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How to return a 2D String

I do not fully understand how to return a 2D object. So I wrote a method that takes in an input with a document and I have to return a list of all unique words in it and their number of occurrences, sorted by the number of occurrences in a descending order. It is a requirement that I cannot control that this be returned as a 2-dimensional array of String.
So here is what I have so far:
static String[][] wordCountEngine(String document) {
// your code goes here
if (document == null || document.length() == 0)
return null;
Map<String, String> map = new HashMap<>();
String[] allWords = document.toLowerCase().split("[^a-zA-Z]+");
for (String s : allWords) {
if (map.containsKey(s)) {
int newVersion = (Integer.parseInt(map.get(s).substring(1, map.get(s).length())) + 1);
String sb = Integer.toString(newVersion);
map.put(s, sb);
} else {
map.put(s, "1");
}
}
String[][] array = new String[map.size()][2];
int count = 0;
for (Map.Entry<String, String> entry : map.entrySet()) {
array[count][0] = entry.getKey();
array[count][1] = entry.getValue();
count++;
}
return array;
}
I'm trying to use a HashMap to store the words and their occurrences. What is the best way to store key --> value pairs from a table into a String[][].
If the input is:
input: document = "Practice makes perfect. you'll only
get Perfect by practice. just practice!"
The output should be:
output: [ ["practice", "3"], ["perfect", "2"],
["by", "1"], ["get", "1"], ["just", "1"],
["makes", "1"], ["only", "1"], ["youll", "1"] ]
How do I store data like this in a 2D array?

String[][] simply is the wrong data structure for this task.
You should use a Map<String, Integer> map instead of <String, String> during the method run and simply return exactly that map.
This has multiple reasons:
you store integers as strings, and even do calculations by parsing the String to an integer again, calculating and then parsing back - bad idea.
The returned array does not guarantee the dimensions, there is no way to enforce that each sub-array has exactly two elements.
Note regarding your comment: if (for some reason) you need to convert the map to a String[][] you can certainly do that, but that conversion logic should be separated from the code generating the map itself. That way the code for wordCountEngine remains clean and easily maintainable.

Just because you need to return a particular typed data-structure does not mean you need to create similarly typed map inside your method. Nothing prevents you from using Map<String, Integer> and then converting it to String[][]:
Here is the code that does not use Java8 streeams:
static String[][] wordCountEngine(String document) {
// your code goes here
if (document == null || document.length() == 0)
return null;
Map<String, Integer> map = new HashMap<>();
for ( String s : document.toLowerCase().split("[^a-zA-Z]+") ){
Integer c = map.get(s);
map.put(s, c != null ? c + 1: 1);
}
String[][] result = new String[ map.size() ][ 2 ];
int count = 0;
for ( Map.Entry<String, Integer> e : map.entrySet() ){
result[count][0] = e.getKey();
result[count][1] = e.getValue().toString();
count += 1;
}
return result;
}
And for fun a Java8 version:
static String[][] wordCountEngine(String document) {
// your code goes here
if (document == null || document.length() == 0)
return null;
return Arrays
//convert words into map with word and count
.stream( document.toLowerCase().split("[^a-zA-Z]+") )
.collect( Collectors.groupingBy( s -> s, Collectors.summingInt(s -> 1) ) )
//convert the above map to String[][]
.entrySet()
.stream().map( (e) -> new String[]{ e.getKey(), e.getValue().toString() } )
.toArray( String[][]::new );
}

