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Replacing the number with the number of spaces

I have a String:
String example = "AA5DD2EE3MM";
I want to replace the number with the number of spaces. Example:
String example = "AA DD EE MM"
If the String would be
String anotherExample = "a3ee"
It should turn into:
String anotherExample = "a ee"
I want to do it for any string. Not only for the examples above.

Split your input at digit and non digit chars as a stream, map digits to the corsponding number of spaces using String.repeat, collect to string using Collectors.joining():
String input = "AA5DD2EE3MM";
String regex = "(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)";
String result = Pattern.compile(regex)
.splitAsStream(input)
.map(s -> s.matches("\\d+") ? " ".repeat(Integer.parseInt(s)) : s)
.collect(Collectors.joining());

You could also use this approach, which is simpler but also far less elegant:
String example = "a4aa";
String newString = "";
for (int i = 0; i < example.length(); i++) {
if (Character.isDigit(example.charAt(i))) {
for (int a = 0; a < Character.getNumericValue(example.charAt(i)); a++) {
newString += " ";
}
} else {
newString += example.charAt(i);
}
}
System.out.println(newString);

Using a pattern matcher approach:
String input = "AA5DD2EE3MM";
Pattern pattern = Pattern.compile("\\d+");
Matcher m = pattern.matcher(input);
StringBuffer buffer = new StringBuffer();
while (m.find()) {
m.appendReplacement(buffer,new String(new char[Integer.valueOf(m.group())]).replace("\0", " "));
}
m.appendTail(buffer);
System.out.println(buffer.toString()); // AA DD EE MM
The idea here is to iterate the string, pausing at each digit match. We replace each digit with space replicated the same number of times as the digit.

public static String replaceDigitsWithSpaces(String input) {
String result = "";
int len = input.length(), i =0;
while( i < len) {
if(Character.isLetter(input.charAt(i))) {
result += input.charAt(i);
}else if(Character.isDigit(input.charAt(i))) {
//generate number upto characters
int k = 0, j = i;
String temp = "";
while(j < len) {
if(Character.isDigit(input.charAt(j))) {
temp += input.charAt(j);
j++;
}else {
break;
}
}
k = Integer.parseInt(temp);
while(k != 0) {
result+= " ";
k--;
}
i = j;
continue;
}
i++;
}
return result;
}
input:: "AA23BB1C11C8"<br>
output:: AA BB C C .

StringBuilder is more efficient for concatenation:
public static String spaceIt(String s) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
for (int j = 0; j < Character.digit(c, 10); j++) {
sb.append(' ');
}
} else {
sb.append(c);
}
}
return sb.toString();
}

How do i Reverse Capitalize And Separate Words By Line Using String In Java?

The input is: i love cake
The output needs to be: Cake Love I
But the actual result is: I Love Cake
What I've got:
import java.util.regex.Pattern;
public class yellow {
static String reverseWords(String str){
Pattern pattern = Pattern.compile("\\s");
String[] temp = pattern.split(str);
String result = "";
for (int i = 0; i < temp.length; i++) {
if (i == temp.length - 1)
result = temp[i] + result;
else
result = " " + temp[i] + result;
}
return result;
}
public static void main(String[] args){
String source = "i love cake";
StringBuffer res = new StringBuffer();
String[] strArr = source.split(" ");
for (String str : strArr) {
char[] stringArray = str.trim().toCharArray();
stringArray[0] = Character.toUpperCase(stringArray[0]);
str = new String(stringArray);
res.append(str).append(" ");
}
System.out.print(res.toString());
}
}
What am I doing wrong?

for (String str : strArr) {
}
This loops forward. What you want is to loop backwards, or to place elements into the string backwards. I recommend you loop backwards and print as you go:
for (int i = strArr.length - 1; i >= 0; i--) {
char[] stringArray = strArr[i].trim().toCharArray();
stringArray[0] = Character.toUpperCase(stringArray[0]);
System.out.println(new String(stringArray));
}
Or, you could use that convenient reverseWords method that you never use anywhere... though looping backwards is faster. Probably.

