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count all numbers in range having a specific digit

I want to count how many numbers have digit number 4 in a range of numbers
e.g 1-100 count all numbers having digit number
i.e 4,14,24,34,40,41,42,43,44,45,46,47,48,49,54,64,74,84 and 94 total 19 numbers
I am having problems with counting number of integers with digit 4 in them please help!!
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int count = 0;
while (true) {
Scanner input = new Scanner(System.in);
int num1 = input.nextInt();
int num2 = input.nextInt();
if (num1 != 0 || num2 != 0) {
for (int num = num1; num <= num2; num++) {
while (num != 0) {
int i = num % 10;
if (i == 4) {
count++;
break;
}
num = num / 10;
}
}
System.out.println(count);
}
else
break;
}
}
}

To check if a number contains a specific digit or not, here is a trick you can use :
Convert both number and digit to String and use String::contains (Simple)
Here is a piece of code you can use If you are using Java 8 :
int number = 4, min = 0, max = 100;
String numberToString = String.valueOf(number);
long count = IntStream.rangeClosed(min, max) //Range of numbers between min and max
.filter(n -> String.valueOf(n).contains(numberToString)) // Use the filter
.count();// Then count the result
System.out.println(count); // 19

//a few changes were needed into your code.. here is the solution...
public class MaxOccurance {
public static void main(String[] args) {
int count = 0;
Scanner input = new Scanner(System.in);
int num1 = input.nextInt();
int num2 = input.nextInt(), temp;
if (num1 != 0 || num2 != 0) {
for (int num = num1; num <= num2; num++) {
temp = num;
while (temp != 0) {
int i = temp % 10;
if (i == 4) {
count++;
// break;
}
temp = temp / 10;
}
}
System.out.println(count);
}
}
}

Can't figure out the error Luhn check

Its supose to tell me if a card is valid or invalid using luhn check
4388576018402626 invalid
4388576018410707 valid
but it keeps telling me that everything is invalid :/
Any tips on what to do, or where to look, would be amazing. I have been stuck for a few hours.
It would also help if people tell me any tips on how to find why a code is not working as intended.
im using eclipse and java
public class Task11 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a credit card number as a long integer: ");
long number = input.nextLong();
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
}
public static boolean isValid(long number) {
return (getSize(number) >= 13) && (getSize(number) <= 16)
&& (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37))
&& (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 == 0;
}
public static int sumOfDoubleEvenPlace(long number) {
int result = 0;
long start = 0;
String digits = Long.toString(number);
if ((digits.length() % 2) == 0) {
start = digits.length() - 1;
} else {
start = digits.length() - 2;
}
while (start != 0) {
result += (int) ((((start % 10) * 2) % 10) + (((start % 10) * 2) / 2));
start = start / 100;
}
return result;
}
public static int getDigit(int number) {
return number % 10 + (number / 10);
}
public static int sumOfOddPlace(long number) {
int result = 0;
while (number != 0) {
result += (int) (number % 10);
number = number / 100;
}
return result;
}
public static boolean prefixMatched(long number, int d) {
return getPrefix(number, getSize(d)) == d;
}
public static int getSize(long d) {
int numberOfDigits = 0;
String sizeString = Long.toString(d);
numberOfDigits = sizeString.length();
return numberOfDigits;
}
public static long getPrefix(long number, int k) {
String size = Long.toString(number);
if (size.length() <= k) {
return number;
} else {
return Long.parseLong(size.substring(0, k));
}
}
}

You should modiffy your isValid() method to write down when it doesn't work, like this:
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
System.out.println("Err: Number "+number+" is too long");
return false;
} else if (! (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37)) ){
System.out.println("Err: Number "+number+" prefix doesn't match");
return false;
} else if( (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 != 0){
System.out.println("Err: Number "+number+" doesn't have sum of odd and evens % 10. ");
return false;
}
return true;
}
My guess for your problem is on the getPrefix() method, you should add some logs here too.
EDIT: so, got more time to help you (don't know if it's still necessary but anyway). Also, I corrected the method I wrote, there were some errors (like, the opposite of getSize(number) >= 13 is getSize(number) < 13)...
First it will be faster to test with a set of data instead of entering the values each time yourself (add the values you want to check):
public static void main(String[] args) {
long[] luhnCheckSet = {
0, // too short
1111111111111111111L, // too long (19)
222222222222222l // prefix doesn't match
4388576018402626l, // should work ?
};
//System.out.print("Enter a credit card number as a long integer: ");
//long number = input.nextLong();
for(long number : luhnCheckSet){
System.out.println("Checking number: "+number);
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
System.out.println("-");
}
}
I don't know the details of this, but I think you should work with String all along, and parse to long only if needed (if number is more than 19 characters, it might not parse it long).
Still, going with longs.
I detailed your getPrefix() with more logs AND put the d in parameter in long (it's good habit to be carefull what primitive types you compare):
public static boolean prefixMatched(long number, long d) {
int prefixSize = getSize(d);
long numberPrefix = getPrefix(number, prefixSize);
System.out.println("Testing prefix of size "+prefixSize+" from number: "+number+". Prefix is: "+numberPrefix+", should be:"+d+", are they equals ? "+(numberPrefix == d));
return numberPrefix == d;
}
Still don't know what's wrong with this code, but it looks like it comes from the last test:
I didn't do it but you should make one method from sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 and log both numbers and the sum (like i did in prefixMatched() ). Add logs in both method to be sure it gets the result you want/ works like it should.
Have you used a debugger ? if you can, do it, it can be faster than adding a lot of logs !
Good luck

