You are allowed to use the following methods from the Java API:
class String:
length,charAt
class StringBuilder:
length,charAt,append,toString
class Character: any method
moveAllXsRight takes a char and a String as input and returns a String.
The output string should be the same as the input string except that every occurrence of the input character should be shifted one character to the right. If it is impossible to shift a character to the right (it is at the end of the string), then it is not shifted. Do not use arrays to solve this problem.
HW2.moveAllXsRight('X', "abcXdeXXXfghXiXXjXX")
"abcdXefXXXghiXjXXXX"
Here is my code now:
public static String moveAllXsRight(char a, String b){
StringBuilder sb = new StringBuilder();
String str ="";
for ( int i = 0; i<b.length(); i++){
if(b.charAt(i) != a){
sb.append(b.charAt(i));
}
else if(b.charAt(i) == a){
str = Character.toString(b.charAt(i));
}
else {
if(b.charAt(i+2)>sb.length()){
sb.append(b.charAt(i));
}
}
}
return sb.toString();
}
Try this:
public static String moveAllXsRight(char a, String b) {
StringBuilder sb = new StringBuilder(b);
for(int i = 0; i < sb.length() - 1; i++) {
if(sb.charAt(i) == a) {
// swapping with the right character
sb.setCharAt(i, sb.charAt(i + 1));
sb.setCharAt(i + 1, a);
// skipping next index (because we already know it contains `a`)
i = i + 1;
}
}
return sb.toString();
}
For this example:
moveAllXsRight('X', "abcXdeXXXfghXiXXjXX");
This is the output:
abcdXeXXfXghiXXXjXX
Update:
By slightly changing the for loop (inverting it):
for(int i = sb.length() - 2; i > 0; i--) { // inverse loop
if(sb.charAt(i) == a) {
// swapping
sb.setCharAt(i, sb.charAt(i + 1));
sb.setCharAt(i + 1, a);
}
}
Now this:
moveAllXsRight('X', "abcXdeXXXfghXiXXjXX");
Results in:
abcdXefXXXghiXjXXXX
Now you can decide which version do you want to use.
Here's one way to do it:
public static String moveAllXsRight(char a, String b) {
char[] chars = b.toCharArray();
for (int i = 0; i < chars.length-1; ++i) {
if (chars[i] == a) {
chars[i] = chars[i+1];
chars[i+1] = a;
}
}
return new String(chars);
}
It gives "abcdeXXXfghXiXXjXXX" when the arguments are 'X' and "abcXdeXXXfghXiXXjXX".
Here's another way:
public static String moveAllXsRight(char a, String b) {
char[] chars = b.toCharArray();
boolean wasa = false;
for (int i = 0; i < chars.length; ++i) {
char c = chars[i];
if (wasa) {
chars[i-1] = c;
chars[i] = a;
}
wasa = c == a;
}
return new String(chars);
}
It gives "abcdXeXXfXghiXXjXXX".
UPDATE
Here's yet another way:
public static String moveAllXsRight(char a, String b) {
char[] chars = b.toCharArray();
for (int i = chars.length-1; --i >= 0;) {
char c = chars[i];
if (c == a) {
chars[i] = chars[i+1];
chars[i+1] = a;
}
}
return new String(chars);
}
It gives "abcdXefXXXghiXjXXXX" which is what you expected.
UPDATE 2:
With a StringBuilder to avoid using arrays:
public static String moveAllXsRight(char a, String b) {
StringBuilder buf = new StringBuilder(b);
for (int i = buf.length()-1; --i >= 0;) {
char c = buf.charAt(i);
if (c == a) {
buf.setCharAt(i, buf.charAt(i+1));
buf.setCharAt(i+1, a);
}
}
return buf.toString();
}
public static String moveAllXsRight(char a, String b){
String str ="";
int n = 0;
char lower_a = Character.toLowerCase(a);
char upper_a = Character.toUpperCase(a);
for ( int i = b.length()-2; i>=0; i--){
if(b.charAt(i) == lower_a || b.charAt(i) == upper_a){
n++;
}
}
str = b.substring(b.length()-n) + b.substring(0,b.length()-n);
return str;
}
I have this code that filters a String str, keeping only some chars, resulting in the reduced String fStr. The subtlety is that I only keep a target char, if it is not equal to the last char in fStr:
ArrayList<Character> targetChars = //{a, b, c};
String str = "axxbxxxxbxbxcxa", fStr = "";
for(int i = 0, s = 0 ; i < str.length() ; i++) {
char c = str.charAt(i);
if(targetChars.contains(c)) {
if(s > 0 && fStr.charAt(s-1) != c) {
fStr += c;
s++;
}
}
}
fStr → "abca"
In the innermost if statement, I have to include s > 0 before fStr.charAt(s-1) != c, otherwise the latter will throw an OutOfBounds exception the first time targetChars.contains(c) is true. But only the first time, it annoys me that the loop will always check that I won't be out of bounds, given that it only has to do it once. I know I could do something like that:
ArrayList<Character> targetChars = //{a, b, c};
String str = "auebskrubnbocpa", fStr = "";
int i = 0, s = 0;
for(; i < str.length() ; i++) {
char c = str.charAt(i);
if(targetChars.contains(c)) {
fStr += c;
s++;
i++;
break;
}
}
for(; i < str.length() ; i++) {
char c = str.charAt(i);
if(targetChars.contains(c)) {
if(fStr.charAt(s-1) != c) {
fStr += c;
s++;
}
}
}
But is there a more elegant and less annoying way to dynamically truncate a conditional statement?