[EDITED]
Call this for each line with string s, then print a line break (If you have multiple sentences & expect them in their own lines).
void reverseCamel(String s){
String[] ar = s.split("\\s+");
for(int i = ar.length - 1;i>=0;i--){
ar[i][0] = Character.toUpperCase(ar[i][0]);
System.out.print(ar[i] + " ");
}
}

Here is what i did.
public class Main {
public static void main(String[] args) {
reverse("I Love Cake");
}
public static void reverse( String string){
String[] word =string.split(" "); // split by spaces
int i = word.length-1;
while (i>=0){
// System.out.print(word[i].toUpperCase()+" ");//if you want in upper case
System.out.print(word[i]+" ");
i--;
}
}
}

First of all you have to reverse the String.
String[] words = source.split("\\s");
String reversedString = "";
for(int i = words.length -1; i>=0; i--){
reversedString += words[i] + " ";
}
Then, you know that the ASCII code of 'a' character is 97, 'A' is 65. To convert from lower case to capital you substract 32. All capitals are between 65 and 92. All small letters are between 97 and 124.
You want to capitalize only letters at the beginning of a word (preceded by a space or first letter).
String capitalCase = "";
for (int i = 0; i < reversedString.length(); i++) {
char c = reversedString.charAt(i);
if (c >= 97 && c <= 124) {
if (i == 0) c -= 32;
else if ((reversedString.charAt(i - 1) + "").equals(" ")) c -= 32;
}
capitalCase += c;
}
And here you go now System.out.println(capitalCase);
Overall, you will have the following code:
import java.util.Scanner;
public class yellow {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.println("Enter a string:");
String source = s.nextLine();
String[] words = source.split("\\s");
String reversedString = "";
for (int i = words.length - 1; i >= 0; i--) {
reversedString += words[i] + " ";
}
String capitalCase = "";
for (int i = 0; i < reversedString.length(); i++) {
char c = reversedString.charAt(i);
if (c >= 97 && c <= 124) {
if (i == 0) c -= 32;
else if ((reversedString.charAt(i - 1) + "").equals(" ")) c -= 32;
}
capitalCase += c;
}
System.out.println(capitalCase);
}
}
Output:
Enter a string:
i love cake
Cake Love I

Java 8 * Apache Commons Lang
public static String reverseWordsInString(String str) {
List<String> words = Pattern.compile("\\s+").splitAsStream(str)
.map(StringUtils::capitalize)
.collect(LinkedList::new, LinkedList::addFirst, (a, b) -> a.addAll(0, b));
return words.stream().collect(Collectors.joining(StringUtils.SPACE));
}
Java 8
public static String reverseWordsInString(String str) {
List<String> words = Pattern.compile("\\s+").splitAsStream(str)
.map(word -> Character.toUpperCase(word.charAt(0)) + word.substring(1).toLowerCase())
.collect(LinkedList::new, LinkedList::addFirst, (a, b) -> a.addAll(0, b));
return words.stream().collect(Collectors.joining(" "));
}

Reverse each word of a string without altering their position(Without using inbuilt fucntion)

I want to reverse each individual word of a String in Java in different different sitautions (not the entire string, just each individual word).
Example1: if input String is "This is a test" then the output should be "sihT si a tset".
Example2: if input String is "This is a test" then the output should be "sihT si a tset".
[When there is more than one space between some words]
Please also provide algorithm for understanding
What I have tried so far
class reverseAString
{
public static void main(String args[])
{
String str= "Test the product";
String strArr[]= str.split(" ");
for(int i=0;i<=strArr.length-1;i++)
{
for(int j=strArr[i].length()-1; j>=0;j--)
{
System.out.print(strArr[i].charAt(j));
}
System.out.printf(" ");
}
}
}
---------
public String reverseWordByWord(String str){
int strLeng = str.length()-1;
String reverse = "", temp = "";
for(int i = 0; i <= strLeng; i++){
temp += str.charAt(i);
if((str.charAt(i) == ' ') || (i == strLeng)){
for(int j = temp.length()-1; j >= 0; j--){
reverse += temp.charAt(j);
if((j == 0) && (i != strLeng))
reverse += " ";
}
temp = "";
}
}
return reverse;
}

This approach is much more succinct.
class StringRev{
public static void main(String args[]){
String str[] = "Test the product".split(" ");
String finalStr="";
for(int i = str.length-1; i>= 0 ;i--){
finalStr += str[i]+" ";
}
System.out.println(finalStr);
}
}

public static void main(String arg[]) {
System.out.println(reverseWords("This is a test"));
}
// function to reverse each word
public static String reverseWords(String input) {
String result = null;
StringBuffer strBuffer = new StringBuffer();
String split[] = input.split(" ");
for (String temp : split) {
if (temp != " ") {
strBuffer.append(reverse(temp));
strBuffer.append(" ");
} else {
strBuffer.append(" ");
}
}
result = strBuffer.toString();
return result;
}
// function to reverse individual word
public static String reverse(String input) {
StringBuffer strBuffer = new StringBuffer();
char[] charArray = input.toCharArray();
for (int j = charArray.length - 1; j >= 0; j--) {
strBuffer.append(charArray[j]);
}
return strBuffer.toString();
}
Split complete string on basis of space
Then convert each string to character array and reverse
individual word.

