We Keep Coding

word length in Java

I am having trouble with understanding some of the codes in my assignment of finding the word lengths in a file. NB: the Environment I am using is Blue J with a library from dukelearntoprogram.com, this is the link for downloading Blue J, http://www.dukelearntoprogram.com/course3/index.php. I have created a void method called countWordLength, with parameter FileResource called resource, and an integer array called counts. this method should return the number words at a specific length in a file. For instance: 2 words of length 2: My as.
They have given me the code for this assignment but I am not understanding the following code, in my countWordLength method.
for (String word : resource.words()) {
int finalLength = 0;
int totalLength = word.length();
if (Character.isLetter(word.charAt(0)) == false) {
finalLength = totalLength-1;
}
else {
finalLength = totalLength;
}
if (Character.isLetter(word.charAt(totalLength-1)) == false && totalLength > 1) {
finalLength = finalLength-1;
}
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
else {
counts[finalLength] += 1;
}
The specific part that I don't understand is the meaning or usefulness of
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
else {
counts[finalLength] += 1;
}
if my question is not clear, and you may want more parts of my code please let know. Any help
will be highly appreciated.

Explanation:
int finalLength = 0;
int totalLength = word.length();
if (Character.isLetter(word.charAt(0)) == false) {
finalLength = totalLength-1;
}
else {
finalLength = totalLength;
}
Above lines are initiating lengths. If 1st character is not alphabet, then reducing length.
if (Character.isLetter(word.charAt(totalLength-1)) == false && totalLength > 1) {
finalLength = finalLength-1;
}
If last character is not alphabet, removing one more character but a special condition added here that it is we reduce word length only if it's length > 1
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
This might be bit tricky for you. If calculated length is greater than array we assigned, we are adding count to last one. Ex: Let's say they assigned array with 20 and if got word length as 30, then we will assign this word count to last index (19).
else {
counts[finalLength] += 1;
}
If length is less than size allowed than we will add to that respective index.

Assistance with a java program that counts up to a number, in binary

I've recently taken up a computer organization course in where we learn binary hex etc, I took it upon myself to attempt to create a program that will count from 0 up to an input number, however the counting is done in binary. I've run into some trouble and confused myself beyond belief, some clarification and assistance would be greatly appreciated. Specifically speaking, how can I efficiently and effectively replace the values of a string containing the previous binary number, with 0's and 1's using some sort of for-loop. I'm aware that there is some method for directly converting a string to binary, however; I wanted to do this more complicated method for practice.
package counting;
import java.util.Scanner;
public class counting
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.println("Hello, this is a number counter, please enter the integer you would like to count to");
int number = input.nextInt();
String start = "0000000000";
// 000~etc is used as the start simply because i'm not sure how to calculate how many digit places
//the number input by the user will have
StringBuilder cont = new StringBuilder(start);
System.out.println(start);
/*What i intend to do is have the binary loop counter continue until it reaches
* the number input by the user, afterwards, working in a right to left manner, start counting from
* 0 up to the number given by the user, starting with 0. then using another loop, still using
* the right to left manner, if there is a 0, it should be replaced with a 1, and if there is a
* 1, it should be replaced with a 0, and the character before it should be replaced with a 1, if there
* is no room, continue to the left until there is a space available for a 1 and then reset all values
* after the 1 back to zero, and resume counting. the way i see it is requires a for loop to be used
* as the current position of a cursor used to determine what changes must be made
*/
for(int i = 0; i < number; i++)
{
int l = start.length();
for(int n = 0; n <= number; n++)
{
for(int w = 1; w <= l; w++)
{
if (cont.charAt(l-w) == '0')
{
cont.setCharAt((cont.length()-w), '1');
System.out.println(cont);
}
else if (cont.charAt(l-w) == '1')
{
cont.setCharAt((cont.length()-w), '0');
cont.setCharAt((cont.length()-(w+1)), '1');
System.out.println(cont);
}
}
}
System.out.println(cont);
}
}
}

