We Keep Coding

credit card numbers need to be added without the dash

So one of the conditions for the credit card number to be valid is that "the sum of first 4 digits must be 1 less than the sum of the last 4 digits" I believe the problem could be it's counting the dashes as a digit but not sure. the rule 4 is that the sum of all digits must be divisible by 4, which seems to work, but rule 5 doesn't.
int sum = ccNumber.chars().filter(Character::isDigit).map(Character::getNumericValue).sum();
if(sum%4!=0){
valid = false;
errorCode = 4;
return;
}
// set values and for loop for fifth rule.
String digits = ccNumber.replaceAll("[ˆ0-9]","");
int firstfourdigits = 0;
int lastfourdigits = 0;
for(int i=0; i<4; i++)
firstfourdigits = firstfourdigits + Character.getNumericValue(ccNumber.charAt(i));
for (int i=0, m = ccNumber.length()-1; i<4; i++, m--)
lastfourdigits = lastfourdigits + Character.getNumericValue(ccNumber.charAt(m));
// mutator for fifth rule
if(lastfourdigits!= firstfourdigits -1){
valid = false;
errorCode = 5;
return;
}
sorry I'm lost and new to coding.

Edit since you altered your question. Original anwser to the original question is at the bottom part
Checking if first part and last part have a difference of one
The code you currently have is close, but there are some mistakes here and there.
Filtering out only digits: The code you use to filter out all characters that are not numeric should work, but in your following code you are no longer using this filtered value in your loop.
firstfourdigits + Character.getNumericValue(ccNumber.charAt(i));
This should use the variable with only your numeric values => digits
firstfourdigits = firstfourdigits + Character.getNumericValue(digits.charAt(i));
Difference in first group vs last group: The -1 should be replaced by +1 here. When you are experiencing problems with this type of checks, it's always adviced to try and calculate it on a piece of paper. Lets assume the sum of the first 4 digits is 8 and the sum of the last 4 digits is 9. As per the requirement, this is a valid number, and should result to false in your check if(lastfourdigits!= firstfourdigits -1)
Let's fill it in: 9 != 8-1 => 9 != 7 so this returns false, and marks it as invalid. If we base it on the requirement, you could write the sum of the first 4 digits should be one less then the last 4 digits as: firstfourdigits = lastfourdigits - 1. This is mathmatically the same as lastfourdigits = firstfourdigits + 1. However, in our check we want to know if this check is not correct, so we should change the statement to: if(lastfourdigits != firstfourdigits + 1)
These 2 changes should give you the results you asked for. Combining these changes, we come to the following code example
String digits = ccNumber.replaceAll("[ˆ0-9]", "");
int firstfourdigits = 0;
int lastfourdigits = 0;
for (int i = 0; i < 4; i++)
firstfourdigits = firstfourdigits + Character.getNumericValue(digits.charAt(i));
for (int i = 0, m = ccNumber.length() - 1; i < 4; i++, m--)
lastfourdigits = lastfourdigits + Character.getNumericValue(digits.charAt(m));
if(lastfourdigits!= firstfourdigits + 1){
valid = false;
errorCode = 5;
return;
}
Other recommendations
The above example should work for what you asked, and is based on your code. However there are some optimalisations possible to the code to make everything more readable
Use brackets on your for loop: To make it clearer what is inside the for loop, and what isn't, I would advise you to make use of curly brackets. Though they are not required, they will make it very clear what is and isn't in the for loop and will prevent hard to spot issues when you add something extra in the for loop
Use the short notation for addition: Instead of writing firstfourdigits = firstfourdigits + Character.getNumericValue(digits.charAt(i));, You could use a shorter notation of +=. This will take the value on the left side of your equals, and will calculate the sum of that value on the right side. firstfourdigits += Character.getNumericValue(digits.charAt(i));
The code looks like this then:
String digits = ccNumber.replaceAll("[ˆ0-9]", "");
int firstfourdigits = 0;
int lastfourdigits = 0;
for (int i = 0; i < 4; i++){
firstfourdigits += Character.getNumericValue(digits.charAt(i));
}
for (int i = 0, m = ccNumber.length() - 1; i < 4; i++, m--) {
lastfourdigits += Character.getNumericValue(digits.charAt(m));
}
if(lastfourdigits!= firstfourdigits + 1){
valid = false;
errorCode = 5;
return;
}
Anwser to original question to calculate the sum of all digits
You could make use of Character.isDigit(char). To simplify the for loop, you can even make use of a stream to get the sum
int sum = ccNumber.chars().filter(Character::isDigit).map(Character::getNumericValue).sum();
if (sum % 4 != 0) {
valid = false;
errorCode = 4;
return;
}
.chars(): This will create a stream of all the characters in the provided string so that we can loop over them one by one
.filter(Character::isDigit): This will filter out every character that is not a digit
.map(Character::getNumericValue): This will map the stream from Characters to their numeric values so that we can use those further
sum() will calculate the sum of the numeric values that we currently have in the Stream

