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Java Calculate Sequence with Queue

I have the following sequence of numbers:
S1 = N, S2 = S1 + 1, S3 = 2*S1 + 1, S4 = S1 + 2, S5 = S2 + 1, S6 = 2*S2 + 1, S7 = S2 + 2 ...
Using the ArrayDeque<E> class, I have to write a program to print its first 50 members for given N.
Examples:
input 2
output 2 3 5 4 4 7 5 6 11 7 5 9 6 ...
This is my code. The problem is that I can't update next S
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;
public class p04 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int numN = scanner.nextInt();
scanner.close();
int counter = 1;
int nexS = numN;
Queue<Integer> fifty = new ArrayDeque<>();
for (int i = 0; i < 50; i++) {
if (i == 0){
fifty.add(numN);
}else {
if (counter == 1){
counter++;
numN = nexS + 1;
fifty.add(numN);
}else if (counter == 2){
counter++;
numN = (nexS * 2) + 1;
fifty.add(numN);
}else {
counter = 1;
numN = nexS +2;
fifty.add(numN);
nexS = nexS + 1;
}
}
}
for (Integer integer : fifty) {
System.out.print(integer + " ");
}
}
}

The way you solve this problem it's easier to solve It with ArrayList. I think that my solution is more queue oriented and that was your task. So this is my take:
import java.util.ArrayDeque;
import java.util.Scanner;
public class SequenceQuestion {
public static void constructSequence(int start, int seqLength) {
ArrayDeque<Integer> queue = new ArrayDeque<>();
queue.add(start);
System.out.print(start);
for (int i = 0; i < seqLength - 1; i++) {
int print = 0;
if (i % 3 == 0 && i != 0) queue.remove();
if (i % 3 == 0) {
print = queue.peek() + 1;
queue.add(print);
} else if (i % 3 == 1) {
print = queue.peek() * 2 + 1;
queue.add(print);
} else if (i % 3 == 2) {
print = queue.peek() + 2;
queue.add(print);
}
System.out.print(", " + print);
}
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
constructSequence(s.nextInt(), 50);
}
}
You don't need counters because you already have one (i) and if you always check for mod 3 at the beginning and if equals to 0, remove the first element from queue. I see that this is the place you had trouble with.

Find the longest sequence of same characters in a string

This code supposed to output the longest run on which a character in a string has a consecutive runs of itself. Though the problem is that it outputs: 8 (which should be 5 instead). I just would like to ask what seems to be the problem regarding this code.
public class Sample {
public static void main(String[] args) {
String setofletters = "aaakkcccccczz"; /* 15 */
int output = runLongestIndex(setofletters);
System.out.println("Longest run that first appeared in index:" + output);
}
public static int runLongestIndex(String setofletters) {
int ctr = 0;
int ctrstor = 0;
int ii = 0;
int output = 0;
// loops until the last character in the string
for (int i = 0; i < setofletters.length() - 1; i++) {
// checks if the letter is the same to the next
if (setofletters.charAt(i) == setofletters.charAt(i++)) {
ctr++;
ii = i++;
// loops until the letter in the index is no longer equal
while (setofletters.charAt(i) == setofletters.charAt(ii)) {
ii++;
ctr++;
}
if (ctr > ctrstor) {
output = i;
}
// storing purposes
ctrstor = ctr;
}
// resets the counter
ctr = 0;
}
return output;
}
}

UPDATE Sorry, I misunderstood your question a bit, you need to make the following changes in your code to make it work.(lines with comments)
public static int runLongestIndex(String setofletters){
int ctr = 1; // every character is repeated at least once, so you should initialize it to 1, not 0
int ctrstor = 0;
int ii = 0;
int output = 0;
for (int i = 0; i < setofletters.length() - 1; i++) {
if (i < setofletters.length() - 1 && setofletters.charAt(i) == setofletters.charAt(i+1)) { // i++ is not same as i+1
ctr++;
ii = i+1; // i++ is not same as i+1
while (setofletters.charAt(i) == setofletters.charAt(ii)) {
ii++;
ctr++;
}
if (ctr > ctrstor) {
output = i;
}
ctrstor = ctr;
}
ctr = 1; // for the same reason I mentioned above
}
return output;
}
EDIT : the easiest way to write your code is :
public static int runLongestIndex(String setofletters){
int ctr = 1;
int output = 0;
int j=0;
for(int i=0; i<setofletters.length()-1;i++){
j=i;
while(i <setofletters.length()-1 && setofletters.charAt(i)==setofletters.charAt(i+1)){
i++;
ctr++;
}
if(ctr>output){
output=j;
}
ctr = 1;
}
return output;
}
Why are you assigning i to output? You should assign ctr to output.
change
if(ctr>ctrstor){
output=i;
}
to
if(ctr>ctrstor){
output=ctr;
}
and also I think you should change
if(setofletters.charAt(i)==setofletters.charAt(i++))
to
if(i<setofletters.length()-1 && setofletters.charAt(i)==setofletters.charAt(i+1)){
and you should intialize ctr to 1 but not 0 because every character is repeated at least once.