Here is a little loop that will do what you are looking for. You just have to remember powers of 2 to count in binary.
public static char flip(char c){
if(c == '0')
return '1';
else
return '0';
}
public static void main(String[] args) {
String start = "0000000000";
StringBuilder cont = new StringBuilder(start);
int number = (int)Math.pow(2,10);
for(int i = 0; i < number; i++)
{
if(i != 0){
int val = (int)Math.floor(i/2);
for(int j = 0; j <= val; j++){
// Flip any bit that when modded by 2^j == 0
if(i % Math.pow(2,j) == 0){
cont.setCharAt((cont.length() - (j + 1)), flip(cont.charAt(cont.length() - (j + 1))));
}
}
}
System.out.println(cont);
}
}

Reversal of an Integer to check for Palindromic Number

I've looked at a few different stack questions and googled, but nothing I've read really has dealt with reversal of integers, but just strings.
So right now my code may or may not work at all, and it may be the dumbest thing you've ever seen, and that's okay and corrections are welcomed, but from what I hope my code will be doing is going through 100 - 999 multiplying the two ints and then checking whether it's palindromic or not. The if with reverse.equals(sum) is totally pseudocode and obviously won't work, however I can't figure out how to do a check for a palindromic int. Is there a simple way to do this? I've read some pretty lengthy and complicated ways, but I'm sure there's gotta be a simple way. Maybe not. :/. Anyway, here's my code.
public class PalandromicNum {
public static void main(String[] args){
int numOne = 100;
int numTwo = 100;
int toteVal;
int counter = 1000;
int sum = 0;
int finalSum = 0;
for(int i=0; i<counter; i++){
toteVal = numOne * numTwo;
numTwo++;
if(numTwo == 999){
numOne++;
numTwo = 100;
}
if(toteVal < sum){
sum += toteVal;
if(reverse.equals(sum)){
finalSum = sum;
System.out.println(finalSum);
}
}
}
}
}
Thanks again in advance!

This is on my phone so sorry for any errors.
Convert your number to a String and:
public static boolean isPalindrome(String str)
{
// base recursive case
if (str.length <= 1) {
return true;
}
// test the first and last characters
char firstChar = str.charAt(0);
char lastChar = str.charAt(str.length - 1) // subtract 1 as indexes are 0 based
if (!firstChar.equals(lastChar)) {
return false;
}
// if the string is longer than 2 chars and both are equal then recursively call with a shorter version
// start at 2nd char, end at char before last
return isPalindrome(str.substring(1,str.length);
}

Reversing integers is quite easy. Remember mod 10 gives u last digit. Loop over it, chopping off last digit of the number one at a time and adding it to reverse to new number. Then its matter of simple integer equality
int rev = 0;
int n = sum;
while(n)
{
rev = rev*10 + n%10;
n /= 10;
}
if(sum==rev)
//palindrome
else
//no no no no.

You can create a function named isPalindrome to check whether a number is a palindrome.
Use this function in your code.
You just need to pass the number you want to check into this function.
If the result is true, then the number is a palindrome.
Else, it is not a palindrome.
public static boolean isPalindrome(int number) {
int palindrome = number; // copied number into variable
int reverse = 0;
while (palindrome != 0) {
int remainder = palindrome % 10;
reverse = reverse * 10 + remainder;
palindrome = palindrome / 10;
}
// if original and reverse of number is equal means
// number is palindrome in Java
if (number == reverse) {
return true;
}
return false;
}
}

I believe that this code should help you out if you are trying to find out how many palindromes are between 100-999. Of course, it will count palindromes twice since it looks at both permutations of the palindromes. If I were you I would start creating methods to complete a majority of your work as it makes debugging much easier.
int total = 100;
StringBuilder stringSumForward;
StringBuilder stringSumBackward;
int numberOfPals = 0;
for(int i = 100; i < 999; i++){
for(int j = 100; j < 999; j++){
total = i * j;
stringSumForward = new StringBuilder(String.valueOf(total));
stringSumBackward = new StringBuilder(String.valueOf(total)).reverse();
if(stringSumForward.toString().equals(stringSumBackward.toString())){
numberOfPals++;
}
}
}

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.