EDIT:
Here are the working functions and below I provided a shorter, more efficient solution too:
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. ");
}
in.close();
}
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;
}
}
I also found a solution with less lines of logic. I know you're probably searching for an OO approach with functions, building from this could be of some help.
Similar question regarding error in Luhn algorithm logic:
Check Credit Card Validity using Luhn Algorithm
Link to shorter solution:
https://code.google.com/p/gnuc-credit-card-checker/source/browse/trunk/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java
And here I tested the solution with real CC numbers:
public class CreditCardValidation{
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);
}
public static void main(String[] args){
//String num = "REPLACE WITH VALID NUMBER"; //Valid
String num = REPLACE WITH INVALID NUMBER; //Invalid
num = num.trim();
if(Check(num)){
System.out.println("Valid");
}
else
System.out.println("Invalid");
//Check();
}
}

Why is my method repeating when I run

public class PalindromicPrimes {
public static void main (String[] args) {
userInt();
System.out.println("The palindromic primes less than " + userInt() +
" are:");
for (int i = 0; i <= userInt(); i++) {
if (isPrime() && isPalindrome()) {
System.out.println(i);
}
}
}
private static boolean isPrime() {
if (userInt() == 2 || userInt() == 3) {
return true;
}
if (userInt() % 2 == 0) {
return false;
}
int sqrt = (int) Math.sqrt(userInt()) + 1;
for (int i = 3; i < sqrt; i += 2) {
if (userInt() % i == 0) {
return false;
}
}
return true;
}
private static boolean isPalindrome() {
if (userInt() < 0)
return false;
int div = 1;
while (userInt() / div >= 10) {
div *= 10;
}
while (userInt() != 0) {
int x = userInt();
int l = x / div;
int r = x % 10;
if (l != r)
return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
private static int userInt() {
Scanner s = new Scanner(System.in);
System.out.print("Enter a positive integer: ");
int userInt = s.nextInt();
return userInt;
}
}
is there a different way of getting the user input? or can I keep it this way?
when it runs it just keeps prompting the user input.

rearrange it like this:
public static void main (String[] args) {
//get it and save it here!
int userValue = userInt();
System.out.println("The palindromic primes less than " + userValue +
" are:");
for (int i = 0; i <= userValue; i++) {
if (isPrime(userValue) && isPalindrome(userValue)) {
System.out.println(i);
}
}
}
then also update all the methods that care about this "userInt" value.
Every time you call userInt() you're telling the code to get a new value from the command line.

Try this:
public static void main (String[] args) {
int value = userInt();
System.out.println("The palindromic primes less than " + value +
" are:");
for (int i = 0; i <= value; i++) {
if (isPrime() && isPalindrome()) {
System.out.println(i);
}
}
}
The term userInt() is a function invocation that prompts the user for input. Odds are you only want to do this once. You're doing it multiple times.

You should store the result of userInt() in a variable.
int typed = userInt();
And then use this variable to reference what the user typed instead of calling userInt() again.
System.out.println("The palindromic primes less than " + typed +
" are:");
for(int i = 0; i < typed; i++) ...

You keep calling userInt(). That is the problem.
I don't understand your logic. So I have not modified that code. But the code runs.
import java.util.Scanner;
public class PalindromicPrimes {
public static void main (String[] args) {
int x = userInt();
System.out.println("The palindromic primes less than " + x +
" are:");
for (int i = 0; i <= x; i++) {
if (isPrime(i) && isPalindrome(i)) {
System.out.println(i);
}
}
}
private static boolean isPrime(int a) {
if (a == 2 || a == 3) {
return true;
}
if (a % 2 == 0) {
return false;
}
int sqrt = (int) Math.sqrt(a) + 1;
for (int i = 3; i < sqrt; i += 2) {
if (a % i == 0) {
return false;
}
}
return true;
}
private static boolean isPalindrome(int a) {
if (a < 0)
return false;
int div = 1;
while (a / div >= 10) {
div *= 10;
}
while (a != 0) {
int x = a;
int l = x / div;
int r = x % 10;
if (l != r)
return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
private static int userInt() {
Scanner s = new Scanner(System.in);
System.out.print("Enter a positive integer: ");
int userInteger = s.nextInt();
return userInteger;
}
}
Remember, don't use the same names for variable and function. In the function userInt(), you have used a variable int userInt, to get the result from the scanner. This might be aa recursive call sometimes. Be careful with that.