The difference is always a positive value e.g. the difference between 4 and 5 or between 5 and 4 is the same i.e. 1. In other words, you need to compare the absolute value of the subtraction with 1.
Therefore, replace
if(lastfourdigits!= firstfourdigits -1)
with
if(Math.abs(lastfourdigits - firstfourdigits) != 1)
Another mistake in your code is that you have used ccNumber, instead of digits in your loops.
Some recommendations to make your code easier to understand:
Replace for (int i=0, m = digits.length()-1; i<4; i++, m--) with for (int m = digits.length() - 1; m >= digits.length() - 4; m--). Note that I've already replaced ccNumber, with digits in these statements.
Replace ccNumber.replaceAll("[^0-9]","") with ccNumber.replaceAll("\\D", "").
Replace firstfourdigits = firstfourdigits + Character.getNumericValue(digits.charAt(i)) with firstfourdigits += Character.getNumericValue(digits.charAt(i)). Note that I've already replaced ccNumber, with digits in these statements.
Always enclose the body of if and loop statements within { } even if there is just one statement inside the body.
Demo:
public class Main {
public static void main(String[] args) {
System.out.println(isValidOnDiffCriteria("1234-5678-9101-1213"));
System.out.println(isValidOnDiffCriteria("1234-5678-9101-1235"));
System.out.println(isValidOnDiffCriteria("1235-5678-9101-1234"));
}
static boolean isValidOnDiffCriteria(String ccNumber) {
String digits = ccNumber.replaceAll("\\D", "");
int firstfourdigits = 0;
int lastfourdigits = 0;
for (int i = 0; i < 4; i++) {
firstfourdigits += Character.getNumericValue(digits.charAt(i));
}
for (int m = digits.length() - 1; m >= digits.length() - 4; m--) {
lastfourdigits += Character.getNumericValue(digits.charAt(m));
}
if (Math.abs(lastfourdigits - firstfourdigits) != 1) {
return false;
}
return true;
}
}
Output:
false
true
true

Try the code above. Should be what you asked. You don't need a try catch.
static boolean isCardValid(String creditCard) {
// group digits in a string array
String[] cards = creditCard.split("-");
int sumAll = 0;
// for every group of digits we convert it to char[]
for (String card : cards) {
sumAll += sum(card.toCharArray());
}
int firstGroupOfDigits = sum(cards[0].toCharArray()) ;
int lastGroupOfDigits = sum(cards[cards.length-1].toCharArray());
if( firstGroupOfDigits == lastGroupOfDigits -1){
if (sumAll % 4 == 0) {
return true;
}
}
return false;
}
// sum the group of digits separated by "-"
static int sum(char[] chr) {
int sum = 0;
for (char c : chr) {
sum += Character.getNumericValue(c);
}
return sum;
}