It's been a long time since I've touch Java, but here's the algorithm.
String stringToReverse = "This is a Test";
String finalString = "";
Stack word = new Stack;
for(int i = 0; i < stringToReverse.length; i++){
char chr = stringToReverse[i];
if(chr == " " || i == (stringToReverse.length - 1)){
while(word.count > 0){
finalString += word.pop();
}
finalString += chr;
}else{
word.push(chr);
}
}

Remove repeated characters in a string

I need to write a static method that takes a String as a parameter and returns a new String obtained by replacing every instance of repeated adjacent letters with a single instance of that letter without using regular expressions. For example if I enter "maaaakkee" as a String, it returns "make".
I already tried the following code, but it doesn't seem to display the last character.
Here's my code:
import java.util.Scanner;
public class undouble {
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.println("enter String: ");
String str = console.nextLine();
System.out.println(removeSpaces(str));
}
public static String removeSpaces(String str){
String ourString="";
int j = 0;
for (int i=0; i<str.length()-1 ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
return ourString;
}
}

You could use regular expressions for that.
For instance:
String input = "ddooooonnneeeeee";
System.out.println(input.replaceAll("(.)\\1{1,}", "$1"));
Output:
done
Pattern explanation:
"(.)\\1{1,}" means any character (added to group 1) followed by itself at least once
"$1" references contents of group 1

maybe:
for (int i=1; i<str.length() ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}

The problem is with your condition. You say compare i and i+1 in each iteration and in last iteration you have both i and j pointing to same location so it will never print the last character. Try this unleass you want to use regex to achive this:
EDIT:
public void removeSpaces(String str){
String ourString="";
for (int i=0; i<str.length()-1 ; i++){
if(i==0){
ourString = ""+str.charAt(i);
}else{
if(str.charAt(i-1) != str.charAt(i)){
ourString = ourString +str.charAt(i);
}
}
}
System.out.println(ourString);
}

if you cannot use replace or replaceAll, here is an alternative. O(2n), O(N) for stockage and O(N) for creating the string. It removes all repeated chars in the string put them in a stringbuilder.
input : abcdef , output : abcdef
input : aabbcdeef, output : cdf
private static String remove_repeated_char(String str)
{
StringBuilder result = new StringBuilder();
HashMap<Character, Integer> items = new HashMap<>();
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == null)
items.put(current, 1);
else
items.put(current, ocurrence + 1);
}
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == 1)
result.append(current);
}
return result.toString();
}

import java.util.*;
public class string2 {
public static void main(String[] args) {
//removes repeat character from array
Scanner sc=new Scanner(System.in);
StringBuffer sf=new StringBuffer();
System.out.println("enter a string");
sf.append(sc.nextLine());
System.out.println("string="+sf);
int i=0;
while( i<sf.length())
{
int j=1+i;
while(j<sf.length())
{
if(sf.charAt(i)==sf.charAt(j))
{
sf.deleteCharAt(j);
}
else
{
j=j+1;
}
}
i=i+1;
}
System.out.println("string="+sf);
}
}

Input AABBBccDDD, Output BD
Input ABBCDDA, Outout C
private String reducedString(String s){
char[] arr = s.toCharArray();
String newString = "";
Map<Character,Integer> map = new HashMap<Character,Integer>();
map.put(arr[0],1);
for(int index=1;index<s.length();index++)
{
Character key = arr[index];
int value;
if(map.get(key) ==null)
{
value =0;
}
else
{
value = map.get(key);
}
value = value+1;
map.put(key,value);
}
Set<Character> keyset = map.keySet();
for(Character c: keyset)
{
int value = map.get(c);
if(value%2 !=0)
{
newString+=c;
}
}
newString = newString.equals("")?"Empty String":newString;
return newString;
}