I'll give you a Scala implementation for that problem.
Here it is the automatic test (in BDD style with ScalaTest)
import org.scalatest._
class RichStringSpec extends FlatSpec with MustMatchers {
"A rich string" should "find the longest run of consecutive characters" in {
import Example._
"abceedd".longestRun mustBe Set("ee", "dd")
"aeebceeedd".longestRun mustBe Set("eee")
"aaaaaaa".longestRun mustBe Set("aaaaaaa")
"abcdefgh".longestRun mustBe empty
}
}
Following is the imperative style implementation, with nested loops and mutable variables as you would normally choose to do in Java or C++:
object Example {
implicit class RichString(string: String) {
def longestRun: Set[String] = {
val chunks = mutable.Set.empty[String]
val ilen = string.length
var gmax = 0
for ((ch, curr) <- string.zipWithIndex) {
val chunk = mutable.ListBuffer(ch)
var next = curr + 1
while (next < ilen && string(next) == ch) {
chunk += string(next)
next = next + 1
}
gmax = chunk.length max gmax
if (gmax > 1) chunks += chunk.mkString
}
chunks.toSet.filter( _.length == gmax )
}
}
}
Following is a functional-style implementation, hence no variables, no loops but tail recursion with result accumulators and pattern matching to compare each character with the next one (Crazy! Isn't it?):
object Example {
implicit class RichString(string: String) {
def longestRun: Set[String] = {
def recurse(chars: String, chunk: mutable.ListBuffer[Char], chunks: mutable.Set[String]): Set[String] = {
chars.toList match {
case List(x, y, _*) if (x == y) =>
recurse(
chars.tail,
if (chunk.isEmpty) chunk ++= List(x, y) else chunk += y,
chunks
)
case Nil =>
// terminate recursion
chunks.toSet
case _ => // x != y
recurse(
chars.tail,
chunk = mutable.ListBuffer(),
chunks += chunk.mkString
)
}
}
val chunks = recurse(string, mutable.ListBuffer(), mutable.Set.empty[String])
val max = chunks.map(_.length).max
if (max > 0) chunks.filter( _.length == max ) else Set()
}
}
}
For example, for the given "aeebceeedd" string, both implementations above will build the following set of chunks (repeating characters)
Set("ee", "eee", "dd")
and they will filter those chunks having the maximum length (resulting "eee").

This code should work for any length of string sequence.
public class LongestStringSequqnce {
static String myString = "aaaabbbbcccchhhhiiiiibbbbbbbbbccccccc";
static int largestSequence = 0;
static char longestChar = '\0';
public static void main(String args[]) {
int currentSequence = 1;
char current = '\0';
char next = '\0';
for (int i = 0; i < myString.length() - 1; i++) {
current = myString.charAt(i);
next = myString.charAt(i + 1);
// If character's are in sequence , increase the counter
if (current == next) {
currentSequence += 1;
} else {
if (currentSequence > largestSequence) { // When sequence is
// completed, check if
// it is longest
largestSequence = currentSequence;
longestChar = current;
}
currentSequence = 1; // re-initialize counter
}
}
if (currentSequence > largestSequence) { // Check if last string
// sequence is longest
largestSequence = currentSequence;
longestChar = current;
}
System.out.println("Longest character sequence is of character "
+ longestChar + " and is " + largestSequence + " long");
}
}
Source : http://www.5balloons.info/program-java-code-to-find-longest-character-sequence-in-a-random-string/

if(ctr>ctrstor){
output=i;
}
//storing purposes
ctrstor=ctr;
This looks like the problem. So if you find 8 consecutive characters, it will set output to 8, and proceed. The next time thru, it finds 3 consecutive characters, so doesn't set output, but sets ctrstor. Next time thru it finds 4 consecutive characters, and this will set output to 4

