We Keep Coding

Find the Least Common Multiple (LCM) of exactly N-1 numbers chosen among N numbers

Find the smallest number M, which is divided by exactly n-1 numbers from the input array. If there is no such M then return -1.
Example:
array = [2,3,5]
Answer :
6
Explanation :
6 can be divided by 2 and 3
Example:
array = [2,3,6]
Answer:
-1
Explanation :
It's not possible in this case so return -1.
My code:
As we need to find the smallest M, I am selecting only the elements from 0 to n-2
public int process(int[] arr) {
int answer = 1;
for(int i=0; i<arr.length-1; i++) {
answer *= arr[i];
}
return answer;
}
This program works for these 2 sample test cases but it was failing for multiple hidden test cases. I trying to understand what I am missing here.

Calculation of the Least Common Multiple (LCM)
A problem inside the task is the calculation of the Least Common Multiple of 2 numbers. The method public int lowerCommonMultiple(int x1, int x2) solves this problem and I think it can be used in other context.
List of the class methods
All the code is included in the methods of the BestMultiple class. These methods (excluding the main) are:
public int[] removeElem(int[] tensArray, int rm_index): used to remove an element from an array
public int leastCommonMultiple(int x1, int x2): calculates the Least Common Multiple of 2 numbers
private int getLeastCommonMultipleNnumber(int[] arr): Calculates the least common multiple of N-1 integer contain in an array
public int process(int[] arr): calculates the least multiple of exactly N-1 number of an array of N integer; it manages many test strange cases (array empty, elem<=0, etc.)
May be the code is not optimized, but I hope it is correct (the output added, shows that it works correctly, at least with the test cases chosen).
public class BestMultiple {
/*++++++++++++++++++++++++++++++++++++++++++++
Method: removeElem() remove an element from
an array.
+++++++++++++++++++++++++++++++++++++++++++*/
public int[] removeElem(int[] tensArray, int rm_index) {
// Create a proxy array of size one less than original array
int[] proxyArray = new int[tensArray.length - 1];
// copy all the elements in the original to proxy array
// except the one at index
for (int i = 0, k = 0; i < tensArray.length; i++) {
// check if index is crossed, continue without copying
if (i == rm_index) {
continue;
}
// else copy the element
proxyArray[k++] = tensArray[i];
}
return proxyArray;
}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Method: leastCommonMultiple() Calculates the Least Common
multiple for 2 numbers
++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
public int leastCommonMultiple(int x1, int x2) {
int lcm = 1;
int max = x1;
if ((x1 == 0) || (x2 == 0)) {
lcm = 0;
} else {
if (x2 > x1) {
max = x2;
}
for (int i = 2; i <= max; i++) {
int exp_x1 = 0;
int exp_x2 = 0;
int exp = 0;
if (x1 > 1) {
while ((x1 % i) == 0) {
exp_x1++;
x1 /= i;
}
}
if (x2 > 1) {
while ((x2 % i) == 0) {
exp_x2++;
x2 /= i;
}
}
if ((exp_x1 > 0) || (exp_x2 > 0)) {
exp = exp_x1;
if (exp_x2 > exp) {
exp = exp_x2;
}
while (exp > 0) {
lcm *= i;
exp--;
}
}
}
}
return lcm;
}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Method: getLeastCommonMultipleNnumber()
Calculates the least common multiple of N-1
integer contain in an array
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
public int getLeastCommonMultipleNnumber(int[] arr) {