this is my Solution to Pramp's question although in C# I think it is the same Idea
[TestMethod]
public void PrampWordCountEngineTest()
{
string document = "Practice makes perfect. you'll only get Perfect by practice. just practice!";
string[,] result = WordCountEngine(document);
string[,] expected =
{
{"practice", "3"}, {"perfect", "2"},
{"makes", "1"}, {"youll", "1"}, {"only", "1"},
{"get", "1"}, {"by", "1"}, {"just", "1"}
};
CollectionAssert.AreEqual(expected,result);
}
public string[,] WordCountEngine(string document)
{
Dictionary<string, int> wordMap = new Dictionary<string, int>();
string[] wordList = document.Split(' ');
int largestCount = 0;
foreach (string word in wordList)
{
string lowerWord = word.ToLower(); // can't assing to the same variable
//remove special/punctuation characters
var sb = new StringBuilder();
foreach (var c in lowerWord)
{
if (c >= 'a' && c <= 'z')
{
sb.Append(c);
}
}
string cleanWord = sb.ToString();
if (cleanWord.Length < 1)
{
continue;
}
int count = 0;
if (wordMap.ContainsKey(cleanWord))
{
count = wordMap[cleanWord];
count++;
}
else
{
count = 1;
}
if (count > largestCount)
{
largestCount = count;
}
wordMap[cleanWord] = count;
}
// we have a list of all of the words in the same length in a given cell of the big list
List<List<string>> counterList = new List<List<string>>();
for (int i = 0; i < largestCount + 1; i++)
{
counterList.Add(new List<string>());
}
foreach (var word in wordMap.Keys)
{
int counter = wordMap[word];
counterList[counter].Add(word);
}
string[,] result = new string[wordMap.Keys.Count,2];
int key = 0;
//foreach list of words with the same length we insert the count of that word into the 2D array
for (var index = counterList.Count-1; index > 0; index--)
{
var list = counterList[index];
List<string> wordListCounter = list;
if (wordListCounter == null)
{
continue;
}
foreach (var word in wordListCounter)
{
result[key, 0] = word;
result[key, 1] = index.ToString();
key++;
}
}
return result;
}

Algorithm translate a number to String [closed]

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I need to design an algorithm where each number is encoded to an alphabet, for example:
1=A, 2=B, 3=C...26=Z
Given a set of numbers, I have to translate them to a combination of strings. For example:
123 can be translated to - ABC(123), AW(1 23) and LC(12 3)
Write an algorithm to find the combinations for number - 123123123.
Now here is what I wrote and I find it inefficient because of multiple "for" loops. Is there any better way I can rewrite this algorithm?
public class ValidCombinations {
Map<Integer, String> mapping = new HashMap<Integer, String>();
public void run() {
String s = "123123123";
/*Convert String to int[]*/
char[] cArray = s.toCharArray();
int[] input = new int[cArray.length];
for (int i=0; i<cArray.length; i++) {
input[i] = Character.getNumericValue(cArray[i]);
}
Set<String> output = new HashSet<String>();
for (int i='A'; i<='Z'; i++) {
mapping.put(i - 'A' + 1, String.valueOf((char)i));
}
for (int i=0; i<input.length; i++) {
if (mapping.containsKey(input[i])) {
output.add(precombine(i, input) + mapping.get(input[i]) + postcombine(i, input));
if (i+1<input.length) {
if (mapping.containsKey(input[i]*10 + input[i+1])) {
output.add(precombine(i, input) + mapping.get(input[i]*10 + input[i+1]) + postcombine(i+1, input));
}
}
}
}
System.out.println(output);
}
public String precombine(int i, int[] input) {
String residue="";
for (int m=0; m<i; m++) {
residue += mapping.get(input[m]);
}
return residue;
}
public String postcombine(int i, int[] input) {
String residue="";
for (int k=i+1; k<input.length; k++) {
residue += mapping.get(input[k]);
}
return residue;
}
public static void main(String[] args) {
ValidCombinations v = new ValidCombinations();
v.run();
}
}
For '123' - [ABC, AW, LC]
For '123123123' - [LCABCABC, AWABCABC, ABCAWABC, ABCLCABC, ABCABCLC, ABCABCABC, ABCABCAW]

