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How do I identify 2 separate 2-digit numbers in a String?

I am trying to find a way to identify 1 or 2 digit numbers in a string (there can't be any 3 digit numbers), add them together so they must be between 80 and 95.
For some reason, the code is not working, as it is always returning false, even when it should (in theory) return true.
ex. "Hi 57 how are you 30" returns false
Thank you in advance for your help!
("line" is the name of the String.)
public boolean isDig(){
int total=0;
int h;
int length = line.length();
for(h=0; h < length-1; h++) {
if (Character.isDigit(line.charAt(h))){
if (Character.isDigit(line.charAt(h+1))){
if (Character.isDigit(line.charAt(h+2))){
return false;
}
else {
total= total+(line.charAt(h)+line.charAt(h+1));
h++;
}
}
else {
total= total+(line.charAt(h));
}
}
if (total>=80 && total<=95){
return true;
}
else {
return false;
}
}

The main problem in the code is that line.charAt(h) isn't the numeric value of the digit at position h. It's the codepoint value, for example '0' is 48.
The easiest way to obtain the numeric value is Character.getNumericValue(line.charAt(h)), and similarly in other places.
You're also missing the multiplication by 10 of the first digit in the pair.
Assuming you know that the string is valid, it's easy enough just to add up any numbers in the string. The fact that they are 2 or 3 digits doesn't really matter from the perspective of obtaining the sum.
int total = 0;
for (int i = 0; i < line.length(); ) {
// Skip past non-digits.
while (i < line.length() && !Character.isDigit(line.charAt(i))) {
++i;
}
// Accumulate consecutive digits into a number.
int num = 0;
while (i < line.length() && Character.isDigit(line.charAt(i))) {
num = 10 * num + Character.getNumericValue(line.charAt(i));
}
// Add that number to the total.
total += num;
}

You should use a regex for this kind of parsing :
public class Example {
public static void main(String[] args) {
String input = "Hi 57 how are you 30";
System.out.println(process(input));
}
private static boolean process(String input) {
Pattern pattern = Pattern.compile(".*?(\\d+).*?(\\d+)");
Matcher matcher = pattern.matcher(input);
if (matcher.matches()) {
int one = Integer.parseInt(matcher.group(1));
int other = Integer.parseInt(matcher.group(2));
System.out.println(one);
System.out.println(other);
int total = one + other;
return total >= 80 && total <= 95;
}
return false;
}
}
Output :
57
30
true

One of the possible solution it to use Regual Expression.
public static boolean isValid(String str) {
// regular expression matches 1 or 2 digit number
Matcher matcher = Pattern.compile("(?<!\\d)\\d{1,2}(?!\\d)").matcher(str);
int sum = 0;
// iterate over all found digits and sum it
while (matcher.find()) {
sum += Integer.parseInt(matcher.group());
}
return sum >= 80 && sum <= 95;
}

Let a java.util.Scanner do the work:
public boolean scan(String line) {
Scanner scanner = new Scanner(line);
scanner.useDelimiter("\\D+");
int a = scanner.nextInt();
int b = scanner.nextInt();
int sum = a + b;
return sum >= 80 && sum <= 95;
}
The invocation of .useDelimiter("\\D+") delimits the string on a regular expression matching non-digit characters, so nextInt finds the next integer. You'll have to tweak it a bit if you want to pick up negative integers.

You could convert the String into an Array and test to see if each element in the String (separated by a space) is a Digit by testing the Integer.parseInt() method on each String element. Here is an example below:
public static boolean isDig(String theString) {
String[] theStringArray = theString.split(" ");
ArrayList<Integer> nums = new ArrayList<Integer>();
for(int x = 0; x < theStringArray.length; x++) {
String thisString = theStringArray[x];
try {
int num = Integer.parseInt(thisString);
nums.add(num);
}catch(NumberFormatException e) {
continue;
}
}
int total = 0;
for(int num: nums) {
total += num;
}
if(total >= 80 && total <= 95) {
return true;
}
else {
System.out.println(total);
return false;
}
}
We first split the original String into an Array based on the empty spaces. We then create an ArrayList that will add each digit in the String to it. We then create a for loop to look at each individual String in the Array and we set up a try-catch block. If we can covert the digit into an int using the Integer.parseInt() method, we will add it to the ArrayList. If not, we will catch the exception and continue the loop with a "continue" statement. Once we break out of the loop, we can create a variable called "total" and create another for loop in order to add each digit in the ArrayList to the total amount. If the total is greater than/equal to 80 and less than/equal to 95, we will return True, or else we will return false. Let's test the code:
String digitTest = "There is a digit here: 50 and a digit here 45";
System.out.println(isDig(digitTest));
The numbers 50 and 45 should equal 95 and our result is:
true