Well, your program is not that bad and as far as I can tell there is only one problem and that is the you simply reversed the required test on the first and last groups. I would advise you to ensure the valid is initialized to true as the default. Then if none of the error codes are set, it will return true.
Presently you have the following:
if (lastfourdigits != firstfourdigits - 1) {
valid = false;
errorCode = 5;
}
But what you need is this
if (lastfourdigits != firstfourdigits + 1) {
valid = false;
errorCode = 5;
}
Your also have the following, unnecessary code.
String digits = ccNumber.replaceAll("[ˆ0-9]","");
The reason being is that you are simply using ccNumber starting at the beginning for the first four characters and starting at the end for the last four. In this way you are not encountering dashes so you don't need to get just the digits.
Another recommendation is that as soon as you find an error you set the error code and return immediately. What's the use in continuing to process a card that has already been found to be flawed?
Other considerations and an alterative approach
It may not be a part of the assignment but I would also consider the following:
What if you have more or less than 16 digits?
What if you have more than three dashes giving more than four groups of numbers.
Checking the above would require additional logic and would complicate your effort. But it is something to consider. What follows demonstrates one way to check on those particular format issues and report them. This uses basic techniques and avoids streams so as not to repeat unnecessary operations.
This example throws selective errors based on problems found. Those may be changed or eliminated altogether as explained later. Credit card validation is a task where the most straightforward solution is best and should require low overhead.
First, declare a special exception to catch credit card errors.
class BadCreditCardException extends Exception {
public BadCreditCardException(String message) {
super(message);
}
}
Now declare some test data.
String[] testData = {
"1234-4566-9292-0210",
"1500-4009-2400-1600",
"1500-4009-2400-160000",
"1234-45669292-0210",
"1#34-45-66-9292-0210",
"1234-45B6-9292-0210",
"1234-4566-9292-2234",
"1234-4566-9292-021022",
"1234-4566-9292-0210",
"4567-4566-92!2-6835",
"1234-4566-9292-0210",
"1234-45+6-9292-0210",
"1234-4566-92x2-0210",
"1234-4566-9292-0210",
};
Test the credit cards and report errors. Note that only first encountered errors are reported. There may be multiple errors in the number.
String fmt = "%-23s - %s%n";
for(String card : testData) {
try {
validate(card);
System.out.printf(fmt,card, "Valid");
} catch (BadCreditCardException bce) {
System.out.printf(fmt,card, bce.getMessage());
}
}
The above prints.
1234-4566-9292-0210 - Invalid credit card checksum
1500-4009-2400-1600 - Valid
1500-4009-2400-160000 - Non group of 4 digits
1234-45669292-0210 - Insufficient or too may dashes
1#34-45-66-9292-0210 - Insufficient or too may dashes
1234-45B6-9292-0210 - Non digit found.
1234-4566-9292-2234 - Valid
1234-4566-9292-021022 - Non group of 4 digits
1234-4566-9292-0210 - Invalid credit card checksum
4567-4566-92!2-6835 - Non digit found.
1234-4566-9292-0210 - Invalid credit card checksum
1234-45+6-9292-0210 - Non digit found.
1234-4566-92x2-0210 - Non digit found.
1234-4566-9292-0210 - Invalid credit card checksum
The Explanation
The validate method. The method works as follows.
split the card into groups using the dash (-) as a delimiter.
If there are not four groups, throw an exception.
Otherwise, sum each of the groups as follows each of these is checked during the summation process.
first check that the group is of size four, if not throw an exception.
as the group characters are iterated, if a non-digit is encountered, throw an exception.
otherwise, continue computing the sum for the current group as follows:
If the character is a digit, subtract 0 to convert it to an int
and add to the current sums array element.
when completed, add that group sum to the totalSum of all digits.
if the totalSum is divisible by four and the first group is one less than the last group, it is a valid card. Otherwise, throw an exception.
Alternative error handling modification
If the exceptions are not wanted, but just a pass or fail indication, then make the following changes.
change the void return type to boolean
if an exception was throw, simply return false
if all tests pass, then the last statement should return true
public static void validate(String cardNumber) throws BadCreditCardException {
int [] groupSums = new int[4];
int totalSum = 0;
String [] groups = cardNumber.split("-");
if (groups.length != 4) {
throw new BadCreditCardException("Insufficient or too may dashes");
}
for (int i = 0; i < groupSums.length; i++) {
if (groups[i].length() != 4) {
throw new BadCreditCardException("Non group of 4 digits");
}
for(int digit : groups[i].toCharArray()) {
if (!Character.isDigit(digit)) {
throw new BadCreditCardException("Non digit found.");
}
groupSums[i]+= digit -'0';
}
totalSum += groupSums[i];
}
if (groupSums[0]+1 != groupSums[3] || totalSum % 4 != 0) {
throw new BadCreditCardException("Invalid credit card checksum");
}
}

