Develop Reference

split a string when there is a change in character without a regular expression

There is a way to split a string into repeating characters using a regex function but I want to do it without using it.
for example, given a string like: "EE B" my output will be an array of strings e.g
{"EE", " ", "B"}
my approach is:
given a string I will first find the number of unique characters in a string so I know the size of the array. Then I will change the string to an array of characters. Then I will check if the next character is the same or not. if it is the same then append them together if not begin a new string.
my code so far..
String myinput = "EE B";
char[] cinput = new char[myinput.length()];
cinput = myinput.toCharArray(); //turn string to array of characters
int uniquecha = myinput.length();
for (int i = 0; i < cinput.length; i++) {
if (i != myinput.indexOf(cinput[i])) {
uniquecha--;
} //this should give me the number of unique characters
String[] returninput = new String[uniquecha];
Arrays.fill(returninput, "");
for (int i = 0; i < uniquecha; i++) {
returninput[i] = "" + myinput.charAt(i);
for (int j = 0; j < myinput.length - 1; j++) {
if (myinput.charAt(j) == myinput.charAt(j + 1)) {
returninput[j] += myinput.charAt(j + 1);
} else {
break;
}
}
} return returninput;
but there is something wrong with the second part as I cant figure out why it is not beginning a new string when the character changes.

You question says that you don't want to use regex, but I see no reason for that requirement, other than this is maybe homework. If you are open to using regex here, then there is a one line solution which splits your input string on the following pattern:
(?<=\S)(?=\s)|(?<=\s)(?=\S)
This pattern uses lookarounds to split whenever what precedes is a non whitespace character and what proceeds is a whitespace character, or vice-versa.
String input = "EE B";
String[] parts = input.split("(?<=\\S)(?=\\s)|(?<=\\s)(?=\\S)");
System.out.println(Arrays.toString(parts));
[EE, , B]
^^ a single space character in the middle
Demo

If I understood correctly, you want to split the characters in a string so that similar-consecutive characters stay together. If that's the case, here is how I would do it:
public static ArrayList<String> splitString(String str)
{
ArrayList<String> output = new ArrayList<>();
String combo = "";
//iterates through all the characters in the input
for(char c: str.toCharArray()) {
//check if the current char is equal to the last added char
if(combo.length() > 0 && c != combo.charAt(combo.length() - 1)) {
output.add(combo);
combo = "";
}
combo += c;
}
output.add(combo); //adds the last character
return output;
}
Note that instead of using an array (has a fixed size) to store the output, I used an ArrayList, which has a variable size. Also, instead of checking the next character for equality with the current one, I preferred to use the last character for that. The variable combo is used to temporarily store the characters before they go to output.
Now, here is one way to print the result following your guidelines:
public static void main(String[] args)
{
String input = "EEEE BCD DdA";
ArrayList<String> output = splitString(input);
System.out.print("[");
for(int i = 0; i < output.size(); i++) {
System.out.print("\"" + output.get(i) + "\"");
if(i != output.size()-1)
System.out.print(", ");
}
System.out.println("]");
}
The output when running the above code will be:
["EEEE", " ", "B", "C", "D", " ", "D", "d", "A"]

Finding Unique Words In A Text File Using ArrayList

I'm working on a project where I enter a URL, the file is read and the amount of lines, characters, and words are outputted in a text file. I'm not having an issue with that. Code below will be pretty long, sorry in advance.
I also have to output to the same text file all of the words in the file, and the amount of times each word is displayed in the file. I've been working on it for a while and I've gotten to the point where all the lines/characters/words are outputted to the text file, but I can't figure out how to display the actual words and the amount of times they are in the file.
String[] wordSubstrings = line.replaceAll("\\s+", " ").split(" ");
List<String> uniqueWords = new ArrayList<String>();
for (int i = 0; i < wordSubstrings.length; i++) {
if (!(uniqueWords.contains(wordSubstrings[i]))) {
uniqueWords.add(wordSubstrings[i]);