This problem is crying out for recursion. Here's a quick and dirty implementation that takes the input "number" in as a string and uses substring() to consume the digits. You could adapt it to use numerical methods to get the first (or first two) decimal digits from an integer if you prefer.
If you choose to work directly from an int, it would probably be easier to start at the end (working with the least-significant-digits) than at the beginning -- lastDigit = number % 10; otherDigits = number / 10
public List<String> encodings(String number) {
List<String> out = new ArrayList<>();
addEncodings("", number, out);
return out;
}
private void addEncodings(String prefix, String number, List<String> out) {
if (number.length() == 0) {
out.add(prefix);
} else {
addParsingNDigits(1, prefix, number, out);
addParsingNDigits(2, prefix, number, out);
}
}
private void addParsingNDigits(int digits, String prefix, String number, List<String> out) {
if (number.length() >= digits) {
char encodedChar = parseChars(number, digits);
if (encodedChar >= 'A' && encodedChar <= 'Z') {
addEncodings(prefix + encodedChar, number.substring(digits), out);
}
}
}
private char parseChars(String number, int length) {
int intVal = Integer.parseInt(number.substring(0, length));
return (char) ('A' + intVal - 1);
}
I don't think your solution will find all possible encodings -- I think you need some sort of stack to solve it. The solution above implicitly uses the execution stack, because of recursive method calls. Another solution could explicitly place objects representing "todo" calculations onto a stack data structure in the heap:
private static class StackItem {
public StackItem(String prefix, String number) {
this.prefix = prefix;
this.number = number;
}
public String prefix;
public String number;
}
public List<String> encodings(String number) {
List<String> results = new ArrayList<>();
Stack<StackItem> stack = new Stack<>();
stack.push(new StackItem("", number));
while (!stack.isEmpty()) {
StackItem current = stack.pop();
if (current.number.equals("")) {
results.add(current.prefix);
} else {
addToStackTakingNChars(2, current, stack);
addToStackTakingNChars(1, current, stack);
}
}
return results;
}
private void addToStackTakingNChars(int n, StackItem current, Stack<StackItem> stack) {
if (current.number.length() >= n) {
char c = parseChars(current.number, n);
if (c >= 'A' && c <= 'Z') {
stack.push(new StackItem(current.prefix + c, current.number.substring(n)));
}
}
}
Although "println debugging" is generally a bad habit, it would probably be a good learning exercise to run these examples with some println()s to observe how it works.

I think you could split the String in the middle (recursively), search for all combinations in both substrings and build the cross product. To not miss any combinations we have to also build the cross product for the two substrings you get by splitting in the middle with an offset of one. Something like this:
private static int[] values;
public static final Set<String> solve(String s) {
values = new int[s.length()];
for (int i = 0; i < values.length; i++)
values[i] = s.charAt(i) - '0';
return solve(0, values.length);
}
private static final Set<String> solve(int start, int len) {
Set<String> ret = new HashSet<>();
if (len == 1) {
ret.add("" + ((char)(values[start] - 1 + 'A')));
} else if (len == 2) {
ret.add("" + ((char)(values[start] - 1 + 'A')) +
((char)(values[start + 1] - 1 + 'A')));
int n = values[start] * 10 + values[start + 1];
if (n <= 26)
ret.add("" + ((char)(n - 1 + 'A')));
} else {
int next = start + len / 2;
cross(solve(start, next - start), solve(next, start + len - next), ret);
cross(solve(start, next - start + 1), solve(next + 1, start + len - next - 1), ret);
}
return ret;
}
private static final void cross(Set<String> a, Set<String> b, Set<String> target) {
for (Iterator<String> itr = a.iterator(); itr.hasNext();) {
String s = itr.next();
for (Iterator<String> itr2 = b.iterator(); itr2.hasNext();) {
target.add(s + itr2.next());
}
}
}
Btw. the solution for "123123123" are the following 27 strings: LCABCAW, LCABCLC, ABCLCABC, ABCLCAW, ABCAWLC, AWLCABC, ABCAWAW, ABCAWABC, ABCLCLC, ABCABCABC, LCAWLC, LCAWAW, AWABCLC, LCAWABC, AWABCAW, LCLCAW, AWABCABC, LCLCLC, LCLCABC, LCABCABC, AWAWLC, AWAWABC, AWAWAW, ABCABCLC, ABCABCAW, AWLCAW, AWLCLC.

