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));
I am having trouble with understanding some of the codes in my assignment of finding the word lengths in a file. NB: the Environment I am using is Blue J with a library from dukelearntoprogram.com, this is the link for downloading Blue J, http://www.dukelearntoprogram.com/course3/index.php. I have created a void method called countWordLength, with parameter FileResource called resource, and an integer array called counts. this method should return the number words at a specific length in a file. For instance: 2 words of length 2: My as.
They have given me the code for this assignment but I am not understanding the following code, in my countWordLength method.
for (String word : resource.words()) {
int finalLength = 0;
int totalLength = word.length();
if (Character.isLetter(word.charAt(0)) == false) {
finalLength = totalLength-1;
}
else {
finalLength = totalLength;
}
if (Character.isLetter(word.charAt(totalLength-1)) == false && totalLength > 1) {
finalLength = finalLength-1;
}
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
else {
counts[finalLength] += 1;
}
The specific part that I don't understand is the meaning or usefulness of
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
else {
counts[finalLength] += 1;
}
if my question is not clear, and you may want more parts of my code please let know. Any help
will be highly appreciated.
Explanation:
int finalLength = 0;
int totalLength = word.length();
if (Character.isLetter(word.charAt(0)) == false) {
finalLength = totalLength-1;
}
else {
finalLength = totalLength;
}
Above lines are initiating lengths. If 1st character is not alphabet, then reducing length.
if (Character.isLetter(word.charAt(totalLength-1)) == false && totalLength > 1) {
finalLength = finalLength-1;
}
If last character is not alphabet, removing one more character but a special condition added here that it is we reduce word length only if it's length > 1
if (finalLength >= counts.length) {
counts[counts.length-1] += 1;
}
This might be bit tricky for you. If calculated length is greater than array we assigned, we are adding count to last one. Ex: Let's say they assigned array with 20 and if got word length as 30, then we will assign this word count to last index (19).
else {
counts[finalLength] += 1;
}
If length is less than size allowed than we will add to that respective index.
I am doing codingbat as practice for an upcoming quiz that I have. I am doing the recursion problems using recursion, but my teacher said that I should be able to do them using other loops. I figured that I should use for loops as they achieve are easily able to achieve the same result.
But I am having trouble converting the recursion to a for loop.
This is the problem:
Given a string and a non-empty substring sub, compute recursively the number of times that sub appears in the string, without the sub strings overlapping.
strCount("catcowcat", "cat") → 2
strCount("catcowcat", "cow") → 1
strCount("catcowcat", "dog") → 0
This is the code I am trying to use:
public int strCount(String str, String sub) {
int number = 0;
for (int i = 0; i >= str.length() - 1; i++) {
if (str.substring(i, sub.length()).equals(sub)) {
number += 1;
}
}
return number;
}
When I return, everything returns as 0.
In your for loop when you say
i >= str.length() - 1
the loop is never entered, because you are testing that i is greater than the allowed length (and it isn't). You need something like
i <= str.length() - 1
or
i < str.length()
Also, number += 1; can be written as number++;
One of the details you missed is "without the sub strings overlapping". This problem calls for a while loop, rather than a for loop, since the index will be incremented by different amounts, depending on whether there is a match or not.
Here's executable code to test whether or not the strCount method works correctly.
package com.ggl.testing;
public class StringCount {
public static void main(String[] args) {
StringCount stringCount = new StringCount();
System.out.println(stringCount.strCount("catcowcat", "cat"));
System.out.println(stringCount.strCount("catcowcat", "cow"));
System.out.println(stringCount.strCount("catcowcat", "dog"));
}
public int strCount(String str, String sub) {
int count = 0;
int length = str.length() - sub.length();
int index = 0;
while (index <= length) {
int substringLength = index + sub.length();
if (str.substring(index, substringLength).equals(sub)) {
count++;
index += sub.length();
} else {
index++;
}
}
return count;
}
}
This method is essentially taking an array that contains percentages, and based on whether or not the percentages are above or below 50%, a letter is appended onto a returned String to result in a final type.