You could use a Multiset
Multiset<String> words = HashMultiset.create();
for (String word : wordList)
words.add(word);
for (String word : words.elementSet())
System.out.println(word + ": " + words.count(word));

I've tested something with a HashMap which seems to work pretty well.
Here is my code that I used to test it, I hope it helps:
String[] wordSubstrings = new String[]{"test","stuff","test","thing","test","test","stuff"};
HashMap<String,Integer> uniqueWords = new HashMap<>();
for ( int i = 0; i < wordSubstrings.length; i++)
{
if(!(uniqueWords.containsKey(wordSubstrings[i])))
{
uniqueWords.put(wordSubstrings[i], 1);
}
else
{
int number = uniqueWords.get(wordSubstrings[i]);
uniqueWords.put(wordSubstrings[i],number + 1);
}
}
for (Map.Entry<String, Integer> entry : uniqueWords.entrySet()) {
String key = entry.getKey();
int value = entry.getValue();
//Do Something with the key and value
}

You can use arraylist of class which will contain word and count as member variables.
List <MyClass> uniqueWords = new ArrayList<MyClass> ();
MyClass()
{
String uniqueword;
int count;
}

Issue iterating through two arraylists

EDIT: Thanks so much for all the really quick feedback. Wow. I did just paste it all for you instead of just those two for loops. Thanks.
This may have been totally answered before. I have read SO for the last few years but this is my first post. I have been using the site and others to help solve this so my apologies in advance if this has been answered!
I am iterating through two arraylists. One is derived from user input; the other is a dictionary file converted into an arraylist. I am trying to compare a word in the input with a dictionary word. The input list and the dictionary list are valid and if I simply iterate through them, they contain what they should (so that isn't the issue. I assume my issue is somewhere with how I am handling the iteration. I'm a fairly novice Java programmer so please go easy on me.
Thanks
public String isSub(String x) throws FileNotFoundException, IOException {
//todo handle X
String out = "**********\nFor input \n" + x + "If you're reading this no match was found.\n**********";
String dictionary;
boolean solve = true;
/// Get dictionary
dictMaker newDict = new dictMaker();
dictionary = newDict.arrayMaker();
List<String> myDict = new ArrayList<String>(Arrays.asList(dictionary.split(",")));
List<String> input = new ArrayList<String>(Arrays.asList(x.split(" ")));
List<String> results = new ArrayList<String>();
//results = input;
String currentWord;
String match = "";
String checker = "";
String fail="";
//Everything to break sub needs to happen here.
while (solve) {
for(int n = 0; n < input.size(); n++) { //outside FOR (INPUT)
if(!fail.equals("")) results.add(fail);
checker = input.get(n).trim();
for(int i = 0; i < myDict.size(); i++) { //inside FOR (dictionary)
currentWord = myDict.get(i).trim();
System.out.print(checker + " " + currentWord + "\n");
if(checker.equals(currentWord)) {
match = currentWord;
results.add(currentWord);
fail="";
} //end if
else {
fail = "No match for " + checker;
}
}//end inside FOR (dictionary)
} //END OUTSIDE FOR (input)
solve=false;
} //end while
out = results.toString();
return out;
}
Output results for input "test tester asdasdfasdlfk"
[test, No match for test, tester, No match for tester]

Carl Manaster gave the correct explanation.
Here's an improved version of your code:
for (int n = 0; n < input.size(); n++) { //outside FOR (INPUT)
String checker = input.get(n).trim();
boolean match = false;
for (int i = 0; i < myDict.size(); i++) { //inside FOR (dictionary)
String currentWord = myDict.get(i).trim();
System.out.print(checker + " " + currentWord + "\n");
if (checker.equals(currentWord)) {
match = true;
results.add(currentWord);
break;
} //end if
} //end inside FOR (dictionary)
if (!match) {
results.add("No match for " + checker);
}
} //END OUTSIDE FOR (input)
Also, consider using a HashMap instead of an ArrayList to store the dictionary and trim the words when you store them to avoid doing it in each pass.