Why not just use the ascii value?
All you would need to do would be to convert the number to a String Integer.toString(num) and then run a for-loop through the .length() of the String and pull the .charAt(i) from the String convert that back to an int and then add 16 to it. Then you would just need to cast to a char. like so:
int a = 123;
String str = Integer.toString(a);
char[] chars = new char[str.length()];
for(int i=0,n=str.length();i<n;i++){
chars[i] = (char)(str.charAt(i)+16);
}
String message = String.valueOf(chars);

This problem can be done in o(fib(n+2)) time with a standard DP algorithm.
We have exactly n sub problems and button up we can solve each problem in o(fib(i)) time.
Summing the series gives fib (n+2).
If you consider the question carefullly you see that it is a fibunacci series.
I took a standart code and just changed it a bit to fit our conditions.
The space is obviously bound to the size of all solutions o(fib(n)).
Consider this code:
Map<Integer, String> mapping = new HashMap<Integer, String>();
List<String > iterative_fib_sequence(int input) {
int length = Math.floor(Math.log10(Math.abs(input))) + 1;
if (length <= 1)
{
if (length==0)
{
return "";
}
else//input is a-j
{
return mapping.get(input);
}
}
List<String> b = new List<String>();
List<String> a = new List<String>(mapping.get(input.substring(0,0));
List<String> c = new List<String>();
for (int i = 1; i < length; ++i)
{
int dig2Prefix = input.substring(i-1, i); //Get a letter with 2 digit (k-z)
if (mapping.contains(dig2Prefix))
{
String word2Prefix = mapping.get(dig2Prefix);
foreach (String s in b)
{
c.Add(s.append(word2Prefix));
}
}
int dig1Prefix = input.substring(i, i); //Get a letter with 1 digit (a-j)
String word1Prefix = mapping.get(dig1Prefix);
foreach (String s in a)
{
c.Add(s.append(word1Prefix));
}
b = a;
a = c;
c = new List<String>();
}
return a;
}

Arranging word to formpalindrome

Hello i´ve been trying to form palindromes from this input:
String[] text ={"ivcci", "oyotta", "cecarar","bbb","babbbb"};
getPalindrome(text);
and i need to rearrange all words in array to produce this output
civic
-1
rececar
bbb
bbabb
the method expects to receive an array of Strings like
public static String getPalindrome(String[] text){}
"returning -1 means i.g "oyotta" in array can´t form a palíndrome
i´ve been testing this code and it works but i.g "cecarar" is not producing "racecar", as im a bit new in java i used an String intead an array of Strings, can anybody help to write this code properly please?
Thanks a lot!
public static String getPalindrome(String s) {
if (s == null)
return null;
Map<Character, Integer> letters = new HashMap<Character, Integer>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!letters.containsKey(c))
letters.put(c, 1);
else
letters.put(c, letters.get(c) + 1);
}
char[] result = new char[s.length()];
int i = 0, j = result.length - 1;
Character middle = null;
for (Entry<Character, Integer> e : letters.entrySet()) {
int val = e.getValue();
char c = e.getKey();
if (val % 2 != 0) {
if (middle == null && s.length() % 2 != 0) {
middle = c;
val--;
} else
return "-1";
}
for (int k = 0; k < val / 2; k++) {
result[i++] = c;
result[j--] = c;
}
}
if (middle != null)
result[result.length / 2] = middle;
return new String(result);
}

