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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 can I reprint a string starting from the first character till the last?

Given a non-empty string str like "Code" print a string like "CCoCodCode". Where at each index in the string you have to reprint the string up to that index.
I know there is DEFINITELY something wrong with this code that I wrote because the answer should be CCoCodCode, but instead it's giving me the alphabet! I don't know how I should change it.
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
String str = scan.next();
int x = str.length();
for(char i = str.charAt(0); i <= str.charAt(x-1); i++)
{
System.out.print(i);
}
}

The char datatype can be treated as a number; you can increment it and manipulate it as a number.
What you really want is successive substrings of str to be printed. Loop over an int that will represent the ending position of the substring to be printed.
for (int i = 0; i < str.length(); i++)
{
System.out.print(str.substring(0, i + 1));
}
The end index argument to substring is exclusive, which is why I added 1.

Let's say that str is "Code". We can perform some mental substitutions to see what happens to your loop.
str is "Code"
x is 4
str.charAt(0) is 'C'
str.charAt(x-1) is 'e'
Making these substitutions, your loop is:
for(char i = 'C'; i <= 'e'; i++)
{
System.out.print(i);
}
Does this help you see the problem? I would think you'd have a loop from 0 to 3, not from 'C' to 'e'...

Many ways to get it done, suppose we have the input from user stored in a string named "c"... then...
String c = "Code";
for (int i = 0; i < c.length(); i++) {
System.out.print(c.substring(0, i));
}
System.out.print(c);
And this will print the sequence you are looking for.

It is outputting the alphabet because you are printing the counter instead of the characters in the string!
As it is, the first iteration of the for loop will set i to the first character, print that, then the operation i++ will increment i by one. Wait, so if the first character is "C", so i = 'C', what is i++?
Well it turns out characters can be represented by numbers. For example, 'C' has a value of 67. So incrementing it makes it 68, which represents 'D'. So if you run the loop on "Code", it will increment your counter 4 times, giving "CDEF". If you run on "Codecodecode", that will make the loop run 12 times, giving "CDEFGHIJKLMN".
What you really want is to loop through the string by its index instead:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String str = scan.next();
int length = str.length();
for (int i = 0; i < length; i++) {
System.out.print(str.substring(0, i + 1));
}
}

Vowel check - array out of bounds error

I'm trying to write a program which accepts a word in lowercase, converts it into uppercase and changes the vowels in the word to the next alphabet. So far, I've done this:
import java.util.*;
class prg11
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter a word in lowercase.");
String word = sc.next();
word = word.toUpperCase();
int length = word.length();
char ch[] = new char[length+1];
for (int i = 0; i<=length; i++)
{
ch[i] = word.charAt(i);
if("aeiou".indexOf(ch[i]) == 0)
{
ch[i]+=1;
}
}
String str = new String(ch);
System.out.println(str);
}
}
The code compiles fine. But, when I run the program and enter a word, say 'hey', the word is printed in uppercase only. The vowels in it (in this case, 'e'), do not get changed to the next alphabet.
How do I resolve this? TIA.

Need to change three places, according to the code in the question.
word = word.toUpperCase();
int length = word.length();
// yours: char ch[] = new char[length + 1];
// resulting array needs to be as same length as the original word
// if not, there will be array index out of bound issues
char ch[] = new char[length];
// yours: for (int i = 0; i<=length; i++)
// need to go through valid indexes of the array - 0 to length-1
for (int i = 0; i < length; i++) {
ch[i] = word.charAt(i);
// yours: if ("aeiou".indexOf(ch[i]) == 0) {
// two problems when used like that
// 1. indexOf() methods are all case-sensitive
// since you've uppercased your word, need to use AEIOU
// 2. indexOf() returns the index of the given character
// which would be >= 0 when that character exist inside the string
// or -1 if it does not exist
// so need to see if the returned value represents any valid index, not just 0
if ("AEIOU".indexOf(ch[i]) >= 0) {
ch[i] += 1;
}
}
Here's a little concise version. Note the changes I've done.
String word = sc.next().toUpperCase();
char ch[] = word.toCharArray();
for (int i = 0; i < ch.length; i++) {
if ("AEIOU".indexOf(ch[i]) >= 0) {
ch[i] += 1;
}
}
Java doc of indexOf().
public int indexOf(int ch)
Returns the index within this string of the first occurrence of the specified character.
If a character with value ch occurs in the character sequence represented by this String object,
then the index (in Unicode code units) of the first such occurrence is returned.
For values of ch in the range from 0 to 0xFFFF (inclusive), this is the smallest value k such that:
this.charAt(k) == ch
is true. For other values of ch, it is the smallest value k such that:
this.codePointAt(k) == ch
is true. In either case, if no such character occurs in this string, then -1 is returned.
Parameters:
ch - a character (Unicode code point).
Returns:
the index of the first occurrence of the character in the character sequence represented by this object,
or -1 if the character does not occur.

