Develop Reference

Splitting a string in an array of strings of limited size

I have a string of a random address like
String s = "H.N.-13/1443 laal street near bharath dental lab near thana qutubsher near modern bakery saharanpur uttar pradesh 247001";
I want to split it into array of string with two conditions:
each element of that array of string is of length less than or equal to 20
No awkward ending of an element of array of string
For example, splitting every 20 characters would produce:
"H.N.-13/1443 laal st"
"reet near bharath de"
"ntal lab near thana"
"qutubsher near moder"
"n bakery saharanpur"
but the correct output would be:
"H.N.-13/1443 laal"
"street near bharath"
"dental lab near"
"thana qutubsher near"
"modern bakery"
"saharanpur"
Notice how each element in string array is less than or equal to 20.
The above is my output for this code:
static String[] split(String s,int max){
int total_lines = s.length () / 24;
if (s.length () % 24 != 0) {
total_lines++;
}
String[] ans = new String[total_lines];
int count = 0;
int j = 0;
for (int i = 0; i < total_lines; i++) {
for (j = 0; j < 20; j++) {
if (ans[count] == null) {
ans[count] = "";
}
if (count > 0) {
if ((20 * count) + j < s.length()) {
ans[count] += s.charAt (20 * count + j);
} else {
break;
}
} else {
ans[count] += s.charAt (j);
}
}
String a = "";
a += ans[count].charAt (0);
if (a.equals (" ")) {
ans[i] = ans[i].substring (0, 0) + "" + ans[i].substring (1);
}
System.out.println (ans[i]);
count++;
}
return ans;
}
public static void main (String[]args) {
String add = "H.N.-13/1663 laal street near bharath dental lab near thana qutubsher near modern bakery";
String city = "saharanpur";
String state = "uttar pradesh";
String zip = "247001";
String s = add + " " + city + " " + state + " " + zip;
String[]ans = split (s);
}

Find all occurrences of up to 20 chars starting with a non-space and ending with a word boundary, and collect them to a List:
List<String> parts = Pattern.compile("\\S.{1,19}\\b").matcher(s)
.results()
.map(MatchResult::group)
.collect(Collectors.toList());
See live demo.

The code is not very clear, but at first glance it seems you are building character by character that is why you are getting the output you see. Instead you go word by word if you want to retain a word and overflow it to next String if necessary. A more promising code would be:
static String[] splitString(String s, int max) {
String[] words = s.split("\s+");
List<String> out = new ArrayList<>();
int numWords = words.length;
int i = 0;
while (i <numWords) {
int len = 0;
StringBuilder sb = new StringBuilder();
while (i < numWords && len < max) {
int wordLength = words[i].length();
len += (i == numWords-1 ? wordLength : wordLength + 1);//1 for space
if (len <= max) {
sb.append(words[i]+ " ");
i++;
}
}
out.add(sb.toString().trim());
}
return out.toArray(new String[] {});
}
Note: It works on your example input, but you may need to tweak it so it works for cases like a long word containing more than 20 characters, etc.

Put a space every 5 characters without using regex or StringBuilder

I want a user to put in a sentence with the scanner class.
Make sure to filter out all the spaces (for example the sentence: this is a test becomes thisisatest)
And then print out that sentence with a for loop with a space every 5 characters
(for example thisi sates t).
This is what i have so far
import java.util.Scanner;
public class BlockText {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.println("Give your sentence: ");
String sentence = s.nextLine();
String nospace = sentence.replace(" ", "");
String out = "";
for (int i = 5; i < nospace.length(); i+=5) {
out += nospace.replaceAll( nospace.substring(i) , " ");
}
System.out.println("Sentence without spaces: " + nospace);
System.out.println("This gives: " + out);
}
}
but I have the issue that he repeats certain characters and removes others.
Like you can see underneath after "this gives:"
run:
Give your sentence:
this is a test
Sentence without spaces: thisisatest
This gives: thisi hisisa es
BUILD SUCCESSFUL (total time: 8 seconds)
Can someone help me out? Like I said in the titel, I want to accomplish this with a for loop and without using regex or StringBuilder.