There are few traps in the code that your logic felt in:
Code incorrectly assumes that there is always next character to compare current one.
This fails for string like "a" or the last character in any string.
Code does not store the max count of characters but only the max index (i).
MaxCount is needed to compare the next chars sequence size.
Loop for and loop while repeat the same subset of characters.
Also variable name style makes it harder to understand the code.
After correcting above
public static int runLongestIndex(String setofletters) {
int maxCount = 0;
int maxIndex = 0;
// loops each character in the string
for (int i = 0; i < setofletters.length() - 1; ) {
// new char sequence starts here
char currChar = setofletters.charAt(i);
int count = 1;
int index = i;
while ( (index < setofletters.length() - 1) &&
(currChar == setofletters.charAt(++index)) ) {
count++;
}
if (count > maxCount) {
maxIndex = i;
maxCount = count;
}
i = index;
}
return maxIndex;
}
See Java DEMO

I think you don't need an internal loop:
public static int runLongestIndex(String setofletters) {
if (setofletters == null || setofletters.isEmpty()) {
return -1;
}
int cnt = 1;
char prevC = setofletters.charAt(0);
int maxCnt = 1;
//char maxC = prevC;
int maxRunIdx = 0;
int curRunIdx = 0;
for (int i = 1; i < setofletters.length(); i++){
final char c = setofletters.charAt(i);
if (prevC == c) {
cnt++;
} else {
if (cnt > maxCnt) {
maxCnt = cnt;
//maxC = prevC;
maxRunIdx = curRunIdx;
}
cnt = 1;
curRunIdx = i;
}
prevC = c;
}
if (setofletters.charAt(setofletters.length() - 1) == prevC) {
if (cnt > maxCnt) {
//maxC = prevC;
maxCnt = cnt;
maxRunIdx = curRunIdx;
}
}
return maxRunIdx;
}
and this code:
System.out.println(runLongestIndex("aaakkcccccczz"));
gives you
5

This is how a "colleague" of mine is understanding to write readable code in order to solve this problem, even if this is working :)
public static int count (String str) {
int i = 0;
while(i < str.length()-1 && str.charAt(i)==str.charAt(i+1))
i ++;
return ++i;
}
public static int getLongestIndex(String str){
int output = 0;
for(int i=0, cnt = 1, counter = 0 ; i<str.length() - 1;i += cnt, cnt = count(str.substring(i)), output = (counter = (cnt > counter ? cnt : counter)) == cnt ? i : output);
return output;
}

int indexOfLongestRun(String str) {
char[] ar = str.toCharArray();
int longestRun = 0;
int lastLongestRun = 0;
int index = 0;
for(int i = ar.length-1; i>0; i--){
if(ar[i] == ar[i-1]){
longestRun++;
}else{
if(longestRun > lastLongestRun){
lastLongestRun = longestRun;
longestRun = 0;
index = i;
}
}
}
return index;

Well, the solution a bit depends on the additional requirements. Here is the code which returns the FIRST longest sequence of a repeated character int the given string, meaning if you have a second sequence with the same length you never get it out :(. But still, this is a simple and clear solution here, so good news - it works! :)
string = 'abbbccddddddddeehhhfffzzzzzzzzdddvyy'
longest_sequence = ''
for i in range(len(string)):
is_sequence = True
ch_sequence = ''
while is_sequence:
ch_sequence += string[i]
if i+1 < len(string) and string[i]==string[i+1]:
i += 1
else:
is_sequence = False
if len(ch_sequence) > len(longest_sequence):
longest_sequence = ch_sequence
print (longest_sequence)

#Paolo Angioletti already provided an answer using Scala, but it's more complicated than it needs to be. The idea is not very different from Run-length encoding. Time complexity O(n).
def longestConsecutive(s: String): (Char, Int) = {
Iterator.iterate(('\u0000', 0, 0)) { case (ch, longestRun, i) =>
val run = (i until s.length)
.takeWhile(s(_) == s(i))
.size
if (run > longestRun) (s(i), run, i + run)
else (ch, longestRun, i + run)
}
.dropWhile(i => s.isDefinedAt(i._3))
.take(1)
.map(x => (x._1, x._2))
.next()
}
Tested with:
("s", "ch", "n")
----------------
("", '\u0000', 0),
("a", 'a', 1),
("aabcddbbbea", 'b', 3),
("abcddbbb", 'b', 3),
("cbccca", 'c', 3)