public class RemoveDuplicateCharecterInString {
static String input = new String("abbbbbbbbbbbbbbbbccccd");
static String output = "";
public static void main(String[] args)
{
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
char temp = input.charAt(i);
boolean check = false;
for (int j = 0; j < output.length(); j++) {
if (output.charAt(j) == input.charAt(i)) {
check = true;
}
}
if (!check) {
output = output + input.charAt(i);
}
}
System.out.println(" " + output);
}
}
Answer : abcd

public class RepeatedChar {
public static void main(String[] args) {
String rS = "maaaakkee";
String outCome= rS.charAt(0)+"";
int count =0;
char [] cA =rS.toCharArray();
for(int i =0; i+1<cA.length; ++i) {
if(rS.charAt(i) != rS.charAt(i+1)) {
outCome += rS.charAt(i+1);
}
}
System.out.println(outCome);
}
}

TO WRITE JAVA PROGRAM TO REMOVE REPEATED CHARACTERS:
package replace;
public class removingrepeatedcharacters
{
public static void main(String...args){
int i,j=0,count=0;
String str="noordeen";
String str2="noordeen";
char[] ch=str.toCharArray();
for(i=0;i<=5;i++)
{
count=0;
for(j=0;j<str2.length();j++)
{
if(ch[i]==str2.charAt(j))
{
count++;
System.out.println("at the index "+j +"position "+ch[i]+ "+ count is"+count);
if(count>=2){
str=str2;
str2=str.replaceFirst(Character.toString(ch[j]),Character.toString(' '));
}
System.out.println("after replacing " +str2);
}
}
}
}
}

String outstr = "";
String outstring = "";
for(int i = 0; i < str.length() - 1; i++) {
if(str.charAt(i) != str.charAt(i + 1)) {
outstr = outstr + str.charAt(i);
}
outstring = outstr + str.charAt(i);
}
System.out.println(outstring);

public static void remove_duplicates(String str){
String outstr="";
String outstring="";
for(int i=0;i<str.length()-1;i++) {
if(str.charAt(i)!=str.charAt(i+1)) {
outstr=outstr+str.charAt(i);
}
outstring=outstr+str.charAt(i);
}
System.out.println(outstring);
}

More fun with java 7:
System.out.println("11223344445555".replaceAll("(?<nums>.+)\\k<nums>+","${nums}"));
No more cryptic numbers in regexes.

public static String removeDuplicates(String str) {
String str2 = "" + str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i - 1) == str.charAt(i) && i != 0) {
continue;
}
str2 = str2 + str.charAt(i);
}
return str2;
}

Write a method to replace all spaces in a string with '%20'

I have a question about a programming problem from the book Cracking The Code Interview by Gayl Laakmann McDowell, 5th Edition.
The problem states: Write a method to replace all spaces in a string with '%20'. Assume string has sufficient space at end of string to hold additional characters, and that you're given a true length of a string. I used the books code, implementing the solution in Java using a character array (given the fact that Java Strings are immutable):
public class Test {
public void replaceSpaces(char[] str, int length) {
int spaceCount = 0, newLength = 0, i = 0;
for(i = 0; i < length; i++) {
if (str[i] == ' ')
spaceCount++;
}
newLength = length + (spaceCount * 2);
str[newLength] = '\0';
for(i = length - 1; i >= 0; i--) {
if (str[i] == ' ') {
str[newLength - 1] = '0';
str[newLength - 2] = '2';
str[newLength - 3] = '%';
newLength = newLength - 3;
}
else {
str[newLength - 1] = str[i];
newLength = newLength - 1;
}
}
System.out.println(str);
}
public static void main(String[] args) {
Test tst = new Test();
char[] ch = {'t', 'h', 'e', ' ', 'd', 'o', 'g', ' ', ' ', ' ', ' ', ' ', ' '};
int length = 6;
tst.replaceSpaces(ch, length);
}
}
The output I am getting from the replaceSpaces() call is: the%20do which is cutting of the last character of the original array. I have been scratching my head over this, can anyone explain to me why the algorithm is doing this?

public String replace(String str) {
String[] words = str.split(" ");
StringBuilder sentence = new StringBuilder(words[0]);
for (int i = 1; i < words.length; ++i) {
sentence.append("%20");
sentence.append(words[i]);
}
return sentence.toString();
}

You are passing the length as 6, which is causing this. Pass length as 7 including space.
Other wise
for(i = length - 1; i >= 0; i--) {
will not consider last char.