You should really use a StringBuilder here, because appending strings with += in a loop is very inefficient.
Without a string builder, you can do something like this:
private static String addSpacesEvery5(String s) {
String out = "";
for (int i = 0 ; i < s.length() ; i++) {
if (i % 5 == 0 && i != 0) {
out += " "; // this will run once every five iterations, except the first one
}
out += s.charAt(i);
}
return out;
}
Or more efficiently without +=:
private static String addSpacesEvery5(String s) {
// s.length() / 5 is how many spaces we will add
char[] charArray = new char[s.length() + s.length() / 5];
int currentPos = 0;
for (int i = 0 ; i < s.length() ; i++) {
if (i % 5 == 0 && i != 0) {
charArray[currentPos] = ' ';
currentPos++;
}
charArray[currentPos] = s.charAt(i);
currentPos++;
}
return new String(charArray);
}
And then you can use it in your main method like this:
Scanner s = new Scanner(System.in);
System.out.println("Give your sentence: ");
String sentence = s.nextLine();
String nospace = sentence.replace(" ", "");
String out = addSpacesEvery5(nospace);
System.out.println("Sentence without spaces: " + nospace);
System.out.println("This gives: " + out);
With a string builder, the addSpacesEvery5 could be rewritten as:
private static String addSpacesEvery5(String s) {
StringBuilder out = new StringBuilder();
for (int i = 0 ; i < s.length() ; i++) {
if (i % 5 == 0 && i != 0) {
out.append(" ");
}
out.append(s.charAt(i));
}
return out.toString();
}

Here is one relatively simple and faster:
String tmp = new String();
int len = str.length();
int remOdds = len % 5;
int i = 0;
while (i < len - remOdds)
{
tmp = tmp.concat(str.substring(i, i + 5));
i += 5;
if (i < len)
{
tmp += " ";
}
}
while (i < len)
{
tmp += str.charAt(i);
i++;
}
str = tmp;

I want the string pattern aabbcc to be displayed as 2a2b2c

I have somehow got the output with the help of some browsing. But I couldn't understand the logic behind the code. Is there any simple way to achieve this?
public class LetterCount {
public static void main(String[] args)
{
String str = "aabbcccddd";
int[] counts = new int[(int) Character.MAX_VALUE];
// If you are certain you will only have ASCII characters, I would use `new int[256]` instead
for (int i = 0; i < str.length(); i++) {
char charAt = str.charAt(i);
counts[(int) charAt]++;
}
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0)
//System.out.println("Number of " + (char) i + ": " + counts[i]);
System.out.print(""+ counts[i] + (char) i + "");
}
}
}

There are 3 conditions which need to be taken care of:
if (s.charAt(x) != s.charAt(x + 1) && count == 1) ⇒ print the counter and character;
if (s.charAt(x) == s.charAt(x + 1)) ⇒ increase the counter;
if (s.charAt(x) != s.charAt(x + 1) && count >= 2) ⇒ reset to counter 1.
{
int count= 1;
int x;
for (x = 0; x < s.length() - 1; x++) {
if (s.charAt(x) != s.charAt(x + 1) && count == 1) {
System.out.print(s.charAt(x));
System.out.print(count);
}
else if (s.charAt(x)== s.charAt(x + 1)) {
count++;
}
else if (s.charAt(x) != s.charAt(x + 1) && count >= 2) {
System.out.print(s.charAt(x));
System.out.print(count);
count = 1;
}
}
System.out.print(s.charAt(x));
System.out.println(count);
}

The code is really simple.It uses the ASCII value of a character to index into the array that stores the frequency of each character.
The output is simply got by iterating over that array and which character has frequency greater than 1, print it accordingly as you want in the output that is frequency followed by character.
If the input string has same characters consecutive then the solution can be using space of O(1)
For example in your string aabbcc, the same characters are consecutive , so we can take advantage of this fact and count the character frequency and print it at the same time.
for (int i = 0; i < str.length(); i++)
{
int freq = 1;
while((i+1)<str.length()&&str.charAt(i) == str.charAt(i+1))
{++freq;++i}
System.out.print(freq+str.charAt(i));
}

You are trying to keep count of the number of times each character is found. An array is referenced by an index. For example, the ASCII code for the lowercase letter a is the integer 97. Thus the count of the number of times the letter a is seen is in counts[97]. After every element in the counts array has been set, you print out how many have been found.