It looks as though every word in input gets compared to every word in your dictionary. So for every word that doesn't match, you get a fail (although you only write the last failure in the dictionary to the results). The problem appears to be that you keep looping even after you have found the word. To avoid this, you probably want to add break to the success case:
if (checker.equals(currentWord)) {
match = currentWord;
results.add(currentWord);
fail = "";
break;
} else {
fail = "No match for " + checker;
}

If you are using a dictionary, you should get it with keys not with index. So it should be
if(myDict.containsKey(checker)){
String currentWord =myDict.get(checker);
System.out.print(checker + " " + currentWord + "\n");
match = currentWord;
results.add(currentWord);
fail = "";
}
else {
fail = "No match for " + checker;
}
I think more or less your code should like following.
ArrayList<String> input= new ArrayList<String>();
input.add("ahmet");
input.add("mehmet");
ArrayList<String> results= new ArrayList<String>();
Map<String, String> myDict = new HashMap<String, String>();
myDict.put("key", "ahmet");
myDict.put("key2", "mehmet");
String match="";
String fail="";
for (int n = 0; n < input.size(); n++) { //outside FOR (INPUT)
if (!fail.equals(""))
results.add(fail);
String checker = input.get(n).trim();
for (int i = 0; i < myDict.size(); i++) { //inside FOR (dictionary)
// String currentWord = myDict.get(i).trim();
if(myDict.containsKey(checker)){
String currentWord =myDict.get(checker);
System.out.print(checker + " " + currentWord + "\n");
match = currentWord;
results.add(currentWord);
fail = "";
}
else {
fail = "No match for " + checker;
}
} // end inside FOR (dictionary)
} // end outside FOR (input)
// solve = false; I dont know what is this
//} //end while. no while in my code
return results.toString();

You should place the dictionary to a HashSet and trim while add all words. Next you just need to loop the input list and compare with dict.conatins(inputWord). This saves the possible huge dictionary loop processed for all input words.
Untested brain dump:
HashSet<String> dictionary = readDictionaryFiles(...);
List<String> input = getInput();
for (String inputString : input)
{
if (dictionary.contains(inputString.trim()))
{
result.add(inputString);
}
}
out = result.toString()
....
And a solution similar to the original posting. The unnecessary loop index variables are removed:
for (String checker : input)
{ // outside FOR (INPUT)
fail = "No match for " + checker;
for (String currentWord : myDict)
{ // inside FOR (dictionary)
System.out.print(checker + " " + currentWord + "\n");
if (checker.equals(currentWord))
{
match = currentWord;
results.add(currentWord);
fail = null;
break;
}
} // end inside FOR (dictionary)
if (fail != null)
{
results.add(fail);
}
} // end outside FOR (input)
solve = false;
return results.toString();
The trim should be made while add the elements to the list. Trim the dictionary values each time is overhead. And the inner loop itself too. The complexity of the task can be reduced if the dictionary data structure is changed from List to Set.
Adding the result of "fail" is moved to the end of the outer loop. Otherwise the result of the last input string is not added to the result list.
The following code is terrible:
else {
fail = "No match for " + checker;
}
The checker does not change within the dictionary loop. But the fail string is constructed each time the checker and the dictionary value does not match.

Split string into key-value pairs

I have a string like this:
pet:cat::car:honda::location:Japan::food:sushi
Now : indicates key-value pairs while :: separates the pairs.
I want to add the key-value pairs to a map.
I can achieve this using:
Map<String, String> map = new HashMap<String, String>();
String test = "pet:cat::car:honda::location:Japan::food:sushi";
String[] test1 = test.split("::");
for (String s : test1) {
String[] t = s.split(":");
map.put(t[0], t[1]);
}
for (String s : map.keySet()) {
System.out.println(s + " is " + map.get(s));
}
But is there an efficient way of doing this?
I feel the code is inefficient because I have used 2 String[] objects and called the split function twice.
Also, I am using t[0] and t[1] which might throw an ArrayIndexOutOfBoundsException if there are no values.