With these two changes I got the output: the%20dog
1) Change space count to 2 [since length already includes 1 of the 3 characters you need for %20]
newLength = length + (spaceCount * 2);
2) Loop should start on length
for(i = length; i >= 0; i--) {

Here is my solution. I check for the ascii code 32 then put a %20 instead of it.Time complexity of this solution is O(N)
public String replace(String s) {
char arr[] = s.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == 32)
sb.append("%20");
else
sb.append(arr[i]);
}
return sb.toString();
}

This is my code for this question. Seems like working for me. If you're interested, please have a look. It's written in JAVA
public class ReplaceSpaceInString {
private static char[] replaceSpaceInString(char[] str, int length) {
int spaceCounter = 0;
//First lets calculate number of spaces
for (int i = 0; i < length; i++) {
if (str[i] == ' ')
spaceCounter++;
}
//calculate new size
int newLength = length + 2*spaceCounter;
char[] newArray = new char[newLength+1];
newArray[newLength] = '\0';
int newArrayPosition = 0;
for (int i = 0; i < length; i++) {
if (str[i] == ' ') {
newArray[newArrayPosition] = '%';
newArray[newArrayPosition+1] = '2';
newArray[newArrayPosition+2] = '0';
newArrayPosition = newArrayPosition + 3;
}
else {
newArray[newArrayPosition] = str[i];
newArrayPosition++;
}
}
return newArray;
}
public static void main(String[] args) {
char[] array = {'a','b','c','d',' ','e','f','g',' ','h',' ','j'};
System.out.println(replaceSpaceInString(array, array.length));
}
}

You can also use substring method and the ascii for space (32).
public String replaceSpaceInString(String s){
int i;
for (i=0;i<s.length();i++){
System.out.println("i is "+i);
if (s.charAt(i)==(int)32){
s=s.substring(0, i)+"%20"+s.substring(i+1, s.length());
i=i+2;
}
}
return s;
}
To test:
System.out.println(cc.replaceSpaceInString("mon day "));
Output:
mon%20day%20

You could just do this.
No need to calculate the length or whatever.
Strings are immutable anyways.
import java.util.*;
public class ReplaceString {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
String str=in.nextLine();
String n="";
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)==' ')
n=n+"%20";
else
n=n+str.charAt(i);
}
str=n;
System.out.println(str);
}
}

void Rep_Str(char *str)
{
int j=0,count=0;
int stlen = strlen(str);
for (j = 0; j < stlen; j++)
{
if (str[j]==' ')
{
count++;
}
}
int newlength = stlen+(count*2);
str[newlength--]='\0';
for (j = stlen-1; j >=0 ; j--)
{
if (str[j]==' ')
{
str[newlength--]='0';
str[newlength--]='2';
str[newlength--]='%';
}
else
{
str[newlength--]=str[j];
}
}
}
This code works :)

We can use a regular expression to solve this problem. For example:
public String replaceStringWithSpace(String str){
return str.replaceAll("[\\s]", "%20");
}

This works correctly. However, using a StringBuffer object increases space complexity.
Scanner scn = new Scanner(System.in);
String str = scn.nextLine();
StringBuffer sb = new StringBuffer(str.trim());
for(int i = 0;i<sb.length();i++){
if(32 == (int)sb.charAt(i)){
sb.replace(i,i+1, "%20");
}
}

public static String replaceAllSpaces(String s) {
char[] c = s.toCharArray();
int spaceCount = 0;
int trueLen = s.length();
int index = 0;
for (int i = 0; i < trueLen; i++) {
if (c[i] == ' ') {
spaceCount++;
}
}
index = trueLen + spaceCount * 2;
char[] n = new char[index];
for (int i = trueLen - 1; i >= 0; i--) {
if (c[i] == ' ') {
n[index - 1] = '0';
n[index - 2] = '2';
n[index - 3] = '%';
index = index - 3;
} else {
n[index - 1] = c[i];
index--;
}
}
String x = new String(n);
return x;
}