#include <iostream>
#include<algorithm>
using namespace std;
int main() {
string s="abbcccccbbffffffffff";
//cin>>s;
int count=1;
int maxcount=1;
int start=0;
int ps=0;
for (int i=0;i<s.size()-1;i++)
{
if(s.at(i)==s.at(i+1))
{
count +=1;
maxcount=max(maxcount,count);
}
else
{
ps=max(ps,start+count);
count =1;
start=i;
}
}
for(int i=1;i<=maxcount;i++)
{
cout<<s.at(i+ps);
}
// your code goes here
return 0;
}

This is the simplest I can think of and it will print the number of the longest sequenced identical characters in a one line string.
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String s = scanner.nextLine();
scanner.close();
int count = 0;
int curCount = 1;
for (int i = 0; i < s.length() -1; i++) {
if (s.charAt(i) == s.charAt(i + 1)) {
curCount++;
if (curCount > count) {
count = curCount;
}
}else {
if (curCount > count) {
count = curCount;
}
curCount = 1;
}
}
System.out.println(count);
}

Check Credit Card Validity using Luhn Algorithm

I tried to check the validation of credit card using Luhn algorithm, which works as the following steps:
Double every second digit from right to left. If doubling of a digit results in a two-digit number, add up the two digits to get a single-digit number.
2 * 2 = 4
2 * 2 = 4
4 * 2 = 8
1 * 2 = 2
6 * 2 = 12 (1 + 2 = 3)
5 * 2 = 10 (1 + 0 = 1)
8 * 2 = 16 (1 + 6 = 7)
4 * 2 = 8
Now add all single-digit numbers from Step 1.
4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37
Add all digits in the odd places from right to left in the card number.
6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38
Sum the results from Step 2 and Step 3.
37 + 38 = 75
If the result from Step 4 is divisible by 10, the card number is valid; otherwise, it is invalid. For example, the number 4388576018402626 is invalid, but the number 4388576018410707 is valid.
Simply, my program always displays valid for everything that I input. Even if it's a valid number and the result of sumOfOddPlace and sumOfDoubleEvenPlace methods are equal to zero. Any help is appreciated.
import java.util.Scanner;
public class CreditCardValidation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
long array[] = new long [16];
do
{
count = 0;
array = new long [16];
System.out.print("Enter your Credit Card Number : ");
long number = in.nextLong();
for (int i = 0; number != 0; i++) {
array[i] = number % 10;
number = number / 10;
count++;
}
}
while(count < 13);
if ((array[count - 1] == 4) || (array[count - 1] == 5) || (array[count - 1] == 3 && array[count - 2] == 7)){
if (isValid(array) == true) {
System.out.println("\n The Credit Card Number is Valid. ");
} else {
System.out.println("\n The Credit Card Number is Invalid. ");
}
} else{
System.out.println("\n The Credit Card Number is Invalid. ");
}
}
public static boolean isValid(long[] array) {
int total = sumOfDoubleEvenPlace(array) + sumOfOddPlace(array);
if ((total % 10 == 0)) {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return true;
} else {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return false;
}
}
public static int getDigit(int number) {
if (number <= 9) {
return number;
} else {
int firstDigit = number % 10;
int secondDigit = (int) (number / 10);
return firstDigit + secondDigit;
}
}
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i=0; i< array.length; i++)
{
while (array[i] > 0) {
result += (int) (array[i] % 10);
array[i] = array[i] / 100;
}}
System.out.println("\n The sum of odd place is " + result);
return result;
}
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
long temp = 0;
for (int i=0; i< array.length; i++){
while (array[i] > 0) {
temp = array[i] % 100;
result += getDigit((int) (temp / 10) * 2);
array[i] = array[i] / 100;
}
}
System.out.println("\n The sum of double even place is " + result);
return result;
}
}

You can freely import the following code:
public class Luhn
{
public static boolean Check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
}
Link reference: https://github.com/jduke32/gnuc-credit-card-checker/blob/master/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java

Google and Wikipedia are your friends. Instead of long-array I would use int-array. On Wikipedia following java code is published (together with detailed explanation of Luhn algorithm):
public static boolean check(int[] digits) {
int sum = 0;
int length = digits.length;
for (int i = 0; i < length; i++) {
// get digits in reverse order
int digit = digits[length - i - 1];
// every 2nd number multiply with 2
if (i % 2 == 1) {
digit *= 2;
}
sum += digit > 9 ? digit - 9 : digit;
}
return sum % 10 == 0;
}
You should work on your input processing code. I suggest you to study following solution:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
boolean repeat;
List<Integer> digits = new ArrayList<Integer>();
do {
repeat = false;
System.out.print("Enter your Credit Card Number : ");
String input = in.next();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c < '0' || c > '9') {
repeat = true;
digits.clear();
break;
} else {
digits.add(Integer.valueOf(c - '0'));
}
}
} while (repeat);
int[] array = new int[digits.size()];
for (int i = 0; i < array.length; i++) {
array[i] = Integer.valueOf(digits.get(i));
}
boolean valid = check(array);
System.out.println("Valid: " + valid);
}