int multiple = 1;
if (arr.length >= 2) {
multiple = leastCommonMultiple(arr[0], arr[1]);
for (int j = 2; j < arr.length; j++) {
multiple = leastCommonMultiple(multiple, arr[j]);
}
} else {
// array with only 2 elements
multiple = arr[0];
}
return multiple;
}
/*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Method: process()
Calculates the least multiple of EXACTLY N-1
number of an array of N integer
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++*/
public int process(int[] arr) {
int answer;
if (arr.length <= 1) {
// array contains only one element or is empty => return -1
answer = -1;
} else {
int pos_elem_zero = -1;
int prod = 1;
for (int i = 0; i < arr.length; i++) {
if (arr[i] > 0) {
prod *= arr[i];
} else {
if (arr[i] < 0) {
// integer < 0 are not allowed
return -1;
}
if (pos_elem_zero == -1) {
pos_elem_zero = i;
} else {
// there are more element == 0
return -1;
}
}
}
if (pos_elem_zero >= 0) {
// there is one element == 0
arr = this.removeElem(arr, pos_elem_zero);
return getLeastCommonMultipleNnumber(arr);
}
// managing of normal test case
answer = prod;
for (int i = 0; i < arr.length; i++) {
int elem = arr[i];
int[] arr2 = this.removeElem(arr, i);
int multiple = getLeastCommonMultipleNnumber(arr2);
if (multiple > elem) {
if ((multiple % elem) != 0) {
if (multiple < answer) {
answer = multiple;
}
}
} else {
if (multiple < elem) {
answer = multiple;
}
}
}
if (answer == prod) {
answer = -1;
}
}
return answer;
}
/*++++++++++++++++++++++++++++++++++++++++++
Method: main() Executes test of process()
method
+++++++++++++++++++++++++++++++++++++++++*/
public static void main(String[] args) {
BestMultiple bm = new BestMultiple();
int[] arr1 = {6,30,5,3};
int[] arr2 = {1,2,3};
int[] arr3 = {1,2,3,3};
int[] arr4 = {6,7,5,3};
int[] arr5 = {9,14, 21};
int[] arr6 = {2,4};
int[] arr7 = {2,3,5};
int[] arr8 = {2,3,6};
int[] arr9 = {2};
int[] arr10 = {};
int[] arr11 = {2,3,0};
int[] arr12 = {0,2,3,0};
int[] arr13 = {20,3};
int[] arr14 = {0,6,15};
int[] arr15 = {1,6,15,-1};
int[] arr16 = {1,6,15};
int[] arr17 = {2,3,0,6,15};
System.out.println("{6,30,5,3} --> " + bm.process(arr1));
System.out.println("{1,2,3} --> " + bm.process(arr2));
System.out.println("{1,2,3,3} --> " + bm.process(arr3));
System.out.println("{6,7,5,3} --> " + bm.process(arr4));
System.out.println("{9,14,21} --> " + bm.process(arr5));
System.out.println("{2,4} --> " + bm.process(arr6));
System.out.println("{2,3,5} --> " + bm.process(arr7));
System.out.println("{2,3,6} --> " + bm.process(arr8));
System.out.println("{2} --> " + bm.process(arr9));
System.out.println("{} --> " + bm.process(arr10));
System.out.println("{2,3,0} --> " + bm.process(arr11));
System.out.println("{0,2,3,0} --> " + bm.process(arr12));
System.out.println("{20,3} --> " + bm.process(arr13));
System.out.println("{0,6,15} --> " + bm.process(arr14));
System.out.println("{1,6,15,-1} --> " + bm.process(arr15));
System.out.println("{1,6,15} --> " + bm.process(arr16));
System.out.println("{2,3,0,6,15} --> " + bm.process(arr17));
}
}
Output of the program
The output of the program with the test cases chosen is:
{6,30,5,3} --> -1
{1,2,3} --> 2
{1,2,3,3} --> 3
{6,7,5,3} --> 30
{9,14,21} --> 42
{2,4} --> 2
{2,3,5} --> 6
{2,3,6} --> -1
{2} --> -1
{} --> -1
{2,3,0} --> 6
{0,2,3,0} --> -1
{20,3} --> 3
{0,6,15} --> 30
{1,6,15,-1} --> -1
{1,6,15} --> 6
{2,3,0,6,15} --> 30

