("\\s+");
for(int i = 0; i < split.length; i++)
{
if(split[i].length() > 5)
{
if(split[i].matches("[a-zA-Z]+"))
{
result.add(split[i]);
}
}
}
return result.toArray(new String[0]);
}
It prints out the results I want, but the online software I use to turn in the assignment, still says I'm doing something wrong. More specifically, it states:
Edith de Stance states:
⇒ You might want to use: +=
⇒ You might want to use: ==
⇒ You might want to use: +
not really sure what that means....
The main problem is that you can't have an array that makes itself bigger as you add elements.
You have 2 options:
ArrayList (basically a variable-length array).
Make an array guaranteed to be bigger.
Also, some notes:
The definition of an array needs to look like:
int size = ...; // V- note the square brackets here
String[] a = new String[size];
Arrays don't have an add method, you need to keep track of the index yourself.
You're currently only splitting on spaces, so 87,000,000 will also match. You could validate the string manually to ensure it consists of only letters.
It's >=, not =>.
I believe the function needs to return an array:
public static String[] getBigWords(String sentence)
It actually needs to return something:
return result.toArray(new String[0]);
rather than
return null;
The "You might want to use" suggestions points to that you might have to process the array character by character.
First, try and print out all the elements in your split array. Remember, you do only want you look at words. So, examine if this is the case by printing out each element of the split array inside your for loop. (I'm suspecting you will get a false positive at the moment)
Also, you need to revisit your books on arrays in Java. You can not dynamically add elements to an array. So, you will need a different data structure to be able to use an add() method. An ArrayList of Strings would help you here.
split your string on bases of white space, it will return an array. You can check the length of each word by iterating on that array.
you can split string though this way myString.split("\\s+");
Try this...
public static String[] getBigWords(String sentence)
{
java.util.ArrayList<String> result = new java.util.ArrayList<String>();
String[] split = sentence.split("\\s+");
for(int i = 0; i < split.length; i++)
{
if(split[i].length() > 5)
{
if(split[i].matches("[a-zA-Z]+"))
{
result.add(split[i]);
}
if (split[i].matches("[a-zA-Z]+,"))
{
String temp = "";
for(int j = 0; j < split[i].length(); j++)
{
if((split[i].charAt(j))!=((char)','))
{
temp += split[i].charAt(j);
//System.out.print(split[i].charAt(j) + "|");
}
}
result.add(temp);
}
}
}
return result.toArray(new String[0]);
}
Whet you have done is correct but you can't you add method in array. You should set like a[position]= spilt[i]; if you want to ignore number then check by Float.isNumber() method.
Your logic is valid, but you have some syntax issues. If you are not using an IDE like Eclipse that shows you syntax errors, try commenting out lines to pinpoint which ones are syntactically incorrect. I want to also tell you that once an array is created its length cannot change. Hopefully that sets you off in the right directions.
Apart from syntax errors at String array declaration should be like new String[n]
and add method will not be there in Array hence you should use like
a[i] = split[i];
You need to add another condition along with length condition to check that the given word have all letters this can be done in 2 ways
first way is to use Character.isLetter() method and second way is create regular expression
to check string have only letter. google it for regular expression and use matcher to match like the below
Pattern pattern=Pattern.compile();
Matcher matcher=pattern.matcher();
Final point is use another counter (let say j=0) to store output values and increment this counter as and when you store string in the array.
a[j++] = split[i];
I would use a string tokenizer (string tokenizer class in java)
Iterate through each entry and if the string length is more than 4 (or whatever you need) add to the array you are returning.
You said no code, so... (This is like 5 lines of code)