I was told that I should "should create either two Strings or two arrays
that hold the characters, then loop through the dimensions and
test percentageB[i] with the 3 tests"
I don't understand why doing this is necessarily more valid or effective than the method I have already used. Could someone explain this to me?
public static String getPersonality(int[] percentageB) {
char [] types = {'E','I','S','N','T','F','J','P'};
String MBTItype = "";
for (int j = 0; j < types.length; j += 2) {
int i = j/2;
if (percentageB[i] < 50){
MBTItype += types[j];
} else if (percentageB[i] > 50) {
MBTItype += types[j + 1];
} else if (percentageB[i] == 50) {
MBTItype += 'X';
}
}
return MBTItype;
}
The reason you would have been told to use two arrays is that it makes the relationship between the letters clearer, at each position you are assigned the letter from one of the two lists. By combining them into a single array you are not making the mutually exclusive nature of the letters obvious to a reader of your code. Compare this implementation with your own
public static String getPersonality(int[] percentageB) {
char [] lowPercentTypes = {'E','S','T','J'};
char [] highPercentTypes = {'I','N','F','P'};
String MBTItype = "";
for (int i = 0; i < lowPercentTypes.length; i++) {
if (percentageB[i] < 50){
MBTItype += lowPercentTypes[i];
} else if (percentageB[i] > 50) {
MBTItype += highPercentTypes[i];
} else if (percentageB[i] == 50) {
MBTItype += 'X';
}
}
return MBTItype;
}
Reading this you do not need to go through the mental effort of knowing that each type has its compliment in the next item of the array, instead you are explicitly told that it is a high/low percentage type.
We are trying to reduce the cognitive load on the future reader of your code so you want to make things as straightforward as possible
Hey I have to create a method where I have to calculate pi to a passed term (a) using the leibiniz sequence. This is what I have so far
public static double calculatePi(int a){
double oddNum=1;
double pi=0;
for(int x=0; x<a; x++){
if(x%2==0)
pi=pi+(4/oddNum);
else
pi=pi-(4/oddNum);
oddNum=oddNum+2;
}
return pi;
}
I also need help writing a method that accepts a passed string and a (x)term. In the method it will add a "#" every x letters. So if its passed (sundae, 2) it will return su#nd#ae#. I have most of it down but theres a logical error that doesnt allow something like (cats, 3) to compile.
public static String addPounds(String s, int x){
String a="";
for(int i=0; i<s.length(); i=i+x){
a=(a+s.substring(i,i+x)+"#");
}
return a;
}
Thanks so much!
Your pi method is working fine.
You should change the other one to this. I have written it a little bit different so you get the logic easily.
public static String addPounds(String s, int x){
String a = "";
//starting from 1 so i can do mod division easily
for(int i = 1; i <= s.length(); i++){
//check if next # has to be placed at the end
if(i + 1 > s.length() && (i + 1) % x == 0){
a += "#";
//check if next # has to be placed somewhere inside the string
}else if(i % x == 0){
a += s.charAt(i - 1);
a += "#";
//oherwise just add the char at i-1
}else {
a += s.charAt(i - 1 );
}
}
return a;
}
Your addPounds method throws a StringIndexOutOfBoundsException with your given example (cats,3).
for(int i=0; i<s.length(); i=i+x){
a=(a+s.substring(i,i+x)+"#");
}
In the first execution of this for-loop, your variable 'a' will correctly be "Cat#". But now it goes wrong.
The variable 'i' gets increased to 3. And now you want to get a substring starting from the index 3 and ending with the index 6. The String "Cats" is only 4 letters, therefore the IndexOutOfBoundsException.
I guess the easiest way to solve your problem would be inserting a if else statement like this:
for(int i=0; i<s.length(); i=i+x){
if(s.length()>=i+x){
a=(a+s.substring(i,i+x)+"#");
}
else{
a= a+s.substring(i);
}
}