You could do a single call to split() and a single pass on the String using the following code. But it of course assumes the String is valid in the first place:
Map<String, String> map = new HashMap<String, String>();
String test = "pet:cat::car:honda::location:Japan::food:sushi";
// split on ':' and on '::'
String[] parts = test.split("::?");
for (int i = 0; i < parts.length; i += 2) {
map.put(parts[i], parts[i + 1]);
}
for (String s : map.keySet()) {
System.out.println(s + " is " + map.get(s));
}
The above is probably a little bit more efficient than your solution, but if you find your code clearer, then keep it, because there is almost zero chance such an optimization has a significant impact on performance, unless you do that millions of times. Anyway, if it's so important, then you should measure and compare.
EDIT:
for those who wonder what ::? means in the above code: String.split() takes a regular expression as argument. A separator is a substring that matches the regular expression. ::? is a regular expression which means: 1 colon, followed by 0 or 1 colon. It thus allows considering :: and : as separators.

Using Guava library it's a one-liner:
String test = "pet:cat::car:honda::location:Japan::food:sushi";
Map<String, String> map = Splitter.on( "::" ).withKeyValueSeparator( ':' ).split( test );
System.out.println(map);
The output:
{pet=cat, car=honda, location=Japan, food=sushi}
This also might work faster than JDK String.split as it does not create a regexp for "::".
Update it even handles correctly the corner case from the comments:
String test = "pet:cat::car:honda::location:Japan::food:sushi:::cool";
Map<String, String> map = Splitter.on( "::" ).withKeyValueSeparator( ':' ).split( test );
System.out.println(map);
The output is:
{pet=cat, car=honda, location=Japan, food=sushi, =cool}

Your solution is indeed somewhat inefficient.
The person who gave you the string to parse is also somewhat of a clown. There are industry standard serialization formats, like JSON or XML, for which fast, efficient parses exist. Inventing the square wheel is never a good idea.
First question: Do you care? Is it slow enough that it hinders performance of your application? It's likely not to, but there is only one way to find out. Benchmark your code.
That said, more efficient solutions exist. Below is an example
public static void main (String[] args) throws java.lang.Exception
{
String test = "pet:cat::car:honda::location:Japan::food:sushi";
boolean stateiskey = true;
Map<String, String> map = new HashMap<>();
int keystart = 0;
int keyend = 0;
int valuestart = 0;
int valueend = 0;
for(int i = 0; i < test.length(); i++){
char nextchar = test.charAt(i);
if (stateiskey) {
if (nextchar == ':') {
keyend = i;
stateiskey = false;
valuestart = i + 1;
}
} else {
if (i == test.length() - 1 || (nextchar == ':' && test.charAt(i + 1) == ':')) {
valueend = i;
if (i + 1 == test.length()) valueend += 1; //compensate one for the end of the string
String key = test.substring(keystart, keyend);
String value = test.substring(valuestart, valueend);
keystart = i + 2;
map.put(key, value);
i++;
stateiskey = true;
}
}
}
System.out.println(map);
}
This solution is a finite state machine with only two states. It looks at every character only twice, once when it tests it for a boundary, and once when it copies it to the new string in your map. This is the minimum amount.
It doesn't create objects that are not needed, like stringbuilders, strings or arrays, this keeps collection pressure low.
It maintains good locality. The next character probably always is in cache, so the lookup is cheap.
It comes at a grave cost that is probably not worth it though:
It's far more complicated and less obvious
There are all sorts of moving parts
It's harder to debug when your string is in an unexpected format
Your coworkers will hate you
You will hate you when you have to debug something
Worth it? Maybe. How fast do you need that string parsed exactly?
A quick and dirty benchmark at https://ideone.com/8T7twy tells me that for this string, this method is approximately 4 times faster. For longer strings the difference is likely somewhat greater.
But your version is still only 415 milliseconds for 100.000 repetitions, where this one is 99 milliseconds.

