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How to obtain the length of the last word in the string

I have reversed the string and have a for loop to iterate through the reversed string.
I am counting characters and I know I have a logic flaw, but I cannot pinpoint why I am having this issue.
The solution needs to return the length of the last word in the string.
My first thought was to iterate through the string backward (I don't know why I decided to create a new string, I should have just iterated through it by decrementing my for loop from the end of the string).
But the logic should be the same from that point for my second for loop.
My logic is basically to try to count characters that aren't whitespace in the last word, and then when the count variable has a value, as well as the next whitespace after the count has counted the characters of the last word.
class Solution {
public int lengthOfLastWord(String s) {
int count = 0;
int countWhite = 0;
char ch;
String reversed = "";
for(int i = 0; i < s.length(); i++) {
ch = s.charAt(i);
reversed += ch;
}
for(int i = 0; i < reversed.length(); i++) {
if(!Character.isWhitespace(reversed.charAt(i))) {
count++;
if(count > 1 && Character.isWhitespace(reversed.charAt(i)) == true) {
break;
}
}
}
return count;
}
}

Maybe try this,
public int lengthOfLastWord(String s) {
String [] arr = s.trim().split(" ");
return arr[arr.length-1].length();
}

Another option would be to use index of last space and calculate length from it:
public int lengthOfLastWord(String string) {
int whiteSpaceIndex = string.lastIndexOf(" ");
if (whiteSpaceIndex == -1) {
return string.length();
}
int lastIndex = string.length() - 1;
return lastIndex - whiteSpaceIndex;
}
String.lastIndexOf() finds the start index of the last occurence of the specified string. -1 means the string was not found, in which case we have a single word and length of the entire string is what we need. Otherwise means we have index of the last space and we can calculate last word length using lastIndexInWord - lastSpaceIndex.

There are lots of ways to achieve that. The most efficient approach is to determine the index of the last white space followed by a letter.
It could be done by iterating over indexes of the given string (reminder: String maintains an array of bytes internally) or simply by invoking method lastIndexOf().
Keeping in mind that the length of a string that could be encountered at runtime is limited to Integer.MAX_VALUE, it'll not be a performance-wise solution to allocate in memory an array, produced as a result of splitting of this lengthy string, when only the length of a single element is required.
The code below demonstrates how to address this problem with Stream IPA and a usual for loop.
The logic of the stream:
Create an IntStream that iterates over the indexes of the given string, starting from the last.
Discard all non-alphabetic symbols at the end of the string with dropWhile().
Then retain all letters until the first non-alphabetic symbol encountered by using takeWhile().
Get the count of element in the stream.
Stream-based solution:
public static int getLastWordLength(String source) {
return (int) IntStream.iterate(source.length() - 1, i -> i >= 0, i -> --i)
.map(source::charAt)
.dropWhile(ch -> !Character.isLetter(ch))
.takeWhile(Character::isLetter)
.count();
}
If your choice is a loop there's no need to reverse the string. You can start iteration from the last index, determine the values of the end and start and return the difference.
Just in case, if you need to reverse a string that is the most simple and efficient way:
new StringBuilder(source).reverse().toString();
Iterative solution:
public static int getLastWordLength(String source) {
int end = -1; // initialized with illegal index
int start = 0;
for (int i = source.length() - 1; i >= 0; i--) {
if (Character.isLetter(source.charAt(i)) && end == -1) {
end = i;
}
if (Character.isWhitespace(source.charAt(i)) && end != -1) {
start = i;
break;
}
}
return end == -1 ? 0 : end - start;
}
main()
public static void main(String[] args) {
System.out.println(getLastWord("Humpty Dumpty sat on a wall % _ (&)"));
}
output
4 - last word is "wall"

Firstly, as you have mentioned, your reverse string formed is just a copy of your original string. To rectify that,
for (int i = s.length() - 1; i >= 0; i--) {
ch = s.charAt(i);
reversed += ch;
}
Secondly, the second if condition is inside your first if condition. That is why, it will never break ( because you are first checking if character is whitespace, if it is, then you are not going inside the if statement, thus your second condition of your inner if loop will never be satisfied).
public class HW5 {
public static void main(String[] args) {
String s = "My name is Mathew";
int count = lengthOfLastWord(s);
System.out.println(count);
}
public static int lengthOfLastWord(String s) {
int count = 0;
int countWhite = 0;
char ch;
String reversed = "";
System.out.println("original string is----" + s);
for (int i = s.length() - 1; i >= 0; i--) {
ch = s.charAt(i);
reversed += ch;
}
System.out.println("reversed string is----" + reversed);
for (int i = 0; i < reversed.length(); i++) {
if (!Character.isWhitespace(reversed.charAt(i)))
count++;
if (count > 1 && Character.isWhitespace(reversed.charAt(i)) == true) {
break;
}
}
return count;
}
}
=
and the output is :
original string is----My name is Mathew
reversed string is----wehtaM si eman yM
6
Another way to go about is : you use the inbuilt function split which returns an array of string and then return the count of last string in the array.