A separate class for Credit card and its parts
Add a Part class that manages a portion of the credit card
Add a CreditCard class that manages these portions
Valid each portion
In addition to validating each potion individually, validate additional check
Depending on the number of times, the valid & sumDigits method will be called, validation/sum can be added in respective methods or in constructor.
import java.util.Arrays;
public class CreditCard {
private final String input;
private final Part[] parts;
private final boolean valid;
CreditCard(String card) {
this.input = card;
if (card == null || card.length() != 19) {
valid = false;
parts = null;
} else {
parts = Arrays.stream(card.split("-")).map(Part::new).toArray(Part[]::new);
final int totalSum = Arrays.stream(parts).mapToInt(Part::sumDigits).sum();
valid = totalSum % 4 == 0 && parts.length == 4
&& parts[0].sumOfDigits + 1 == parts[3].sumOfDigits
&& Arrays.stream(parts).allMatch(Part::isValid);
}
}
static class Part {
final int num;
final boolean valid;
final int sumOfDigits;
Part(String part) {
int localNum = 0;
try {
localNum = Integer.parseInt(part);
} catch (Throwable ignored) {
}
this.num = localNum;
valid = part.length() == 4 && part.equals(String.format("%04d", num));
if (valid) {
sumOfDigits = part.chars().map(Character::getNumericValue).sum();
} else {
sumOfDigits = -1;
}
}
boolean isValid() {
return valid;
}
int sumDigits() {
return sumOfDigits;
}
}
public static void main(String[] args) {
String[] creditCards = {
"1000-0000-0001-0002",
"0000-0000-0000-0000",
"10000-0000-0001-0002",
"10000000-0001-0002",
"1a00-0000-0001-0002",
"1234-4826-6535-1235",
};
Arrays.stream(creditCards).map(CreditCard::new)
.forEach(c -> System.out.println(c.input + " is " + c.valid));
}
}

Everything is fine except the second for loop and your if condition.
Replace your code with the following changes and it should work fine:
int firstfourdigits = 0, lastfourdigits = 0;
for(int i=0; i<4; i++)
firstfourdigits = firstfourdigits + Character.getNumericValue(ccNumber.charAt(i));
for (int m = ccNumber.length()-1; m>ccNumber.length()-5; m--)
lastfourdigits = lastfourdigits + Character.getNumericValue(ccNumber.charAt(m));
if(firstfourdigits != lastfourdigits - 1){
valid = false;
errorCode = 5;
return;
}
You do not need to extract digits at all.

public boolean ccnCheck(String ccn){
String iccn = ccn.replaceAll("-","");
int length = iccn.length();
int fsum = 0;
int lsum = 0;
int allsum = 0;
for( int i = 0; i < length; i++){
int val = Character.getNumericValue(iccn.charAt(m))
if( i < 4)
fsum += val;
if( i >= length-4)
lsum += val;
allsum += val;
}
if( (allsum % 4) != 0)
return false;
if( fsum != lsum-1 )
return false;
return true;
}

In your rule five check, you're using ccNumber instead of your digits string.
For example, shouldn't
Character.getNumericValue(ccNumber.charAt(i));
be this instead:
Character.getNumericValue(digits.charAt(i));

Java 2d array matrix program

int even = 0;
int odd = 0;
for(i=0;i<5;i++){
for(j=0;j<5;j++){
if(j%2>=i) // I think the problem is this
even += twoD[i][j];
else
if(i%2!>=j) // I think the problem is this
odd += twoD[j][i];
}
}
System.out.println("The sum of the even elements above the diagonal is: "+even);
System.out.println("The sum of the odd elements below the diagonal is: "+odd);
This is the code I am working with. The problem is it doesn't display the real sum of the elements above and below as I'd want in a diagonal. The matrix is basically randomized so I had to check it everytime to verify the sum.