Error with multiplying char with a number and add with char ( Big Java Ex 6.2)

Currently going through an exercise in my book, but i'm stuck ( I havent learned arrays yet, this chapter is on loops nested loops and for loops)
The first part of the problem is to take a credit card number and then sum every other number backwards
Consider 4358 9795, which should output the sum 5+7+8 + 3 = 23.
Heres my solution which isn't working
class Checkit{
private String creditNum;
private int sum;
public Checkit(String creditNum)
{
this.creditNum = creditNum;
sum = 0;
}
public int getSum()
{
for (int i = creditNum.length() ; i > 0 ; i--)
{
char ch = creditNum.charAt(i-1);
if(i%2 == 1 )
{
sum+=ch;
}
}
return sum;
}
}
public class test{
public static void main(String [] args)
{
Checkit sampleNumber = new Checkit("4358 9795");
System.out.println(sampleNumber.getSum());
}
}
I'm not exactly sure whats wrong with my logic. ch is taking all of the values of my credit card number 5,7,8,3. But for some reason the sum is messing up.
BONUS PART
Take Each number that wasn't added and double it, so 9+9 + 5 + 4, double each of those terms ( that becomes 18 + 18 + 10 + 8), and then get the sum of 1 + 8 + 1 + 8 + 1 + 0 + 8.
I tried the bonus part, but for some reason every time I get 9, 9, 5,4 and times that char value by 2, I get letters. I don't think I can multiply chars by integers, so should I do conversions? Note I didn't learn arrays yet
NEW CODE
class Checkit {
private String creditCardNum;
private int sum;
public Checkit(String creditCardNum) {
sum = 0;
this.creditCardNum = creditCardNum;
}
public int getSum() {
creditCardNum = creditCardNum.replaceAll("\\s+", "");
for (int i = creditCardNum.length(); i > 0; i--) {
char ch = creditCardNum.charAt(i - 1);
if (i % 2 == 0) {
sum += Character.getNumericValue(ch);
}
}
return sum;
}
public int doubleDigitSum() {
sum = 0;
creditCardNum = creditCardNum.replaceAll("\\s", "");
for (int i = creditCardNum.length(); i > 0; i--) {
char ch = creditCardNum.charAt(i - 1);
if (i % 2 == 1) {
int newChar = Character.getNumericValue(ch) * 2;
String newCharString = Integer.toString(newChar);
for (int j = 0; j < newCharString.length(); j++) {
char sumThis = newCharString.charAt(j);
sum += Character.getNumericValue(sumThis);
}
}
}
return sum;
}
}
public class DataSet{
public static void main(String [] args) {
Checkit data = new Checkit("4358 9795");
System.out.println(data.getSum());
System.out.println(data.doubleDigitSum());
}
}

You're adding char values rather than int values. Replace sum+=ch; with
sum += ch - '0';
The reason why this works as opposed to the original solution is that when you're adding char values to an int value, the char is converted to a decimal via its ASCII value. For example, '5' has a ASCII dec value of 53. However, '5' - '0' equals 5. More on these ASCII values can be found here, http://www.asciitable.com/