In order for a set of characters to be able to produce a palindrome, only one of the letters can be repeated an odd number of times, so you can first weed that out.
Without writing actual code for you, here is the algorithm I would use:
Create a map of characters to a counter. Possible to do int[] counts = new int[26];
Go through each character in the input string, and increment the count: ++counts[Character.toLower(c)-'a'];
Then go through each character, and see if its odd if (counts[i] & 1 != 0) { if (oddIndex != -1) { return -1; } oddIndex=i; } This will return -1 if there is two or more odd counts.
Then, you can create a StringBuilder, and start with the oddIndex in the middle, if it exists.
Then go through the counts, and add count[i]/2 to the front and back of your string builder.
That'll give you a symmetric string from the original inputs.
Now, if you actually need words, then you'll have to have a dictionary of palindromes. You can actually preprocess all the palindromes to have a map of "sorted character string"=>"palindrome"
class PalindromeChecker
{
final Map<String, String> palindromes = new HashMap<String, String>();
public PalindromeChecker(Iterable<String> allPalindromes) {
for (String palindrome: allPalindromes) {
char[] chars = palindrome.getChars();
Arrays.sort(chars);
palindromes.put(String.valueOf(chars), palindromes);
}
}
public String getPalindrome(String input) {
char[] chars = input.getChars();
Arrays.sort(chars);
return palindromes.get(String.valueOf(chars));
}
}

As other users pointed out, a string can be rearranged as a palindrome only if there is at most one character that appears an odd number of times.
Once you have confirmed that a string can be converted to a palindrome, you can construct the palindrome as follows (this is just one of many methods of course):
place at the sides of the string all the pairs of characters that you can get
place at the middle of the string the single character that is left out, in case there is such a character.
Example:
public class Palindromes {
public static void main(String[] args) {
String[] text = {"ivcci", "oyotta", "cecarar","bbb","babbbb"};
for(String str : text){
evaluatePalindrome(str);
}
}
private static void evaluatePalindrome(String str){
PalindromeCandidate pc = new PalindromeCandidate(str);
if(pc.isPalindrome()){
System.out.println(pc.getPalindrome());
} else {
System.out.println("-1");
}
}
}
public class PalindromeCandidate {
private final CharacterCount characterCount;
public PalindromeCandidate(String originalString) {
this.characterCount = new CharacterCount(originalString);
}
public boolean isPalindrome(){
Collection<Integer> counts = characterCount.asMap().values();
int oddCountOccurrences = 0;
for(Integer count : counts){
oddCountOccurrences += (count%2);
}
return (oddCountOccurrences <= 1);
}
public String getPalindrome(){
if(!isPalindrome()){
throw new RuntimeException("Cannot be rearranged as a palindrome.");
}
Map<Character, Integer> counts = characterCount.asMap();
StringBuilder leftSide = new StringBuilder();
StringBuilder middle = new StringBuilder();
for(Character ch : counts.keySet()){
int occurrences = counts.get(ch);
while(occurrences > 1){
leftSide.append(ch);
occurrences -= 2;
}
if(occurrences > 0){
middle.append(ch);
}
}
StringBuilder rightSide = new StringBuilder(leftSide).reverse();
return leftSide.append(middle).append(rightSide).toString();
}
}
/**
* Thin wrapper around a Map<Character, Integer>. Used for counting occurences
* of characters.
*/
public class CharacterCount {
private final Map<Character, Integer> map;
public CharacterCount(String str) {
this.map = new HashMap<>();
for(Character ch : str.toCharArray()){
increment(ch);
}
}
private void increment(Character ch){
this.map.put(ch, getCount(ch) + 1);
}
private Integer getCount(Character ch){
if(map.containsKey(ch)){
return map.get(ch);
} else {
return 0;
}
}
public Map<Character, Integer> asMap(){
return new HashMap<>(map);
}
}

comparing string arrays in java

I am working on a simple Java program where we have sorted string array named arr
I am trying to compare two adjacent string and calculate frequency for each string present in that array
for(j1=0;j1<arr.length;j1++){
if(j1+1 < arr.length){ // To prevent arrayOutofBoundsException
if(arr[j1].equals(arr[j1+1])){
counter++;
}
else {
System.out.println(arr[j1]+" "+counter);
counter=1;
}
But it's not working right , what's wrong ?
edit:problem is not in comparing , it's not calculating frequency as desired

OK, besides the equals fix, you want to keep the original order of words:
String orig = "hellow hello hello how how he ho" ;
//String orig = "how are you how do you do";
String[] arr = orig.split(" ");
//Arrays.sort(arr);
for(int j1 = 0; j1 < arr.length; j1++){
if (arr[j1] != null) {
int counter = 1;
for(int j2 = j1+1; j2 < arr.length; j2++) {
if(arr[j2] != null && arr[j1].equals(arr[j2])){
counter++;
arr[j2] = null;
}
}
System.out.println(arr[j1]+" "+counter);
}
}
The trick is that I run through the array, count all occurrences, null the occurrences, so they don't count again, and print the count. No need to sort the array.