For the elements above the diagonal j > i (index of column is always greater than that of the row); for the elements below j < i.
Then the loops may be optimized slightly to provide j > i and swapping indexes for elements below the matrix diagonal:
int even = 0;
int odd = 0;
for(int i=0; i<twoD.length; i++) {
for(int j=i + 1; j<twoD[i].length; j++) {
if(twoD[i][j] %2 == 0) {
even += twoD[i][j];
}
if(twoD[j][i] %2 != 0) { // swap indexes below diagonal
odd += twoD[j][i];
}
}
}

Exactly as you've thought, the conditions are incorrect. Actually I'm not sure what was the logic behind them.
What you need to do is to find whether given number is above or below diagonal. You can do that by comparing indicies themselves - j>i and j<i.
Only then you need to check if a given number (not index) is odd or even - twoD[i][j] % 2 == 0.
Something like so should get the job done:
for(i=0;i<5;i++){
for(j=0;j<5;j++){
if(j>i && twoD[i][j]%2==0){
even += twoD[i][j];
}
if(j<i && twoD[i][j]%2!=0){
odd += twoD[i][j];
}
}
}

Brute force: count number of sub strings in a string array

What I am supposed to do is create an algorithm that counts the number of substrings in a piece of text where the substrings can be the letter B followed by C or C followed by B. I'm not sure what to do, but I tried it and came up with the result below. I'd like to know if I did it correctly.
int substrCount(String S[0...n-1]){
int count = 0;
for (int i = 0; i<=n-2; i++) {
for (int j=i+1; j<i+2; j++) {
if ((S[i] == 'B' && S[j] == 'C' ) || (S[i] == 'C' && S[j] == 'B')) {
count = count + 1;
}
}
}
}
I am gonna ignore whether it includes lowercase or uppercase for now. I also need to find the complexity of the algorithm from which I believe it is O(n^(2)). Did I do this correctly? If so, can I make it any more efficient?

This works well for me
static int substrCount( String str) {
int count=0;
for (int i=0; i<str.length()-1; i++)
{
boolean bc = (str.charAt(i) == 'B' && str.charAt(i+1) == 'C');
boolean cb = (str.charAt(i) == 'C' && str.charAt(i+1) == 'B');
if (bc || cb) {
count++;
}
}
return count;
}
You need to loop the sequence of chars in string just once to get the desired result. Check the couple of chars if they are equal "BC" or "CB" and move one index forward to the end of the string.
Output example:
"ACBAA" gives result 1
"ABCBA" gives result 2
"BCBCB" gives result 4
"BBBBB" gives result 0

Java Loops - Break? Continue?

I've read through a bunch of threads on using break and continue and I suspect the problem isn't necessarily my use of those, but the layout of my loops. In the following code, I am trying to iterate through the chars in a string that is input by the user to find any - symbols. If found, it will throw an error to the user that a negative number was found and exit. Otherwise, if it does not find a - symbol, it should print out all of the chars in the string.
I used break at the end of my first loop to find the - symbol, but it is not continuing on to the next loop. I tried continue as well but that didn't work. Loops are new to me so I may have this completely wrong, all I know is my first loop is working OK and will throw the error when it finds a - in the string.
strNum1 = JOptionPane.showInputDialog ("Enter Number String");
for (int i = 0; i < strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c == '-') {
System.out.println("Negative Digit Found - Exiting");
break;
}
}
for (int i = 0; i < strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c <= 9) {
System.out.println(c);
}
}

The break statement breaks you only from the first loop. In order to skip running the second loop in the event of finding a - character, you can use some boolean variable to indicate whether the second loop should run :
strNum1 = JOptionPane.showInputDialog ("Enter Number String");
boolean isValid = true;
for (int i=0; i<strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c == '-'){
System.out.println("Negative Digit Found - Exiting");
isValid = false;
break;
}
}
if (isValid) {
for (int i=0; i<strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c <= '9'){
System.out.println(c);
}
}
}

If you replace the break with a return it will exit the whole method. It sounds like this is probably what you want.

'break;' will stop the loop that it is in from running, where 'continue;' will skip the current 'iteration' in the loop.
for(int x = 0; x < 10; x++)
{
if (x == 5)
break;
// other code
}
// more other code
This will exit the loop once x == 5, and not do the 6th through 10th iterations.
for(int x = 0; x < 10; x++)
{
if (x == 5)
break;
// other code
}
// more other code
This will do every iteration, besides the 6th iteration.
But if you want to skip the '// more other code', then you would need to use a 'return;', provided your code is in a function, and it will skip the rest of the function, which in this case is the '// more other code'.