I took a stab at this with Java 8:
public static boolean luhn(String cc) {
final boolean[] dbl = {false};
return cc
.chars()
.map(c -> Character.digit((char) c, 10))
.map(i -> ((dbl[0] = !dbl[0])) ? (((i*2)>9) ? (i*2)-9 : i*2) : i)
.sum() % 10 == 0;
}
Add the line
.replaceAll("\\s+", "")
Before
.chars()
If you want to handle whitespace.
Seems to produce identical results to
return LuhnCheckDigit.LUHN_CHECK_DIGIT.isValid(cc);
From Apache's commons-validator.

There are two ways to split up your int into List<Integer>
Use %10 as you are using and store it into a List
Convert to a String and then take the numeric values
Here are a couple of quick examples
public static void main(String[] args) throws Exception {
final int num = 12345;
final List<Integer> nums1 = splitInt(num);
final List<Integer> nums2 = splitString(num);
System.out.println(nums1);
System.out.println(nums2);
}
private static List<Integer> splitInt(int num) {
final List<Integer> ints = new ArrayList<>();
while (num > 0) {
ints.add(0, num % 10);
num /= 10;
}
return ints;
}
private static List<Integer> splitString(int num) {
final List<Integer> ints = new ArrayList<>();
for (final char c : Integer.toString(num).toCharArray()) {
ints.add(Character.getNumericValue(c));
}
return ints;
}

I'll use 5 digit card numbers for simplicity. Let's say your card number is 12345; if I read the code correctly, you store in array the individual digits:
array[] = {1, 2, 3, 4, 5}
Since you already have the digits, in sumOfOddPlace you should do something like
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i = 1; i < array.length; i += 2) {
result += array[i];
}
return result;
}
And in sumOfDoubleEvenPlace:
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
for (int i = 0; i < array.length; i += 2) {
result += getDigit(2 * array[i]);
}
return result;
}

this is the luhn algorithm implementation which I use for only 16 digit Credit Card Number
if(ccnum.length()==16){
char[] c = ccnum.toCharArray();
int[] cint = new int[16];
for(int i=0;i<16;i++){
if(i%2==1){
cint[i] = Integer.parseInt(String.valueOf(c[i]))*2;
if(cint[i] >9)
cint[i]=1+cint[i]%10;
}
else
cint[i] = Integer.parseInt(String.valueOf(c[i]));
}
int sum=0;
for(int i=0;i<16;i++){
sum+=cint[i];
}
if(sum%10==0)
result.setText("Card is Valid");
else
result.setText("Card is Invalid");
}else
result.setText("Card is Invalid");
If you want to make it use on any number replace all 16 with your input number length.
It will work for Visa number given in the question.(I tested it)

Here's my implementation of the Luhn Formula.
/**
* Runs the Luhn Equation on a user inputed CCN, which in turn
* determines if it is a valid card number.
* #param c A user inputed CCN.
* #param cn The check number for the card.
* #return If the card is valid based on the Luhn Equation.
*/
public boolean luhn (String c, char cn)
{
String card = c;
String checkString = "" + cn;
int check = Integer.valueOf(checkString);
//Drop the last digit.
card = card.substring(0, ( card.length() - 1 ) );
//Reverse the digits.
String cardrev = new StringBuilder(card).reverse().toString();
//Store it in an int array.
char[] cardArray = cardrev.toCharArray();
int[] cardWorking = new int[cardArray.length];
int addedNumbers = 0;
for (int i = 0; i < cardArray.length; i++)
{
cardWorking[i] = Character.getNumericValue( cardArray[i] );
}
//Double odd positioned digits (which are really even in our case, since index starts at 0).
for (int j = 0; j < cardWorking.length; j++)
{
if ( (j % 2) == 0)
{
cardWorking[j] = cardWorking[j] * 2;
}
}
//Subtract 9 from digits larger than 9.
for (int k = 0; k < cardWorking.length; k++)
{
if (cardWorking[k] > 9)
{
cardWorking[k] = cardWorking[k] - 9;
}
}
//Add all the numbers together.
for (int l = 0; l < cardWorking.length; l++)
{
addedNumbers += cardWorking[l];
}
//Finally, check if the number we got from adding all the other numbers
//when divided by ten has a remainder equal to the check number.
if (addedNumbers % 10 == check)
{
return true;
}
else
{
return false;
}
}
I pass in the card as c which I get from a Scanner and store in card, and for cn I pass in checkNumber = card.charAt( (card.length() - 1) );.