Is there a way to dynamically change conditions in if statement in Java?
No there isn't. The original version of your code is the best from a readability perspective.
However, if you are concerned about efficiency, then you should be using a StringBuilder rather than fStr += c.
Also a char[] and an explicit for loop is likely to be faster than ArrayList<Character>.contains.
Here is how I would do it, but not sure if it suits your needs
public class Example {
public static void main(String[] args) {
char[] targetChars = {'a', 'b', 'c'};
String str = "axxbxxxxbxbxcxa", fStr = " ";
for(int i = 0 ; i < str.length() ; i++) {
char c = str.charAt(i);
if(isAcceptableChar(c, targetChars)) {
if(fStr.charAt(fStr.length() - 1) != c) {
fStr = fStr.trim() + c;
}
}
}
System.out.println(fStr);
}
private static boolean isAcceptableChar(char newChar, char[] targetChars) {
boolean value = false;
for(char ch : targetChars){
if(ch == newChar) {
value = true;
break;
}
}
return value;
}
}
Sure there is, just call a function that returns a boolean value that you use in your if condition. Different functions could be called at different times by using a function pointer.
I'm dealing with logical expressions in strings. So far I have worked on the following method.
public static String modify(String expression)
{
String temp = expression;
String validated = "";
for(int idx=0; idx<temp.length(); idx++)
{
if(idx!=temp.length()-1)
{
if((Character.isAlphabetic(temp.charAt(idx))) && (Character.isAlphabetic(temp.charAt(idx+1))))
{
validated+=temp.substring(idx,idx+1);
validated+="*";
}
else
validated+=temp.substring(idx,idx+1);
}
else
validated+=temp.substring(idx);
}
return validated;
}
The following are examples of supposed input/output
input: AB+BC+ABC / output: (A*B)+(B*C)+(A*B*C)
input: (A+B)+ABC / output: (A+B)+(A*B*C)
input: (A+B)*(B+C)*(AB) / output: (A+B)*(B+C)*(A*B)
One way you can do it is simply keeping track of brackets with a boolean semaphore
public static String modify(String expression)
{
String temp = expression;
StringBuilder validated = new StringBuilder();
boolean inBrackets=false;
for(int idx=0; idx<temp.length()-1; idx++)
{
if((Character.isLetter(temp.charAt(idx))) && (Character.isLetter(temp.charAt(idx+1))))
{
if(!inBrackets){
inBrackets = true;
validated.append("(");
}
validated.append(temp.substring(idx,idx+1));
validated.append("*");
}
else{
validated.append(temp.substring(idx,idx+1));
if(inBrackets){
validated.append(")");
inBrackets=false;
}
}
}
validated.append(temp.substring(temp.length()-1));
if(inBrackets){
validated.append(")");
inBrackets=false;
}
return validated.toString();
}
Also never use string concatenation instead use StringBuilder or its predecessor StringBuffer in case you are seeking thread safe solution.
Here is what I would do, using StringBuilder and a split:
public static String modify(String expression)
{
StringBuilder finalString = new StringBuilder();
String[] subExpressions = expression.split("\\+");
List<String> formattedSubExpressions = new ArrayList<String>();
for (String subExpression : subExpressions) {
if (subExpression.length() > 1) {
StringBuilder formattedSubExpression = new StringBuilder();
formattedSubExpression.append("(");
for (int i=0; i<subExpression.length(); i++) {
formattedSubExpression.append(subExpression.charAt(i));
if (i != subExpression.length() -1 ) {
formattedSubExpression.append("*");
}
}
formattedSubExpression.append(")");
formattedSubExpressions.add(formattedSubExpression.toString());
} else {
formattedSubExpressions.add(subExpression);
}
}
for (String subExpression : formattedSubExpressions) {
finalString.append(subExpression);
finalString.append("+");
}
if (finalString.charAt(finalString.length() - 1) == '+') {
finalString.deleteCharAt(finalString.length() - 1);
}
return finalString.toString();
}
It gives the following sample input/output:
AB+CD: (A*B)+(C*D)
AB+CD+EF: (A*B)+(C*D)+(E*F)
AB+CD+EFGH: (A*B)+(C*D)+(E*F*G*H)
I based this answer on the idea that what you want to do is group repeating alpha characters between parentheses and put an asterisks between them regardless of the operation (add, subtract, divide, etc) being performed between the groups.
private static final Pattern p = Pattern.compile("[a-zA-Z]{2,}");
public String parse(String s){
if(s == null || "".equals(s)) {
return s;
}
char[] chars = s.toCharArray();
StringBuilder sb = new StringBuilder(100);
Matcher m = p.matcher(s);
int i = 0;
while(i<chars.length && m.find()){
int startIdx = m.start();
int endIdx = m.end();
// Need to get the leading part of the string before this matching region
while(i < startIdx){
sb.append(chars[i]);
i++;
}
sb.append('('); // Start getting the match region
while(i < endIdx){
sb.append(chars[i]);
if(i < endIdx - 1){
sb.append('*');
}
i++;
}
sb.append(')'); // end the match region
}
// If there is a region beyond the last match, append it
if(i < chars.length -1){
for(; i < chars.length; i++){
sb.append(chars[i]);
}
}
return sb.toString();
}
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;
}`