Use the return statement instead of break if you dont want to execcute the second loop after the first one.

You don't say if the number should be an integer, so I'm assuming yes.
If so, instead of using loops, a better way of validating the input would be:
int num1;
try {
num1 = Integer.parseInt(strNum1);
catch (NumberFormatException e) {
//not a valid number, complain about it
}
if (num1<0) {
//negative number not allowed, complain about it
}

for (int i=0; i<strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c == '-'){
System.out.println("Negative Digit Found - Exiting");
break;
}
}
can be modified as
if(strNum1.charAt(0) != '-'){
for (int i=0; i<strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c <= 9){
System.out.println(c);
}
}
}
else {
//negative number found...
}
In This way, unnecessary for loop and break statements can be avoided

All answers are good, but if you want to repeat prompt until you get a valid value,
you will need another loop for that, using labels:
negative: while(true) {
strNum1 = JOptionPane.showInputDialog ("Enter Number String");
for (int i=0; i<strNum1.length(); i++) {
char c = strNum1.charAt(i);
if (c == '-'){
System.out.println("Negative Digit Found - Exiting");
continue negative;
}
break negative;
}
}

Nested loops/ASCII?

So, for an assignment in my computer science class, we've got to use loops, either for or while, depending on preference. Now, the assignment IS to use said loops and a given input to draw a beauteous ASCII diamond made of '$' and '-'. Say, an input of 5 would look like:
____$
___$-$
__$-$-$
_$-$-$-$
$-$-$-$-$
_$-$-$-$
__$-$-$
___$-$
____$
The underscores are to denote spaces. Now, anytime I try using
public static void main(String[] args) {
String input=JOptionPane.showInputDialog(null, "Input a number between three and ten here: ");
double length=Double.parseDouble(input);
int i=0; int j=0;
for(i=1; i<length; i++)
{
System.out.print(" ");
for(j=1; j<=i; j++)
{
if(j<i){System.out.print("-$");
}
else if(j==i){System.out.println("");}
}
}
I come out with something like, say, for input=7:
-$
-$-$
-$-$-$
-$-$-$-$
-$-$-$-$-$
And yes, the two too few in the center is true with any input. Any help?

Since this is your homework, I'm just going to point you towards the correct answer and leave you to figure out the rest. Let's try formatting your code so you can see what's going on:
public static void main(String[] args) {
String input=JOptionPane.showInputDialog(null, "Input a number between three and ten here: ");
double length=Double.parseDouble(input);
int i=0; int j=0;
for(i=1; i<length; i++){
System.out.print(" ");
for(j=1; j<=i; j++){
if(j<i){
System.out.print("-$");
}
else if(j==i){System.out.println("");
}
}
}
Now, you've got an outer loop for i ranging from 1..length-1, and for each i you're going to print a space, then you're going to count from 1 to 1 and print "-$" that many times. Then, you're going to print a newline and repeat the outer loop, incrementing i
So, the first time through the outer loop, you print one space, followed by one "-$", followed by a newline. Then on the second time through the outer loop, you print one space, followed by "-$" twice, followed by a newline. And so forth, until i=length, and then you stop.
You want to print a few more spaces before you print dollar signs - a loop here will probably be useful.

try to go trough this code and see how it works... it might be useful for your homework
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("input a number");
int number = input.nextInt();
input.close();
boolean add = true;
int counter = 0;
do {
if (add) {
counter++;
print(' ', number - counter);
print('$', counter);
if (counter == number)
add = false;
System.out.println();
} else {
counter--;
if (counter == 0)
break;
print(' ', number - counter);
print('$', counter);
System.out.println();
}
} while (true);
}
private static void print(char c, int times) {
if (c == '$')
times = (times * 2) - 1;
for (int i = 0; i < times; i++)
System.out.print(c);
}
if you edit the print method you will get your desired result
private static void print(char c, int times) {
if (c == '$')
times = (times * 2) - 1;
for (int i = 0; i < times; i++)
if (i % 2 == 1 && c == '$')
System.out.print('-');
else
System.out.print(c);
}