Okay, this can be solved with a type conversions to string and some Java 8
stuff. Don't forget numbers and the characters representing numbers are not the same. '1' != 1
public static int[] longToIntArray(long cardNumber){
return Long.toString(cardNumber).chars()
.map(x -> x - '0') //converts char to int
.toArray(); //converts to int array
}
You can now use this method to perform the luhn algorithm:
public static int luhnCardValidator(int cardNumbers[]) {
int sum = 0, nxtDigit;
for (int i = 0; i<cardNumbers.length; i++) {
if (i % 2 == 0)
nxtDigit = (nxtDigit > 4) ? (nxtDigit * 2 - 10) + 1 : nxtDigit * 2;
sum += nxtDigit;
}
return (sum % 10);
}

private static int luhnAlgorithm(String number){
int n=0;
for(int i = 0; i<number.length(); i++){
int x = Integer.parseInt(""+number.charAt(i));
n += (x*Math.pow(2, i%2))%10;
if (x>=5 && i%2==1) n++;
}
return n%10;
}

public class Creditcard {
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
String cardno = sc.nextLine();
if(checkType(cardno).equals("U")) //checking for unknown type
System.out.println("UNKNOWN");
else
checkValid(cardno); //validation
}
private static String checkType(String S)
{
int AM=Integer.parseInt(S.substring(0,2));
int D=Integer.parseInt(S.substring(0,4)),d=0;
for(int i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
d++;
}
if((AM==34 || AM==37) && d==15)
System.out.println("AMEX");
else if(D==6011 && d==16)
System.out.println("Discover");
else if(AM>=51 && AM<=55 && d==16)
System.out.println("MasterCard");
else if(((S.charAt(0)-'0')==4)&&(d==13 || d==16))
System.out.println("Visa");
else
return "U";
return "";
}
private static void checkValid(String S) // S--> cardno
{
int i,d=0,sum=0,card[]=new int[S.length()];
for(i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
card[d++]=S.charAt(i)-'0';
}
for(i=0;i<d;i++)
{
if(i%2!=0)
{
card[i]=card[i]*2;
if(card[i]>9)
sum+=digSum(card[i]);
else
sum+=card[i];
}
else
sum+=card[i];
}
if(sum%10==0)
System.out.println("Valid");
else
System.out.println("Invalid");
}
public static int digSum(int n)
{
int sum=0;
while(n>0)
{
sum+=n%10;
n/=10;
}
return sum;
}
}

Here is the implementation of Luhn algorithm.
public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "4071690065031703";
System.out.println(checkLuhn(cardNum));
}
}

public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "8112189875";
System.out.println(checkLuhn(cardNum));
}
}
Hope it may works.

const options = {
method: 'GET',
headers: {Accept: 'application/json', 'X-Api-Key': '[APIkey]'}
};
fetch('https://api.epaytools.com/Tools/luhn?number=[CardNumber]&metaData=true', options)
.then(response => response.json())
.then(response => console.log(response))
.catch(err => console.error(err));

java: how to loop that i get exactly 3^10 (disjunct)values in 3 columns?