You start by writing a method process that computes the minimum number for each subarray with one element excluded:
public static int process(int... arr) {
int min = -1;
for (int i = 0; i < arr.length; ++i) {
int r = process(arr, i);
if (r != -1) {
if (min == -1) {
min = r;
} else {
min = Math.min(min, r);
}
}
}
return min;
}
Where the second process method looks like this:
private static int process(int[] arr, int exclude) {
int result = 0;
for (int i = 0; i < arr.length; ++i) {
if (i != exclude) {
if (result == 0) {
result = arr[i];
} else {
result = lcm(result, arr[i]);
}
}
}
if (result%arr[exclude] == 0) {
return -1;
}
return result;
}
You need a method that computes the LCM of two numbers. Here, I'll use a second method that computes the GCD:
private static int lcm(int a, int b) {
return a*b/gcd(a,b);
}
private static int gcd(int a, int b) {
if (a == 0) {
return b;
} else if (b == 0) {
return a;
} else {
while (a != b) {
if (a > b) {
a -= b;
} else {
b -= a;
}
}
return a;
}
}
Examples:
System.out.println(process(2, 3, 5)); // prints 6
System.out.println(process(2, 3, 6)); // prints -1
System.out.println(process(9, 14, 21)); // prints 42 (divisible by 14 and 21, but not by 9
System.out.println(process(6, 7, 5, 3)); // prints 30 (divisible by 6, 5, 3, but not by 7

The way you implemented process() assumes the input array is sorted. But anyway, I don't think sorting will help here. Note that the number satisfying the given conditions can be divided by the biggest number. For [2, 3, 5, 6] it is 6. Dividing the product of all the elements by consecutive elements from the biggest to the lowest and stopping at the first that is not a divisor is also not correct. In the example [2, 4, 5, 6] this would give 2 * 4 * 5 = 40 when the correct answer is 20.
My idea is to use an algorithm inspired by Sieve of Eratosthenes. Note that the number that satisfies the conditions can't be bigger than the product of all the elements. So create a table divisors[] with indices from 0 through the product of the elements of array where divisors[i] indicates how many elements from array divide i. Iterate over elements of array and increment all elements in divisors[i] where i is divided by the element. Then find the first i for which divisors[i] == n - 1.
The limitation is that divisors can be quite big depending on what is the product of array, so applicability will be limited to relatively small values in array.

How to encode a number using its prime factors in java using arrays?

I have this question I am trying to solve
I wrote this code
public static int[] encodeNumber(int n) {
int count = 0, base = n, mul = 1;
for (int i = 2; i < n; i++) {
if(n % i == 0 && isPrime(i)) {
mul *= i;
count++;
if(mul == n) {
break;
}
n /= i;
}
}
System.out.println("count is " + count);
int[] x = new int[count];
int j = 0;
for (int i = 2; i < base; i++) {
if(n % i == 0 && isPrime(i)) {
mul *= i;
x[j] = i;
j++;
if(mul == n) break;
n /= i;
}
break;
}
return x;
}
public static boolean isPrime(int n) {
if(n < 2) return false;
for (int i = 2; i < n; i++) {
if(n % i == 0) return false;
}
return true;
}
I am trying to get the number of its prime factors in a count variable and create an array with the count and then populate the array with its prime factors in the second loop.
count is 3
[2, 0, 0]
with an input of 6936. The desired output is an array containing all its prime factors {2, 2, 2, 3, 17, 17}.

Your count is wrong, because you count multiple factors like 2 and 17 of 6936 only once.
I would recommend doing it similar to the following way, recursively:
(this code is untested)
void encodeNumberRecursive(int remainder, int factor, int currentIndex, Vector<Integer> results) {
if(remainder<2) {
return;
}
if(remainder % factor == 0) {
results.push(factor);
remainder /= factor;
currentIndex += 1;
encodeNumberRecursive(remainder , factor, currentIndex, results);
} else {
do {
factor += 1;
} while(factor<remainder && !isPrime(factor));
if(factor<=remainder) {
encodeNumberRecursive(remainder , factor, currentIndex, results);
}
}
}
Finally, call it with
Vector<Integer> results = new Vector<Integer>();
encodeNumberRecursive(n, 2, 0, results);
You can also do it without recursion, I just feel it is easier.