Another way of doing this.
I am assuming the trailing spaces don't need to be converted to %20 and that the trailing spaces provide enough room for %20s to be stuffed in between
public class Main {
public static void main(String[] args) {
String str = "a sd fghj ";
System.out.println(replacement(str));//a%20sd%20fghj
}
private static String replacement(String str) {
char[] chars = str.toCharArray();
int posOfLastChar = 0;
for (int i = 0; i < chars.length; i++) {
if (chars[i] != ' ') {
posOfLastChar = i;
}
}
int newCharPosition = chars.length - 1;
//Start moving the characters to th end of the array. Replace spaces by %20
for (int i = posOfLastChar; i >= 0; i--) {
if (chars[i] == ' ') {
chars[newCharPosition] = '0';
chars[--newCharPosition] = '2';
chars[--newCharPosition] = '%';
} else {
chars[newCharPosition] = chars[i];
}
newCharPosition--;
}
return String.valueOf(chars);
}
}

public class ReplaceChar{
public static void main(String []args){
String s = "ab c de ";
System.out.println(replaceChar(s));
}
public static String replaceChar(String s){
boolean found = false;
StringBuilder res = new StringBuilder(50);
String str = rev(s);
for(int i = 0; i <str.length(); i++){
if (str.charAt(i) != ' ') { found = true; }
if (str.charAt(i) == ' '&& found == true) { res.append("%02"); }
else{ res.append(str.charAt(i)); }
}
return rev(res.toString());
}
// Function to reverse a string
public static String rev(String s){
StringBuilder result = new StringBuilder(50);
for(int i = s.length()-1; i>=0; i-- ){
result.append(s.charAt(i));
}
return result.toString();
}}
A simple approach:
Reverse the given string and check where the first character appears.
Using string builder to append "02%" for spaces - since the string is reversed.
Finally reverse the string once again.
Note: We reverse the string so as to prevent an addition of "%20" to the trailing spaces.
Hope that helps!

The question in the book mentions that the replacement should be in place so it is not possible to assign extra arrays, it should use constant space. You should also take into account many edge cases, this is my solution:
public class ReplaceSpaces {
public static void main(String[] args) {
if ( args.length == 0 ) {
throw new IllegalArgumentException("No string");
}
String s = args[0];
char[] characters = s.toCharArray();
replaceSpaces(characters);
System.out.println(characters);
}
static void replaceSpaces(char[] s) {
int i = s.length-1;
//Skipping all spaces in the end until setting `i` to non-space character
while( i >= 0 && s[i] == ' ' ) { i--; }
/* Used later to check there is enough extra space in the end */
int extraSpaceLength = s.length - i - 1;
/*
Used when moving the words right,
instead of finding the first non-space character again
*/
int lastNonSpaceCharacter = i;
/*
Hold the number of spaces in the actual string boundaries
*/
int numSpaces = 0;
/*
Counting num spaces beside the extra spaces
*/
while( i >= 0 ) {
if ( s[i] == ' ' ) { numSpaces++; }
i--;
}
if ( numSpaces == 0 ) {
return;
}
/*
Throw exception if there is not enough space
*/
if ( extraSpaceLength < numSpaces*2 ) {
throw new IllegalArgumentException("Not enough extra space");
}
/*
Now we need to move each word right in order to have space for the
ascii representation
*/
int wordEnd = lastNonSpaceCharacter;
int wordsCounter = 0;
int j = wordEnd - 1;
while( j >= 0 ) {
if ( s[j] == ' ' ){
moveWordRight(s, j+1, wordEnd, (numSpaces-wordsCounter)*2);
wordsCounter++;
wordEnd = j;
}
j--;
}
replaceSpacesWithAscii(s, lastNonSpaceCharacter + numSpaces * 2);
}
/*
Replaces each 3 sequential spaces with %20
char[] s - original character array
int maxIndex - used to tell the method what is the last index it should
try to replace, after that is is all extra spaces not required
*/
static void replaceSpacesWithAscii(char[] s, int maxIndex) {
int i = 0;
while ( i <= maxIndex ) {
if ( s[i] == ' ' ) {
s[i] = '%';
s[i+1] = '2';
s[i+2] = '0';
i+=2;
}
i++;
}
}
/*
Move each word in the characters array by x moves
char[] s - original character array
int startIndex - word first character index
int endIndex - word last character index
int moves - number of moves to the right
*/
static void moveWordRight(char[] s, int startIndex, int endIndex, int moves) {
for(int i=endIndex; i>=startIndex; i--) {
s[i+moves] = s[i];
s[i] = ' ';
}
}
}

Any reason not to use 'replace' method?
public String replaceSpaces(String s){
return s.replace(" ", "%20");}