I want to make a list of 3 values of the following kind (binary notation!):
0;0;0
1;0;0
0;1;0
11;0;0
10;1;0
.
.
.
1111111111;0;0
0;1111111111;0
0;0;1111111111
and all missing values in between
this list means: all columns have to have all values(permutations?), but only if the bit is not set in another column
this is the problem of putting 10 markable things into 3 different boxes
i tried 3 loops but i always mess it up :(
thats what i have so far:
import java.io.FileNotFoundException;
import java.math.BigInteger;
public class create_referencevalues {
/**
* #param args
*/
public static void main(String[] args) {
Long[] list = { 10L, 40L, 90L, 160L, 250L, 350L, 500L, 650L,800L,1000L };
try {
java.io.PrintStream p = new java.io.PrintStream(
new java.io.BufferedOutputStream(
new java.io.FileOutputStream(new java.io.File(
"C:/users/djdeejay/listall.csv"), false)));
for (Integer i = 0; i < 1024; i++) {
Long sum1 = 0L;
for (Integer j = 0; j < 10; j++) {
if (BigInteger.valueOf(i).testBit(j)) {
sum1 += (list[j]);
}
}
sum1 *= Integer.bitCount(i);
Long sum2 = 0L;
for (int j = 0; j < 10; j++) {
if (BigInteger.valueOf(1023 - i).testBit(j)) {
sum2 += (list[j]);
}
}
sum2 *= 10-Integer.bitCount(i);
p.println(i +";"+ Long.toBinaryString(i)+";" + sum1+";"+ Long.toBinaryString(1023-i)+";"+sum2);
}
p.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
}
how do i integrate a loop wich gives me the 3rd row and all permutations between the rows ?
thanks for any help

Try this :
public class Permutation {
private static final int NB_DIGITS = 10;
private int[] DIGIT_NUMBER = {1,10,100,1000,10000,100000,1000000,10000000,100000000, 1000000000};
public void dump(PrintStream printStream) {
int[] counter = new int[NB_DIGITS];
Arrays.fill(counter, 0);
do {
int column1 = 0;
int column2 = 0;
int column3 = 0;
for (int i = 0; i < NB_DIGITS; i++) {
int columnIdx = counter[i];
switch (columnIdx) {
case 0 : column1+=DIGIT_NUMBER[i];break;
case 1 : column2+=DIGIT_NUMBER[i];break;
case 2 : column3+=DIGIT_NUMBER[i];break;
default:
assert false;
}
}
printStream.format("%d;%d;%d%n", column1, column2, column3);
} while (increase(counter));
}
public boolean increase(int[] counter) {
int idx = 0;
while (idx < counter.length && counter[idx] == 2) {
counter[idx] = 0;
idx++;
}
if (idx == counter.length) {
return false;
}
counter[idx]++;
return true;
}
public static void main(String[] args) {
new Permutation().dump(System.out);
}
}
The counter array contains the column index where to put your digits. This is actually a loop by hand in base 3 between 0 and 3^10-1 so you reach all the possible positions.

I found this solution:
loop all three rows by all values and mask values you dont want:
if (((k & j) == 0) && ((k & i) == 0) && ((j & i) == 0) && (k^j^i)==max)
that means: a bit at a position must be set only in one row and the rows together must have set all bits
package com.djdeejay.cowTrade.client.standaloneplayer.application;
import java.io.FileNotFoundException;
import java.math.BigInteger;
public class create_referencevalues {
/**
* #param args
*/
public static void main(String[] args) {
Long[] list = { 10L, 40L, 90L, 160L, 250L, 350L, 500L, 650L, 800L,
1000L };
try {
java.io.PrintStream p = new java.io.PrintStream(
new java.io.BufferedOutputStream(
new java.io.FileOutputStream(new java.io.File(
"C:/users/djdeejay/listall.csv"), false)));
Long i = 0l;
Long j = 0l;
Long k = 0l;
Long max = 1023L;
Long count = 0l;
for (i = 0l; i < max; i++) {
for (j = 0l; j < max; j++) {
for (k = 0l; k < max; k++) {
String col1 = Long.toBinaryString(i);
String col2 = Long.toBinaryString(j);
String col3 = Long.toBinaryString(k);
if (((k & j) == 0) && ((k & i) == 0)
&& ((j & i) == 0) && (k^j^i)==max) {
count++;
Long sum1 = 0L, sum2 = 0L, sum3 = 0L;
for (int x = 0; x < 10; x++) {
if (BigInteger.valueOf(i).testBit(x)) {
sum1 += (list[x]);
}
if (BigInteger.valueOf(j).testBit(x)) {
sum2 += (list[x]);
}
if (BigInteger.valueOf(k).testBit(x)) {
sum3 += (list[x]);
}
}
sum1 *= Long.bitCount(i);
sum2 *= Long.bitCount(j);
sum3 *= Long.bitCount(k);
p.println(count + ";" + i + ";" + j + ";" + k);
System.out.println(count + ";" + col1 + ";" + col2 + ";" + col3);
}
}
}
}
System.out.println(count + ";" + i + ";" + j + ";" + k);
p.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
}

how to factor a number java

I need to factorize a number like 24 to 1,2,2,2,3. My method for that:
static int[] factorsOf (int val) {
int index = 0;
int []numArray = new int[index];
System.out.println("\nThe factors of " + val + " are:");
for(int i=1; i <= val/2; i++)
{
if(val % i == 0)
{
numArray1 [index] = i;
index++;
}
}
return numArray;
}
however, it is not working. Can anyone help me for that?