Well here is a piece of code I would start with. It is not finished yet and I did not test it, but that's the way you should go basically.
// First find the number of prime factors
int factorsCount = 0;
int originalN = n;
while (n > 1) {
int p = findLowestPrimeFactor(n);
n /= p;
factorsCount++;
}
// Now create the Array of the appropriate size
int[] factors = new int[factorsCount];
// Finally do the iteration from the first step again, but now filling the array.
n = originalN;
int k = 0;
while (n > 1) {
int p = findLowestPrimeFactor(n);
factors[k] = p;
k++;
n = n / p;
}
return factors;

Having found a factor (on increasing candidates), you can assume it is prime,
if you divide out the factor till the candidate no longer is a factor.
Your problem is not repeatedly dividing by the factor.
public static int[] encodeNumber(int n) {
if (n <= 1) {
return null;
}
List<Integer> factors = new ArrayList<>();
for (int i = 2; n != 1; i += 1 + (i&1)) {
while (n % i == 0) { // i is automatically prime, as lower primes done.
factors.add(i);
n /= i;
}
}
return factors.stream().mapToInt(Integer::intValue).toArray();
}
Without data structures, taking twice the time:
public static int[] encodeNumber(int n) {
if (n <= 1) {
return null;
}
// Count factors, not storing them:
int factorCount = 0;
int originalN = n;
for (int i = 2; n != 1; i += 1 + (i&1)) {
while (n % i == 0) {
++factorCount;
n /= i;
}
}
// Fill factors:
n = originalN;
int[] factors = new int[factorCount];
factorCount = 0;
for (int i = 2; n != 1; i += 1 + (i&1)) {
while (n % i == 0) {
factors[factorCount++] = i;
n /= i;
}
}
return factors;
}

Prime number generator between 2 nums

I cant figure out which test cases the code provided below fails.
problem:
All submissions for this problem are available.
Shridhar wants to generate some prime numbers for his cryptosystem. Help him!
Your task is to generate all prime numbers between two given numbers.
Input
The first line contains t, the number of test cases (less then or equal to 10).
Followed by t lines which contain two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n,
one number per line. Separate the answers for each test case by an empty line.
Example
Input:
2
1 10
3 5
Output:
2
3
5
7
3
5
import java.util.Scanner;
public class Main implements Runnable {
public static void main(String args[]) {
new Main().run();
}
#Override
public void run() {
Scanner sc = new Scanner(System.in);
Integer d;
try {
d = sc.nextInt();
boolean isPrime[] = new boolean[100000];
for (int i = 0; i < d; i++) {
int m = sc.nextInt();
int n = sc.nextInt();
if (n <= 0 || m > n) {
continue;
}
if (m <= 0) {
m = 2;
if (m > n) {
continue;
}
}
if (m == 1) {
m = 2;
}
if (m == 2 && n - m == 0) {
System.out.println(2);
} else {
for (int k = 0; k <= n - m; k++) {
isPrime[k] = true;
}
int sqrt = (int) Math.sqrt(n);
for (int j = 2; j <= sqrt; j++) {
int k = (m % j == 0) ? m / j : (m + j) / j;
for (; k <= n / j; k++) {
if (!(m == 2 && (j * k == 2)) && k != 1) {
isPrime[j * k - m] = false;
}
}
}
for (int a = m; a <= n; a++) {
if (isPrime[a - m]) {
System.out.println(a);
}
}
}
System.out.println();
}
} catch (Exception e) {
e.printStackTrace();
} finally {
sc.close();
}
}
}

Your problem is with this:
boolean isPrime[] = new boolean[100000];
and this:
for (int k = 0; k <= n - m; k++) {
isPrime[k] = true;
}
because some time n - m = 100000 then you need isPrime[100000] which you didnt allocate so you need to declare isPrime like this:
boolean isPrime[] = new boolean[100001];

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));

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.