Hm... I am wondering about this problem as well. Considering what I have seen in here. The book solution does not fit Java because it uses in-place
char []
modification and solutions in here that use char [] or return void don't fit as well because Java does not use pointers.
So I was thinking, the obvious solution would be
private static String encodeSpace(String string) {
return string.replcaceAll(" ", "%20");
}
but this is probably not what your interviewer would like to see :)
// make a function that actually does something
private static String encodeSpace(String string) {
//create a new String
String result = new String();
// replacement
final String encodeSpace = "%20";
for(char c : string.toCharArray()) {
if(c == ' ') result+=encodeString;
else result+=c;
}
return result;
}
this looks fine I thought, and you only need one pass through the string, so the complexity should be O(n), right? Wrong! The problem is in
result += c;
which is the same as
result = result + c;
which actually copies a string and creates a copy of it. In java strings are represented as
private final char value[];
which makes them immutable (for more info I would check java.lang.String and how it works). This fact will bump up the complexity of this algorithm to O(N^2) and a sneaky recruiter can use this fact to fail you :P Thus, I came in with a new low-level solution which you will never use in practice, but which is good in theory :)
private static String encodeSpace(String string) {
final char [] original = string != null? string.toCharArray() : new char[0];
// ASCII encoding
final char mod = 37, two = 50, zero = 48, space = 32;
int spaces = 0, index = 0;
// count spaces
for(char c : original) if(c == space) ++spaces;
// if no spaces - we are done
if(spaces == 0) return string;
// make a new char array (each space now takes +2 spots)
char [] result = new char[string.length()+(2*spaces)];
for(char c : original) {
if(c == space) {
result[index] = mod;
result[++index] = two;
result[++index] = zero;
}
else result[index] = c;
++index;
}
return new String(result);
}

But I wonder what is wrong with following code:
private static String urlify(String originalString) {
String newString = "";
if (originalString.contains(" ")) {
newString = originalString.replace(" ", "%20");
}
return newString;
}

Question : Urlify the spaces with %20
Solution 1 :
public class Solution9 {
public static void main(String[] args) {
String a = "Gini Gina Protijayi";
System.out.println( urlencode(a));
}//main
public static String urlencode(String str) {
str = str.trim();
System.out.println("trim =>" + str);
if (!str.contains(" ")) {
return str;
}
char[] ca = str.toCharArray();
int spaces = 0;
for (char c : ca) {
if (c == ' ') {
spaces++;
}
}
char[] newca = new char[ca.length + 2 * spaces];
// a pointer x has been added
for (int i = 0, x = 0; i < ca.length; i++) {
char c = ca[i];
if (c == ' ') {
newca[x] = '%';
newca[x + 1] = '2';
newca[x + 2] = '0';
x += 3;
} else {
newca[x] = c;
x++;
}
}
return new String(newca);
}
}//urlify

My solution using StringBuilder with time complexity O(n)
public static String url(String string, int length) {
char[] arrays = string.toCharArray();
StringBuilder builder = new StringBuilder(length);
for (int i = 0; i < length; i++) {
if (arrays[i] == ' ') {
builder.append("%20");
} else {
builder.append(arrays[i]);
}
}
return builder.toString();
}
Test case :
#Test
public void testUrl() {
assertEquals("Mr%20John%20Smith", URLify.url("Mr John Smith ", 13));
}

Can you use StringBuilder?
public String replaceSpace(String s)
{
StringBuilder answer = new StringBuilder();
for(int i = 0; i<s.length(); i++)
{
if(s.CharAt(i) == ' ')
{
answer.append("%20");
}
else
{
answer.append(s.CharAt(i));
}
}
return answer.toString();
}

I am also looking at that question in the book. I believe we can just use String.trim() and String.replaceAll(" ", "%20) here

I updated the solution here. http://rextester.com/CWAPCV11970
If we are creating new array and not in-place trasition, then why do we need to walk backwards?
I modified the real solution slightly to walk forward to create target Url-encoded-string.
Time complexity:
O(n) - Walking original string
O(1) - Creating target string incrementally
where 'n' is number of chars in original string
Space complexity:
O(n + m) - Duplicate space to store escaped spaces and string.
where 'n' is number of chars in original string and 'm' is length of escaped spaces
public static string ReplaceAll(string originalString, char findWhat, string replaceWith)
{
var newString = string.Empty;
foreach(var character in originalString)
newString += findWhat == character? replaceWith : character + string.Empty;
return newString;
}