You have a few errors, you cannot create int array without size. I used array list instead.
static Integer[] factorsOf(int val) {
List<Integer> numArray = new ArrayList<Integer>();
System.out.println("\nThe factors of " + val + " are:");
for (int i = 2; i <= Math.ceil(Math.sqrt(val)); i++) {
if (val % i == 0) {
numArray.add(i);
val /= i;
System.out.print(i + ", ");
}
}
numArray.add(val);
System.out.print(val);
return numArray.toArray(new Integer[numArray.size()]);
}
Full program using int[] according to your request.
public class Test2 {
public static void main(String[] args) {
int val = 5;
int [] result = factorsOf(val);
System.out.println("\nThe factors of " + val + " are:");
for(int i = 0; i < result.length && result[i] != 0; i ++){
System.out.println(result[i] + " ");
}
}
static int[] factorsOf(int val) {
int limit = (int) Math.ceil(Math.sqrt(val));
int [] numArray = new int[limit];
int index = 0;
for (int i = 1; i <= limit; i++) {
if (val % i == 0) {
numArray[index++] = i;
val /= i;
}
}
numArray[index] = val;
return numArray;
}
}

public int[] primeFactors(int num)
{
ArrayList<Integer> factors = new ArrayList<Integer>();
factors.add(1);
for (int a = 2; num>1; )
if (num%a==0)
{
factors.add(a);
num/=a;
}
else
a++;
int[] out = new int[factors.size()];
for (int a = 0; a < out.length; a++)
out[a] = factors.get(a);
return out;
}

Are you looking a more faster way?:
static int[] getFactors(int value) {
int[] a = new int[31]; // 2^31
int i = 0, j;
int num = value;
while (num % 2 == 0) {
a[i++] = 2;
num /= 2;
}
j = 3;
while (j <= Math.sqrt(num) + 1) {
if (num % j == 0) {
a[i++] = j;
num /= j;
} else {
j += 2;
}
}
if (num > 1) {
a[i++] = num;
}
int[] b = Arrays.copyOf(a, i);
return b;
}

Most of the approaches suggested here have 0(n) time complexity. This can be easily resolved using binary search approach with 0(log n) time complexity.
What do you basically are looking for is called Prime Factors (although 1 is not considered among prime factors).
//finding any occurrence of the no by binary search
static int[] primeFactors(int number) {
List<Integer> al = new ArrayList<Integer>();
//since you wanted 1 in the res adn every no will be divided by 1;
al.add(1);
for(int i = 2; i< number; i++) {
while(number%i == 0) {
al.add(i);
number = number/i;
}
}
if(number >2)
al.add(number);
int[] res = new int[al.size()];
for(int i=0; i<al.size(); i++)
res[i] = al.get(i);
return res;
}
Say input is 24, we keep dividing the input by 2 till all multiples of 2 are gone, the increase i to 3
Here is a link to the working code: http://tpcg.io/AWH2TJ

A Working example
public class Main
{
public static void main(String[] args)
{
System.out.println(factorsOf(24));
}
static List<Integer> factorsOf (int val) {
List<Integer> factors = new ArrayList<Integer>();
for(int i=1; i <= val/2; i++)
{
if(val % i == 0)
{
factors.add(i);
}
}
return factors;
}
}

Rework an algorithm for working with big numbers using BigInteger class. Try this:
import java.math.BigInteger;
class NumbersFactorization {
public void printPrimeNumbers(String bigNumber) {
BigInteger number = new BigInteger(bigNumber);
for (BigInteger i = BigInteger.TWO; i.compareTo(number) <= 0; i = i.add(BigInteger.ONE)) {
while(number.remainder(i) == BigInteger.ZERO) {
System.out.print(i + " ");
number = number.divide(i);
}
}
if (number.compareTo(BigInteger.TWO) > 0) System.out.println(number);
}
}

You just missed one step in if. Following code would be correct:
System.out.println("\nThe factors of " + val + " are:");
You can take a square root of val for comparison and start iterator by value 2
if(val % i == 0)
{
numArray1 [index] = i;
val=val/i; //add this
index++;
}
but here you need to check if index is 2,it is prime.