I'm trying to write a program for Palindromes but my output is not as expected

this is my script
public class palindrome {
public static void main(String[] args) {
int rev = 0;
Scanner r = new Scanner(System.in);
System.out.println("enter any value");
int n = r.nextInt();
while (n != 0) {
rev = rev * 10 + n % 10;
n = n / 10;
}
System.out.println(rev);
if (n == rev) {
System.out.println("number is palindrome");
} else {
System.out.println("not palindrome");
}
}

In your solution n is always 0.
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;
}
http://java67.blogspot.co.il/2012/09/palindrome-java-program-to-check-number.html

You should make a copy of n into another variabile
public static void main(String[] args) {
int rev = 0;
Scanner r = new Scanner(System.in);
System.out.println("enter any value");
int n = r.nextInt();
int original = n;
while (n != 0) {
rev = rev * 10 + n % 10;
n = n / 10;
}
System.out.println(rev);
if (original == rev) {
System.out.println("number is palindrome");
} else {
System.out.println("not palindrome");
}

Count of most occurring digit... Find the digit that occurs most in a given number

the following s the code to
Find the number of occurrences of a given digit in a number.wat shall i do in order to Find the digit that occurs most in a given number.(should i create array and save those values and then compare)
can anyone please help me ..
import java.util.*;
public class NumOccurenceDigit
{
public static void main(String[] args)
{
Scanner s= new Scanner(System.in);
System.out.println("Enter a Valid Digit.(contaioning only numerals)");
int number = s.nextInt();
String numberStr = Integer.toString(number);
int numLength = numberStr.length();
System.out.println("Enter numer to find its occurence");
int noToFindOccurance = s.nextInt();
String noToFindOccuranceStr = Integer.toString(noToFindOccurance);
char noToFindOccuranceChar=noToFindOccuranceStr.charAt(0);
int count = 0;
char firstChar = 0;
int i = numLength-1;
recFunNumOccurenceDigit(firstChar,count,i,noToFindOccuranceChar,numberStr);
}
static void recFunNumOccurenceDigit(char firstChar,int count,int i,char noToFindOccuranceChar,String numberStr)
{
if(i >= 0)
{
firstChar = numberStr.charAt(i);
if(firstChar == noToFindOccuranceChar)
//if(a.compareTo(noToFindOccuranceStr) == 0)
{
count++;
}
i--;
recFunNumOccurenceDigit(firstChar,count,i,noToFindOccuranceChar,numberStr);
}
else
{
System.out.println("The number of occurance of the "+noToFindOccuranceChar+" is :"+count);
System.exit(0);
}
}
}
/*
* Enter a Valid Digit.(contaioning only numerals)
456456
Enter numer to find its occurence
4
The number of occurance of the 4 is :2*/

O(n)
keep int digits[] = new int[10];
every time encounter with digit i increase value of digits[i]++
the return the max of digits array and its index. that's all.
Here is my Java code:
public static int countMaxOccurence(String s) {
int digits[] = new int[10];
for (int i = 0; i < s.length(); i++) {
int j = s.charAt(i) - 48;
digits[j]++;
}
int digit = 0;
int count = digits[0];
for (int i = 1; i < 10; i++) {
if (digits[i] > count) {
count = digits[i];
digit = i;
}
}
System.out.println("digit = " + digit + " count= " + count);
return digit;
}
and here are some tests
System.out.println(countMaxOccurence("12365444433212"));
System.out.println(countMaxOccurence("1111111"));

declare a count[] array
and change your find function to something like
//for (i = 1 to n)
{
count[numberStr.charAt(i)]++;
}
then find the largest item in count[]

public class Demo{
public static void main(String[] args) {
System.out.println("Result: " + maxOccurDigit(327277));
}
public static int maxOccurDigit(int n) {
int maxCount = 0;
int maxNumber = 0;
if (n < 0) {
n = n * (-1);
}
for (int i = 0; i <= 9; i++) {
int num = n;
int count = 0;
while (num > 0) {
if (num % 10 == i) {
count++;
}
num = num / 10;
}
if (count > maxCount) {
maxCount = count;
maxNumber = i;
} else if (count == maxCount) {
maxNumber = -1;
}
}
return maxNumber;
}}
The above code returns the digit that occur the most in a given number. If there is no such digit, it will return -1 (i.e.if there are 2 or more digits that occurs same number of times then -1 is returned. For e.g. if 323277 is passed then result is -1). Also if a number with single digit is passed then number itself is returned back. For e.g. if number 5 is passed then result is 5.