How to separate a string into substrings of two characters

I am trying to split strings in substrings of two chararters for example for the input: "ABCDE" i want to get the substrings "AB" "BC" "CD" "DE".
I tried with this:
String route = "ABCDE";
int i = 0;
while(i < route.length()) {
String sub = route.substring(i,i+2);
System.out.println(sub);
i++;
}
but the index (i) gets out of range int the last iteration and causes an error.
is there any way to do this without getting the index (i) out of range ?

You need to change the loop condition.
while(i < route.length()-1)
In your code i goes till (length-1) and than in the substring(i,i+2) function you gives end index i+2. It is higher than largest index of string.
Also, As far as I know calling a library function in a loop condition is not considered a good practice.
In each iteration you call this function which is time consuming.
control goes to that subroutine in each iteration.
A good alternative to this would be to store the length in a variable and use that in a condition.
int temp = route.length()-1;
while(i<temp){

This should work fine
String route = "ABCDE";
if( route.length() > 2){
int i = 0;
do {
String res = route.substring(i,i+2);
System.out.println(res);
i++;
} while (i + 1 < route.length());
}
else{
System.out.println(route);
}
Edit: Added boundary case for the string has length less than 2

Add check for the size of the string to trap the error:
String route = "ABCDE";
int i = 0;
while(i < route.length()) {
if(i < route.length() - 1) {
String sub = route.substring(i,i+2);
System.out.println(sub);
} else {
String sub = route.substring(i,i+1);
System.out.println(sub);
i++;
}
So whenever the i counter almost close to string size, get the last char.

You are getting an StringIndexOutOfBoundsException because you are trying to access an index of the String that doesn't exist.
To fix this, change your loop condition from
while(i < route.length())
to
while(i < route.length() - 1)
Without the -1 on the last iteration of the while loop i + 2 is equal to 71 which is out of the Strings bounds.
Another (cleaner) solution to this problem is a for loop:
for(int i = 0; i < route.length() - 1; i++) {
System.out.println(route.substring(j, j + 2));
}
The for loop in this situation is just shorter as the declaration, conditional, and increment statements are all in one line.
1: This 7 reduces to 6 since the endIndex of substring is exclusive.

As denis already pointed out, the bug in the code is in the loop condition.
Should be: while(i < route.length() - 1)
. However, how about simplifying this logic to use a for loop.
String route = "ABCDE";
for (int i=0; i+2<=route.length(); i++)
System.out.println(route.substring(i,i+2));

you can not use
i < route.length(),because when i = 5, String sub = route.substring(i,i+2); the i+2=7,is out of index,so use i<route.length instead

Java Code output If input is "aabbba" then output should be "a2b3a1" [closed]

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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
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I am doing one simple java code where if
input is : "aabbba"
then
output should be: "a2b3a1"
I have done the below coding but missing somewhere. So let me know my mistake.
public class Test {
public static void main(String[] args) {
String str = "aabbba";
int count = 1;
for (int i = 0; i < str.length(); i = i + count) {
count = 1;
for (int j = i + 1; j < str.length(); j++) {
if (str.charAt(i) == str.charAt(j)) {
count = count + 1;
} else {
System.out.println(str.charAt(i) + "" + count);
break;
}
}//end of inner for
}//end of outer for
}//end of main
}//end of class

Actually you have too much code, You only need one loop, and you should be comparing the letter to the previous one, not attempting to compare each letter to every letter after it.
If you are confused about what your program is doing, the best place to start is to use your debugger to step through the code.
for(int i = 0, count = 1; i < str.length(); i++, count++) {
char ch = str.charAt(i);
char next = i + 1 < str.length() ? str.charAt(i + 1) : (char) -1;
if (ch != next) {
System.out.print("" + ch + count);
count = 0;
}
}