This should help you understand the basic idea behind how to approach the string compression problem
import java.util.*;
public class LetterCount {
public static void main(String[] args) {
//your input string
String str = "aabbcccddd";
//split your input into characters
String chars[] = str.split("");
//maintain a map to store unique character and its frequency
Map<String, Integer> compressMap = new LinkedHashMap<String, Integer>();
//read every letter in input string
for(String s: chars) {
//java.lang.String.split(String) method includes empty string in your
//split array, so you need to ignore that
if("".equals(s))
continue;
//obtain the previous occurances of the character
Integer count = compressMap.get(s);
//if the character was previously encountered, increment its count
if(count != null)
compressMap.put(s, ++count);
else//otherwise store it as first occurance
compressMap.put(s, 1);
}
//Create a StringBuffer object, to append your input
//StringBuffer is thread safe, so I prefer using it
//you could use StringBuilder if you don't expect your code to run
//in a multithreaded environment
StringBuffer output = new StringBuffer("");
//iterate over every entry in map
for (Map.Entry<String, Integer> entry : compressMap.entrySet()) {
//append the results to output
output.append(entry.getValue()).append(entry.getKey());
}
//print the output on console
System.out.println(output);
}
}

class Solution {
public String toFormat(String input) {
char inChar[] = input.toCharArray();
String output = "";
int i;
for(i=0;i<input.length();i++) {
int count = 1;
while(i+1<input.length() && inChar[i] == inChar[i+1]) {
count+=1;
i+=1;
}
output+=inChar[i]+String.valueOf(count);
}
return output;
}
public static void main(String[] args) {
Solution sol = new Solution();
String input = "aaabbbbcc";
System.out.println("Formatted String is: " + sol.toFormat(input));
}
}

def encode(Test_string):
count = 0
Result = ""
for i in range(len(Test_string)):
if (i+1) < len(Test_string) and (Test_string[i] == Test_string[i+1]):
count += 1
else:
Result += str((count+1))+Test_string[i]
count = 0
return Result
print(encode("ABBBBCCCCCCCCAB"))

If you want to get the correct count considering the string is not in alphabetical order. Sort the string
public class SquareStrings {
public static void main(String[] args) {
SquareStrings squareStrings = new SquareStrings();
String str = "abbccddddbd";
System.out.println(squareStrings.manipulate(str));
}
private String manipulate(String str1) {
//convert to charArray
char[] charArray = str1.toCharArray();
Arrays.sort(charArray);
String str = new String(charArray);
StringBuilder stbuBuilder = new StringBuilder("");
int length = str.length();
String temp = "";
if (length > 1) {
for (int i = 0; i < length; i++) {
int freq = 1;
while (((i + 1) < length) && (str.charAt(i) == str.charAt(i + 1))) {
++freq;
temp = str.charAt(i) + "" + freq;
++i;
}
stbuBuilder.append(temp);
}
} else {
return str + "" + 1;
}
return stbuBuilder.toString();
}
}

Kotlin:
fun compressString(input: String): String {
if (input.isEmpty()){
return ""
}
var result = ""
var count = 1
var char1 = input[0]
for (i in 1 until input.length) {
val char2 = input[i]
if (char1 == char2) {
count++
} else {
if (count != 1) {
result += "$count$char1"
count = 1
} else {
result += "$char1"
}
char1 = char2
}
}
result += if (count != 1) {
"$count$char1"
} else {
"$char1"
}
return result
}

Is substring or replace faster to remove the last character in a string?

I have lots of words that need processing and all of them end with .,
Which option has the best time complexity?
word.substring(0, word.length()-1)
word.replaceAll("\\."，"")
word.replace(".", "")
Or, is there a better way?

A simple test (with JDK1.7.0_75) can show the difference:
private static final int LENGTH = 10000;
public static void main(String[] args) {
String[] strings = new String[LENGTH];
for (int i = 0; i < LENGTH; i++) {
strings[i] = "abc" + i + ".";
}
long start = System.currentTimeMillis();
for (int i = 0; i < strings.length; i++) {
String word = strings[i];
word = word.substring(0, word.length()-1);
}
long end = System.currentTimeMillis();
System.out.println("substring: " + (end - start) + " millisec.");
start = System.currentTimeMillis();
for (int i = 0; i < strings.length; i++) {
String word = strings[i];
word = word.replaceAll(".", "");
}
end = System.currentTimeMillis();
System.out.println("replaceAll: " + (end - start) + " millisec.");
start = System.currentTimeMillis();
for (int i = 0; i < strings.length; i++) {
String word = strings[i];
word = word.replace(".", "");
}
end = System.currentTimeMillis();
System.out.println("replace: " + (end - start) + " millisec.");
}
Output:
substring: 0 millisec.
replaceAll: 78 millisec.
replace: 16 millisec.
As expected, the substring is fastest because:
It avoids compiling a regular expression.
It is constant-time: creating a new String based on the specified begin and end indices.