== compares object identity in terms of memory address - to compare objects in terms of equality, use the equals-method.

This should work:
public static void main(String[] args)
{
String text = "how are you how do you do";
String[] keys = {"how", "are", "you", "how", "do", "you", "do"};
Arrays.sort(keys);
String[] uniqueKeys;
int count = 0;
System.out.println(text);
uniqueKeys = getUniqueKeys(keys);
for(String key: uniqueKeys)
{
if(null == key)
{
break;
}
for(String s : keys)
{
if(key.equals(s))
{
count++;
}
}
System.out.println("Count of ["+key+"] is : "+count);
count=0;
}
}
private static String[] getUniqueKeys(String[] keys)
{
String[] uniqueKeys = new String[keys.length];
uniqueKeys[0] = keys[0];
int uniqueKeyIndex = 1;
boolean keyAlreadyExists = false;
for(int i=1; i<keys.length ; i++)
{
for(int j=0; j<=uniqueKeyIndex; j++)
{
if(keys[i].equals(uniqueKeys[j]))
{
keyAlreadyExists = true;
}
}
if(!keyAlreadyExists)
{
uniqueKeys[uniqueKeyIndex] = keys[i];
uniqueKeyIndex++;
}
keyAlreadyExists = false;
}
return uniqueKeys;
}
Output:
how are you how do you do
Count of [how] is : 2
Count of [are] is : 1
Count of [you] is : 2
Count of [do] is : 2

Are you looking some sort of this
public static void main(String[] args){
String[] arr = new String[5];
arr[0] = "One";
arr[1] = "Two";
arr[2] = "One";
arr[3] = "Three";
arr[4] = "Two";
List<String> lstString = Arrays.asList(arr);
Collections.sort(lstString);
for(String eachString : arr){
System.out.println("Frequency of " + eachString + " is " + getFrequency(eachString,lstString));
}
}
private static int getFrequency(String word, List lstOfString){
int frequency = 1;
if(lstOfString != null && lstOfString.size() > 0){
int firstIndex = lstOfString.indexOf(word);
int lastIndex = lstOfString.lastIndexOf(word);
frequency += lastIndex - firstIndex;
}
return frequency;
}
Result :
Frequency of One is 2
Frequency of One is 2
Frequency of Three is 1
Frequency of Two is 2
Frequency of Two is 2

Java: How to split a string by a number of characters?

I tried to search online to solve this question but I didn't found anything.
I wrote the following abstract code to explain what I'm asking:
String text = "how are you?";
String[] textArray= text.splitByNumber(4); //this method is what I'm asking
textArray[0]; //it contains "how "
textArray[1]; //it contains "are "
textArray[2]; //it contains "you?"
The method splitByNumber splits the string "text" every 4 characters. How I can create this method??
Many Thanks

I think that what he wants is to have a string split into substrings of size 4. Then I would do this in a loop:
List<String> strings = new ArrayList<String>();
int index = 0;
while (index < text.length()) {
strings.add(text.substring(index, Math.min(index + 4,text.length())));
index += 4;
}

Using Guava:
Iterable<String> result = Splitter.fixedLength(4).split("how are you?");
String[] parts = Iterables.toArray(result, String.class);

What about a regexp?
public static String[] splitByNumber(String str, int size) {
return (size<1 || str==null) ? null : str.split("(?<=\\G.{"+size+"})");
}
See Split string to equal length substrings in Java

Try this
String text = "how are you?";
String array[] = text.split(" ");
Or you can use it below
List<String> list= new ArrayList<String>();
int index = 0;
while (index<text.length()) {
list.add(text.substring(index, Math.min(index+4,text.length()));
index=index+4;
}