Using your effort and code, you simply did put the print to the wrong place
String str = "aabbba";
int count = 1;
for(int i = 0; i <str.length();i=i+count){
count =1;
for(int j = i+1; j<str.length();j++){
if(str.charAt(i) == str.charAt(j)){
count = count+1;
}
else{
break;
}
}
// Print here otherwise you will miss the last group of letters
// Also if you just want one line use .print instead of println
System.out.print(str.charAt(i)+""+count);
}

Using Java-8 and my StreamEx library it's a one-liner:
String input = "aabbba";
String result = IntStreamEx.ofChars(input).mapToObj(ch -> (char)ch)
.runLengths().join("").joining();
Step-by step:
IntStreamEx.ofChars(input): create IntStreamEx (enhanced IntStream) where each element is the corresponding character of input line.
.mapToObj(ch -> (char)ch): transform to StreamEx<Character> (enhanced Stream<Character>) where each element is the Character object.
.runLengths(): convert to EntryStream<Character, Long> (enhanced Stream<Map.Entry<Character, Long>>) where keys are Character objects and values are counts of equal adjacent characters.
.join(""): convert to StreamEx<String>, joining keys (characters) and values (counts) via given empty separator.
.joining(): final reduction to the resulting string without additional separators.

You're just missing the print of the last group of letters. you only print inside the loop once you found a different letter, you should take into account the last group of letters that has no "different letter" after it

I would suggest using a StringBuilder:
public String myOutput(String str) {
if (str == null || str.length() == 0)
return str;
StringBuilder sb = new StringBuilder();
int count = 1;
char currentChar;
for (int i = 0; i < str.length() - 1; i++) {
currentChar = str.charAt(i);
if (currentChar == str.charAt(i+1)) {
count++;
} else {
sb.append(currentChar);
sb.append(String.valueOf(count));
count = 1;
}
}
sb.append(str.charAt(str.length()-1));
sb.append(String.valueOf(count));
return sb.toString();
}

You only need 1 loop
System.out.println() will cause your output to have line break. You better use System.out.print(). Now your currrent code is resulting :
a2
b3

Comparing two unknown strings and finding match [closed]

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If I have
String x = "test";
String s = "tastaegasghet;
you can find t e s t inside the string s. The naive way of doing this with a known string would be something like this:
.*t+.*e+.*s+.*t+.*
This will return true if we can find the letters t e s t in order and any characters inbetween. I want to do the same thing but with two unknown Strings x and s or in otherwords, String s and x can be anything. I don't want something hard coded but something for general use instead.

This is the pseudocode for looping solution:
function (
needle, // x
haystack // s
) {
j = 0
for (i = 0; i < haystack.length && j < needle.length; i++) {
if (haystack[i] == needle[j]) {
j++
}
}
return j == needle.length
}
You only need to loop through each character in haystack string and advance the pointer in needle string when you find a matching character. If the pointer reaches the end of the needle string, it means the needle string can be found as a subsequence of the haystack string.
A small optimization you can do is checking needle.length <= haystack.length before starting the loop.
Just for fun
If you want to go the Cthulhu's way, you can use this construction:
(?>.*?t)(?>.*?e)(?>.*?s)(?>.*?t).*+
This doesn't have the risk of catastrophic backtracking, and should work similar to the loop above (linear complexity), except that it has a lot of overhead compiling and matching the regex.

It's not that hard to just use a loop.
String x = "test";
String s = "tastaegasghet";
int index = 0;
for(int i = 0; i < s.length() && index < x.length(); i++){
if(s.charAt(i) == x.charAt(index)) index++;
}
boolean exists = index == x.length();
System.out.println(exists);
This should be significantly faster than a regex, at least for longer input.

I'd make a simple function, with a loop, instead of a regex. Something like the following:
public boolean containsLetters(string a, string b)
{
char[] aArray = a.toCharArray();
char[] bArray = b.toCharArray();
int lettersFound = 0, lastLocation = 0;
for (int i = 0; i < aArray.length; i++)
{
for (lastLocation=lastLocation; lastLocation < bArray.length; lastLocation++)
{
if (aArray[i] == bArray[lastLocation])
{
lettersFound++;
break;
}
}
}
return lettersFound == aArray.length;
}
The inner for loop stops the first time it finds the letter. It doens't need to determine if it appears more than once since the function returns a boolean, so this saves some time for large strings. It will only return true if it finds them in order. It remembers the index of the last letter it found, and searches through for the next letter from that location.