Try this code - see the comments for an explanation:
HashMap<String,String> hmap = new HashMap<>();
String str="abc:1::xyz:2::jkl:3";
String straraay[]= str.split("::?");
for(int i=0;i<straraay.length;i+=2) {
hmap.put(straraay[i],straraay[i+1]);
}
for(String s:straraay){
System.out.println(hmap.values()); //for Values only
System.out.println(hmap.keySet()); //for keys only if you want to more clear
}

I don't know this is best approach or not but i think this is another way of doing same thing without using split method twice
Map<String, String> map = new HashMap<String, String>();
String test = "pet:cat::car:honda::location:Japan::food:sushi";
String[] test1 = test.replaceAll("::",":").split(":");
for(int i=0;i<test1.length;i=i+2)
{
map.put(test1[i], test1[i+1]);
}
for (String s : map.keySet()) {
System.out.println(s + " is " + map.get(s));
}
Hope it will help :)

This might be useful.
*utm_source=test_source&utm_medium=test_medium&utm_term=test_term&
utm_content=test_content&utm_campaign=test_name&referral_code=DASDASDAS
String str[] = referrerString.split("&");
HashMap<String,String> stringStringHashMap= new HashMap<>();
List<String> al;
al = Arrays.asList(str);
String[] strkey ;
for (String s : al) {
strkey= s.split("=");
stringStringHashMap.put(strkey[0],strkey[1]);
}
for (String s : stringStringHashMap.keySet()) {
System.out.println(s + " is " + stringStringHashMap.get(s));
}

Your program is absolutely fine.
Just because you asked for a more optimal code.
I reduced your memory by taking few variables instead of taking arrays and storing in them.
Look at your string it follows a patter.
key : value :: key : value ::....
What can we do from this?
get the key till it is : , once it reaches : get value until it reaches '::'.
package qwerty7;
import java.util.HashMap;
public class Demo {
public static void main(String ar[])
{
StringBuilder s = new StringBuilder("pet:cat::car:honda::location:Japan::food:sushi");
boolean isKey = true;
String key = "", value = "";
HashMap<String, String> hm = new HashMap();
for(int i = 0; i < s.length(); i++)
{
char ch = s.charAt(i);
char nextChar = s.charAt(i+1);
if(ch == ':' && nextChar != ':')
{
isKey = false;
continue;
}
else if(ch == ':' && nextChar == ':')
{
hm.put(key, value);
isKey = true;
key = "";
value = "";
i+=1;
continue;
}
if(isKey)
{
key += ch;
}
else
{
value += ch;
}
if(i == s.length() - 1)
{
hm.put(key, value);
}
}
for (String x : hm.keySet()) {
System.out.println(x + " is " + hm.get(x));
}
}
}
Doing so doesn't take up much iterations on splitting each time.
Doesn't take up much memory.
Time complexity O(n)
Output:
car is honda
location is Japan
pet is cat
food is sushi

String joining with delimiter not working in Android

I am trying to convert string array into string and join all values with delimiter (,) but it is adding only first two values, what is wrong in conversion code. Please see inline comments below
String[] array = new String[20];
for (int i = 0; i <= count; i++) {
Log.d(TAG, "arrayvalue : " + array[i]); //Here I will get 5 values which is exact value count, but in next converted log I will have concatenation of only first two values, what is wrong in conversion code.
// Joining:
StringBuilder buffer = new StringBuilder();
for (String each : array)
buffer.append(",").append(each);
String joined = buffer.deleteCharAt(0).toString();
Log.d("Prefs", "Converted Array to String : " + joined);
}

There is standard method for String join defined in Android SDK:
final String joined = TextUtils.join(",", array);

Use in dynamic array
String[] partno = new String[part.size()];
for(int i = 0; i < part.size(); i++){
partno[i]=part.get(i).getText().toString();
Hpartno= TextUtils.join(",",partno);
}