Finding specific number in prime number array

I'm trying to find prime numbers with a specific condition in Java.
The challenge is to show all the prime numbers (under 100.000) which contain a '3' four times.
I already have a code which shows all the prime numbers under 100.000, but I can't seem to figure out how to count the ones that contain the number '3' four times.
I can however count all the prime numbers.
Can someone help me with this?
Here's the code I have, where am I going to put the numbers into strings?
package Proeftentamen;
import java.util.regex.*;
/**
*
* #author Stefan
*/
public class Vraag_6 {
/// priemgetallen waar 4x een 3 in voor komt???? wtf...
public static void main(String[] args) {
boolean[] lijst = new boolean[1000000]; // hoeveelheid getallen
vularray(lijst);
lijst = zeef(lijst);
drukaf(lijst);
}
public static void vularray(boolean[] lijst) {
for (int i = 2; i < lijst.length; i++) {
lijst[i] = true;
}
}
public static boolean[] zeef(boolean[] lijst) {
for (int i = 2; i < lijst.length / 2; i++) {
if (lijst[i]) {
for (int j = 2 * i; j < lijst.length; j += i) {
lijst[j] = false;
}
}
}
return lijst;
}
public static void drukaf(boolean[] lijst) {
int count = 0;
for (int i = 2; i < lijst.length; i++) {
if (lijst[i] == true) {
System.out.println(i + " " + lijst[i]);
count++;
}
}
System.out.println("Aantal priemgetallen: " + count);
}
}

This question really sounds like a homework, so you should write down what you have come up with and what you tried so far.
There are a lot of ways to count numbers. Just to give you a clue, you can use the reminder operation (in Java - %):
56 % 10 = 6
25 % 5 = 0
So, when you divide by 10 and use a reminder operation you can get the last digit of your number. Now use a loop and counter and you'll be fine.
Another option (very ugly, so don't really use it :) ) - to turn your number into a String and iterate (loop) over its characters.
Hope this helps and good luck!

This code generate 50 permutation of numbers that has four '3' in it's digits
so check each number that is prime or not
public void generateNumbers() {
StringBuilder s = new StringBuilder();
s.append("3333");
for (int i = 0; i < 5; i++) {
for (int j = 0; j <= 9; j++) {
if (j%3==0) continue;
s.insert(i,String.valueOf(j));
int number=Integer.parseInt(s.toString());
System.out.println(number);
s.delete(i,i+1);
}
}
}

Iterate across each prime number.
For each prime number, convert it to a string using the Integer.toString(int) static method.
With this string, iterate over every character (use a for loop and the non-static method String.charAt(int index)) and count the number of times that method returns '3'. (The character '3', not the String "3").
Unless you have some other purpose for an array of prime-number Strings, don't bother to store them anywhere outside the loop.

Please refer below code to validate all such prime numbers.
void getPrimes(int num ,int frequency,char digit) {
int count = 0;
String number=Integer.toString(num);
for (int i = 0; i < number.length(); i++) {
if (count < frequency) {
if (number.charAt(i) == digit)
count++;
}
if (count == frequency)
{
System.out.println(number);
return ;
}
}
}

Using the primes function from an exercise on the Sieve of Eratosthenes, as well as the digits and filter functions from the Standard Prelude, this Scheme expression finds the seven solutions:
(filter
(lambda (n)
(= (length
(filter
(lambda (d) (= d 3))
(digits n)))
4))
(primes 100000))
The outer filter runs over all the primes less than 100000 and applies the test of the outer lambda to each. The inner filter computes the digits of each prime number and keeps only the 3s, then the length function counts them and the equality predicate keeps only those that have 4 3s. You can run the program and see the solution at http://codepad.org/e98fow2u.

you only have at most five digits, four of which must be 3. So what can you say about the remaining digit?
It's not hard to just write out the resulting numbers by hand, and then test each one for primality. Since there are no more than 50 numbers to test, even the simplest trial division by odds will do.
But if you want to generate the numbers programmatically, just do it with 5 loops: add 10,000 to 03333 9 times; add 1,000 to 30333 9 times; add 100 to 33033 9 times; etc. In C++:
int results[50];
int n_res = 0;
int a[5] = {13333, 31333, 33133, 33313, 33331};
for( int i=0, d=10000; i<5; ++i, d/=10)
for( int j=1; j<9; ++j, a[i]+=d )
if( is_prime(a[i]) )
results[n_res++] = a[i];