class Program
{
static void Main(string[] args)
{
string name = "Stack Over Flow ";
StringBuilder stringBuilder = new StringBuilder();
char[] array = name.ToCharArray(); ;
for(int i = 0; i < name.Length; i++)
{
if (array[i] == ' ')
{
stringBuilder.Append("%20");
}
else
stringBuilder.Append(array[i]);
}
Console.WriteLine(stringBuilder);
Console.ReadLine();
}
}

public class Sol {
public static void main(String[] args) {
String[] str = "Write a method to replace all spaces in a string with".split(" ");
StringBuffer sb = new StringBuffer();
int count = 0;
for(String s : str){
sb.append(s);
if(str.length-1 != count)
sb.append("%20");
++count;
}
System.out.println(sb.toString());
}
}

public class Test {
public static void replace(String str) {
String[] words = str.split(" ");
StringBuilder sentence = new StringBuilder(words[0]);
for (int i = 1; i < words.length; i++) {
sentence.append("%20");
sentence.append(words[i]);
}
sentence.append("%20");
System.out.println(sentence.toString());
}
public static void main(String[] args) {
replace("Hello World "); **<- Hello<3spaces>World<1space>**
}
}
O/P:: Hello%20%20%20World%20

Remember that you only to want replace ' ' with '%20' when the latter is not a leading or trailing space. Several answers above do not account for this. For what it's worth, I get "index out of bounds error" when I run Laakmann's solution example.
Here's my own solution, which runs O(n) and is implemented in C#:
public static string URLreplace(string str, int n)
{
var len = str.Length;
if (len == n)
return str;
var sb = new StringBuilder();
var i = 0;
while (i < len)
{
char c = str[i];
if (c == ' ')
{
while (i < len && str[i] == ' ')
{
i++; //skip ahead
}
}
else
{
if (sb.Length > 0 && str[i - 1] == ' ')
sb.Append("%20" + c);
else
sb.Append(c);
i++;
}
}
return sb.ToString();
}
Test:
//Arrange
private string _str = " Mr John Smith ";
private string _result = "Mr%20John%20Smith";
private int _int = 13;
[TestMethod]
public void URLified()
{
//Act
var cleaned = URLify.URLreplace(_str, _int);
//Assert
Assert.IsTrue(cleaned == _result);
}

One line code
System.out.println(s.trim().replace(" ","%20"));

// while passing the input make sure you use the .toCharArray method becuase strings are immutable
public static void replaceSpaces(char[] str, int length) {
int spaceCount = 0, newLength = 0, i = 0;
for (i = 0; i < length; i++) {
if (str[i] == ' ')
spaceCount++;
}
newLength = length + (spaceCount * 2);
// str[newLength] = '\0';
for (i = length - 1; i >= 0; i--) {
if (str[i] == ' ') {
str[newLength - 1] = '0';
str[newLength - 2] = '2';
str[newLength - 3] = '%';
newLength = newLength - 3;
} else {
str[newLength - 1] = str[i];
newLength = newLength - 1;
}
}
System.out.println(str);
}
public static void main(String[] args) {
// Test tst = new Test();
char[] ch = "Mr John Smith ".toCharArray();
int length = 13;
replaceSpaces(ch, length);
}

`// Maximum length of string after modifications.
const int MAX = 1000;
// Replaces spaces with %20 in-place and returns
// new length of modified string. It returns -1
// if modified string cannot be stored in str[]
int replaceSpaces(char str[])
{
// count spaces and find current length
int space_count = 0, i;
for (i = 0; str[i]; i++)
if (str[i] == ' ')
space_count++;
// Remove trailing spaces
while (str[i-1] == ' ')
{
space_count--;
i--;
}
// Find new length.
int new_length = i + space_count * 2 + 1;
// New length must be smaller than length
// of string provided.
if (new_length > MAX)
return -1;
// Start filling character from end
int index = new_length - 1;
// Fill string termination.
str[index--] = '\0';
// Fill rest of the string from end
for (int j=i-1; j>=0; j--)
{
// inserts %20 in place of space
if (str[j] == ' ')
{
str[index] = '0';
str[index - 1] = '2';
str[index - 2] = '%';
index = index - 3;
}
else
{
str[index] = str[j];
index--;
}
}
return new_length;
}
// Driver code
int main()
{
char str[MAX] = "Mr John Smith ";
// Prints the replaced string
int new_length = replaceSpaces(str);
for (int i=0; i<new_length; i++)
printf("%c", str[i]);
return 0;
}`