I think this should do it, let me know if it doesn't
public class prg11 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter a word.");
String word = sc.next();
sc.close();
word = word.toUpperCase();
int length = word.length();
char ch[] = new char[length+1];
for (int i = 0; i<length; i++) {
ch[i] = word.charAt(i);
if("AEIOU".indexOf(ch[i]) > -1) {
ch[i]+=1;
}
}
String str = new String(ch);
System.out.println(str);
}
}
Let me know if it works.
Happy coding ;) -Charlie

Use:
for (int i = 0; i<length; i++)
instead as the last index is length-1.

use for (int i = 0; i<=length-1; i++) instead of for (int i = 0; i<=length; i++) and if("AEIOU".indexOf(ch[i]) != -1) instead of if("aeiou".indexOf(ch[i]) == 0)
reason
1.array index starts from 0 that's why length-1
2. As you already made your string in upper case so check condition on "AEIOU"
3. every non-vowel character will return -1 so use if("AEIOU".indexOf(ch[i]) != -1)

"aeiou".indexOf(ch[i]) == 0 will only match 'a' characters (since that is the character at index 0). You should be looking for any index that is greater than -1. Additionally, since you've already converted the string to uppercase, you should be checking against "AEIOU" instead of "aeiou".

Why is my forloop not working?

This is my first question on this site so I'm not sure how to do this, but my question is as follows:
This is just a small piece of a code with multiple methods.
I need to print the ASCII codes of all the characters in a String (input from the user). Now I am trying to use a for-loop which scans the first character prints the ASCII code of it, then scans the next one etc. But at the moment its just printing the first character's ASCII code a few times. Obviously there's something wrong with my for-loop but I've been trying to figure it out and I really can't find it.
static String zin(String zin) {
int length = zin.length();
char letter = zin.charAt(0);
int ascii = (int) letter;
for (int i = 0; i < zin.length(); i++ ) {
System.out.println((int) ascii);
}
return zin;
}

The reason is because you don't re-assign ascii. Try this:
static String zin(String zin) {
int i = 0;
int length = zin.length();
for ( i = 0; i < zin.length(); i++ ) {
int ascii = (int)zin.charAt(i);
System.out.println(ascii);
}
return zin;
}

The problem in your code is though you have a for loop you are not iterating through the strung using that for loop. You only get the 1st char of that string. Instead of that use
static String zin(String zin) {
for (int i = 0; i < zin.length(); i++) {
System.out.println((int) zin.charAt(i));
}
return zin;
}

storing string in an integer array.

I am trying to store a string into an integer array with the following code:
public LargeInteger(String s) {
for (int i = 0; i < s.length(); i++) {
intArray[i] = Integer.parseInt( s.charAt(i));
}
}
eclipse is giving me an error saying: the method parseInt(string) is not applicable for the arguments (char)
What am I doing wrong?

You need to parse the char, or convert it to a String.
If you're trying to get one digit at a time, and you know your input is a digit, then the easiest way to convert a single digit to an int is just
intArray[i] = Character.digit(s.charAt(i), 10); // in base 10
If you want to keep using Integer.parseInt, then just do
intArray[i] = Integer.parseInt(String.valueOf(s.charAt(i)));
// or
intArray[i] = Integer.parseInt(s.substring(i, i+1));

That's because Integer.parseInt() expects a String as parameter, while you are passing a char (s.charAt() returns a char).
Since you are creating the array one digit at a time, a better way to get the decimal representation would be:
intArray[i] = s.charAt(i) - '0';

char is not a String so use a substring function s.substring(i, i+1) or better intArray[i] = s.charAt(i)

s.charAt returns a char.
+ parseInt take a String
= Eclipse gives you the compilation error
You may create a String from the char if really needed:
s.charAt(i)+""

String[] split = s.split("");
int[] nums = new int[split.length];
for(int i = 0; i < split.length; i++){
nums[i] = Integer.parseInt(split[i]);
}

public class Storing_String_to_IntegerArray
{
public static void main(String[] args)
{
System.out.println(" Q.37 Can you store string in array of integers. Try it.");
String str="I am Akash";
int arr[]=new int[str.length()];
char chArr[]=str.toCharArray();
char ch;
for(int i=0;i<str.length();i++)
{
arr[i]=chArr[i];
}
System.out.println("\nI have stored it in array by using ASCII value");
for(int i=0;i<arr.length;i++)
{
System.out.print(" "+arr[i]);
}
System.out.println("\nI have stored it in array by using ASCII value to original content");
for(int i=0;i<arr.length;i++)
{
ch=(char)arr[i];
System.out.print(" "+ch);
}
}
}