Quick Hack
private String[] splitByNumber(String s, int size) {
if(s == null || size <= 0)
return null;
int chunks = s.length() / size + ((s.length() % size > 0) ? 1 : 0);
String[] arr = new String[chunks];
for(int i = 0, j = 0, l = s.length(); i < l; i += size, j++)
arr[j] = s.substring(i, Math.min(l, i + size));
return arr;
}

Using simple java primitives and loops.
private static String[] splitByNumber(String text, int number) {
int inLength = text.length();
int arLength = inLength / number;
int left=inLength%number;
if(left>0){++arLength;}
String ar[] = new String[arLength];
String tempText=text;
for (int x = 0; x < arLength; ++x) {
if(tempText.length()>number){
ar[x]=tempText.substring(0, number);
tempText=tempText.substring(number);
}else{
ar[x]=tempText;
}
}
return ar;
}
Usage : String ar[]=splitByNumber("nalaka", 2);

I don't think there's an out-of-the-box solution, but I'd do something like this:
private String[] splitByNumber(String s, int chunkSize){
int chunkCount = (s.length() / chunkSize) + (s.length() % chunkSize == 0 ? 0 : 1);
String[] returnVal = new String[chunkCount];
for(int i=0;i<chunkCount;i++){
returnVal[i] = s.substring(i*chunkSize, Math.min((i+1)*chunkSize-1, s.length());
}
return returnVal;
}
Usage would be:
String[] textArray = splitByNumber(text, 4);
EDIT: the substring actually shouldn't surpass the string length.

This is the simplest solution i could think off.. try this
public static String[] splitString(String str) {
if(str == null) return null;
List<String> list = new ArrayList<String>();
for(int i=0;i < str.length();i=i+4){
int endindex = Math.min(i+4,str.length());
list.add(str.substring(i, endindex));
}
return list.toArray(new String[list.size()]);
}

Here's a succinct implementation using Java8 streams:
String text = "how are you?";
final AtomicInteger counter = new AtomicInteger(0);
Collection<String> strings = text.chars()
.mapToObj(i -> String.valueOf((char)i) )
.collect(Collectors.groupingBy(it -> counter.getAndIncrement() / 4
,Collectors.joining()))
.values();
Output:
[how , are , you?]

Try this solution,
public static String[]chunkStringByLength(String inputString, int numOfChar) {
if (inputString == null || numOfChar <= 0)
return null;
else if (inputString.length() == numOfChar)
return new String[]{
inputString
};
int chunkLen = (int)Math.ceil(inputString.length() / numOfChar);
String[]chunks = new String[chunkLen + 1];
for (int i = 0; i <= chunkLen; i++) {
int endLen = numOfChar;
if (i == chunkLen) {
endLen = inputString.length() % numOfChar;
}
chunks[i] = new String(inputString.getBytes(), i * numOfChar, endLen);
}
return chunks;
}

My application uses text to speech!
Here is my algorithm, to split by "dot" and conconate string if string length less then limit
String[] text = sentence.split("\\.");
ArrayList<String> realText = sentenceSplitterWithCount(text);
Function sentenceSplitterWithCount: (I concanate string lf less than 100 chars lenght, It depends on you)
private ArrayList<String> sentenceSplitterWithCount(String[] splittedWithDot){
ArrayList<String> newArticleArray = new ArrayList<>();
String item = "";
for(String sentence : splittedWithDot){
item += DataManager.setFirstCharCapitalize(sentence)+".";
if(item.length() > 100){
newArticleArray.add(item);
item = "";
}
}
for (String a : newArticleArray){
Log.d("tts", a);
}
return newArticleArray;
}
function setFirstCharCapitalize just capitalize First letter: I think, you dont need it, anyway
public static String setFirstCharCapitalize(String input) {
if(input.length()>2) {
String k = checkStringStartWithSpace(input);
input = k.substring(0, 1).toUpperCase() + k.substring(1).toLowerCase();
}
return input;
}