Wrap the string after a number of characters word-wise in Java

I have this code:
String s = "A very long string containing " +
"many many words and characters. " +
"Newlines will be entered at spaces.";
StringBuilder sb = new StringBuilder(s);
int i = 0;
while ((i = sb.indexOf(" ", i + 20)) != -1) {
sb.replace(i, i + 1, "\n");
}
System.out.println(sb.toString());
The output of the code is:
A very long string containing
many many words and
characters. Newlines
will be entered at spaces.
The above code is wrapping the string after the next space of every 30 characters, but I need to wrap the string after the previous space of every 30 characters, like for the first line it will be:
A very long string
And the 2nd line will be
containing many
Please give some proper solution.

You can use Apache-common's WordUtils.wrap().

Use lastIndexOf instead of indexOf, e.g.
StringBuilder sb = new StringBuilder(s);
int i = 0;
while (i + 20 < sb.length() && (i = sb.lastIndexOf(" ", i + 20)) != -1) {
sb.replace(i, i + 1, "\n");
}
System.out.println(sb.toString());
This will produce the following output:
A very long string
containing many
many words and
characters.
Newlines will be
entered at spaces.

You can try the following:
public static String wrapString(String s, String deliminator, int length) {
String result = "";
int lastdelimPos = 0;
for (String token : s.split(" ", -1)) {
if (result.length() - lastdelimPos + token.length() > length) {
result = result + deliminator + token;
lastdelimPos = result.length() + 1;
}
else {
result += (result.isEmpty() ? "" : " ") + token;
}
}
return result;
}
call as wrapString("asd xyz afz","\n",5)

I know it's an old question, but . . . Based on another answer I found here, but can't remember the posters name. Kuddos to him/her for pointing me in the right direction.
public String truncate(final String content, final int lastIndex) {
String result = "";
String retResult = "";
//Check for empty so we don't throw null pointer exception
if (!TextUtils.isEmpty(content)) {
result = content.substring(0, lastIndex);
if (content.charAt(lastIndex) != ' ') {
//Try the split, but catch OutOfBounds in case string is an
//uninterrupted string with no spaces
try {
result = result.substring(0, result.lastIndexOf(" "));
} catch (StringIndexOutOfBoundsException e) {
//if no spaces, force a break
result = content.substring(0, lastIndex);
}
//See if we need to repeat the process again
if (content.length() - result.length() > lastIndex) {
retResult = truncate(content.substring(result.length(), content.length()), lastIndex);
} else {
return result.concat("\n").concat(content.substring(result.length(), content.length()));
}
}
//Return the result concatenating a newline character on the end
return result.concat("\n").concat(retResult);;
//May need to use this depending on your app
//return result.concat("\r\n").concat(retResult);;
} else {
return content;
}
}

public static void main(String args[]) {
String s1="This is my world. This has to be broken.";
StringBuffer buffer=new StringBuffer();
int length=s1.length();
int thrshld=5; //this valueis threshold , which you can use
int a=length/thrshld;
if (a<=1) {
System.out.println(s1);
}else{
String split[]=s1.split(" ");
for (int j = 0; j < split.length; j++) {
buffer.append(split[j]+" ");
if (buffer.length()>=thrshld) {
int lastindex=buffer.lastIndexOf(" ");
if (lastindex<buffer.length()) {
buffer.subSequence(lastindex, buffer.length()-1);
System.out.println(buffer.toString());
buffer=null;
buffer=new StringBuffer();
}
}
}
}
}
this can be one way to achieve

"\n" makes a wordwrap.
String s = "A very long string containing \n" +
"many many words and characters. \n" +
"Newlines will be entered at spaces.";
this will solve your problem