You can use the Pattern class for this.
String x= "yourValue";
Pattern pattern = Pattern.compile(Mention your pattern here);
Matcher matcher = pattern.matcher(x);
if (matcher.find()) {
System.out.println(matcher.group(0)); //prints /{item}/
} else {
System.out.println("Match not found");
}

Java code without using any built-in function
String s1 = "test";
String s2 = "tastaegasghets";
char ch[] = new char[s1.length()];
char ch1[] = new char[s2.length()];
int k = 0;
for (int i = 0; i < s1.length(); i++) {
ch[i] = s1.charAt(i);
for (int j = 0; j < s2.length(); j++) {
ch1[j] = s2.charAt(j);
if (ch[i] == ch1[j]) {
k++;
break;
}
}
}
if (k == s1.length()) {
System.out.println("true");
} else {
System.out.println("false");
}

Using for loop to get the Hamming distance between 2 strings

In this task i need to get the Hamming distance (the Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different - from Wikipedia) between the two strings sequence1 and sequence2.
First i made 2 new strings which is the 2 original strings but both with lowered case to make comparing easier. Then i resorted to using the for loop and if to compare the 2 strings. For any differences in characters in these 2 pair of string, the loop would add 1 to an int x = 0. The returns of the method will be the value of this x.
public static int getHammingDistance(String sequence1, String sequence2) {
int a = 0;
String sequenceX = sequence1.toLowerCase();
String sequenceY = sequence2.toLowerCase();
for (int x = 0; x < sequenceX.length(); x++) {
for (int y = 0; y < sequenceY.length(); y++) {
if (sequenceX.charAt(x) == sequenceY.charAt(y)) {
a += 0;
} else if (sequenceX.charAt(x) != sequenceY.charAt(y)) {
a += 1;
}
}
}
return a;
}
So does the code looks good and functional enough? Anything i could to fix or to optimize the code? Thanks in advance. I'm a huge noob so pardon me if i asked anything silly

From my point the following implementation would be ok:
public static int getHammingDistance(String sequence1, String sequence2) {
char[] s1 = sequence1.toCharArray();
char[] s2 = sequence2.toCharArray();
int shorter = Math.min(s1.length, s2.length);
int longest = Math.max(s1.length, s2.length);
int result = 0;
for (int i=0; i<shorter; i++) {
if (s1[i] != s2[i]) result++;
}
result += longest - shorter;
return result;
}
uses array, what avoids the invocation of two method (charAt) for each single char that needs to be compared;
avoid exception when one string is longer than the other.

your code is completely off.
as you said yourself, the distance is the number of places where the strings differ - so you should only have 1 loop, going over both strings at once. instead you have 2 nested loops that compare every index in string a to every index in string b.
also, writing an if condition that results in a+=0 is a waste of time.
try this instead:
for (int x = 0; x < sequenceX.length(); x++) { //both are of the same length
if (sequenceX.charAt(x) != sequenceY.charAt(x)) {
a += 1;
}
}
also, this is still a naive approach which will probbaly not work with complex unicode characters (where 2 characters can be logically equal yet not have the same character code)

public static int getHammingDistance(String sequenceX, String sequenceY) {
int a = 0;
// String sequenceX = sequence1.toLowerCase();
//String sequenceY = sequence2.toLowerCase();
if (sequenceX.length() != sequenceY.length()) {
return -1; //input strings should be of equal length
}
for (int i = 0; i < sequenceX.length(); i++) {
if (sequenceX.charAt(i) != sequenceY.charAt(i)) {
a++;
}
}
return a;
}

Your code is OK, however I'd suggest you the following improvements.
do not use charAt() of string. Get char array from string using toCharArray() before loop and then work with this array. This is more readable and more effective.
The structure
if (sequenceX.charAt(x) == sequenceY.charAt(y)) {
a += 0;
} else if (sequenceX.charAt(x) != sequenceY.charAt(y)) {
a += 1;
}
looks redundant. Fix it to:
if (sequenceX.charAt(x) == sequenceY.charAt(y)) {
a += 0;
} else {
a += 1;
}
Moreover taking into account that I recommended you to work with array change it to something like:
a += seqx[x] == seqY[x] ? 0 : 1
less code less bugs...
EDIT: as mentionded by #radai you do not need if/else structure at all: adding 0 to a is redundant.