Your issue is that you are not converting the char to an integer:
public int getSum()
{
for (int i = creditNum.length() ; i > 0 ; i--)
{
char ch = creditNum.charAt(i-1);
if(i%2 == 1 )
{
sum+=ch;
}
}
return sum;
}
Fun fact that's causing your error - chars are numbers! They're simply a number that represents the ascii character code of the letter. So when you add them to a sum, Java is OK with that and just adds the corresponding number for the char '8' for example, which is 56.
To make your code work, you need to properly convert to an integer:
public int getSum()
{
for (int i = creditNum.length() ; i > 0 ; i--)
{
char ch = creditNum.charAt(i-1);
if(i%2 == 1)
{
sum+=Character.getNumericValue(ch);
}
}
return sum;
}

for (int i = creditNum.length() ; i > 0 ; i--) {
char ch = creditNum.charAt(i-1);
First iteration thru loop gets IndexOutOfBounds exception. And then you never check the initial character of the credit card number, since your termination criterion is ( i > 0 ), not (i >= 0)
And why not just use i -= 2 for the increment instead of the if-check?

Too many Exception out of bounds - JAVA

Code is supposed to do this: Return the number of times that the string "code" appears anywhere in the given string, except we'll accept any letter for the 'd', so "cope" and "cooe" count.
The problem: Ran across Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 11 (line number:10)
public int countCode(String str){
int a = 0; // counter goes thru string
int b = str.length()-1;
int counter = 0; //counts code;
if(str.length() < 4) return 0;
else{
while (a <=b){
if(str.charAt(a) == 'c'){
if(str.charAt(a+1) == 'o'){
if(str.charAt(a+3) == 'e'){
counter++;
a= a+3;
} // checks e
else a++;
} // checks o
else a++;
} // checks c
else a++;
}
return counter;
}
}
Here's what I tried to evaluate to get said exception:
countCode("xxcozeyycop") --> EXPECTED RESULT 1
countCode("cozcop") --> EXPECTED RESULT

Your loop goes from 0 to the length of the string (excluded). But inside the loop, you're doing
str.charAt(a+3)
Obviously, if a is length - 1, a + 3 is length + 2, and you're thus trying to access an element outside the bounds of the string.
Side note: you would understand your own code better if you indented it correctly.

Instead of
while (a <=b){
use
while (a <= b - 3){
Reason: Your end sign in the while is the condition that the start of the String "code" is inside the String. However, if a = b - 2, then a + 3 = b + 1 = (str.length() - 1 + 1) = str.length() which is just outside the String.

public int countCode(String str) {
int count = 0;
for(int i = 0; i < str.length()-3; i++)
if(str.substring(i, i+2).equals("co") && str.charAt(i+3) == 'e')
count++;
return count;
}

How to count in Java the trailing zeros from an Integer? (Ex: 234000 => 3 zeros)

The title is pretty much self explanatory. :)
1232 => 0
1231030 => 1
2000 => 3
34444400000 => 5

If it fits into an int/long, just check if the number modulo 10 is 0 and keep a counter:
long x = ...
if (x == 0) {
return 0;
}
int counter = 0;
while (x % 10 == 0) {
counter++;
x /= 10;
}
If it's too big to fit in long, store it in a String and count zeroes from the last char:
String s = ...
int counter = 0;
while(counter < s.length() && s.charAt(s.length() - 1 - counter) == '0') {
counter++;
}

Integer class has an inbuilt function to count the trailing zeros.
javadocs
int trailingZeroes = Integer.numberOfTrailingZeros(int i);

Three lines:
int zeroes = 0
while(num%10 == 0 && num != 0) {
zeroes++;
num /= 10;
}
This uses the modulus operator. As long as we can divide by ten without remainder, increment the counter.

Here is another solution using Java 8 Streams:
int trailingZeros = String.valueOf(number).chars()
.reduce(0, (count, ch) -> (ch == '0') ? count + 1 : 0);
This transforms the number to an IntStream. This stream is then reduced using a lambda which resets a counter each time a non-zero char comes up.

You could always just use a regular expression:
Pattern pattern = Pattern.compile("(0+)$");
Matcher matcher = pattern.matcher(String.valueOf(123140000));
Integer trailingZeroes = 0;
if (matcher.find()) {
trailingZeroes = matcher.group(1).length();
}
System.out.println(trailingZeroes);

you can turn the int to a String and iterate in reverse, counting the zeros until you find a char that is not zero:
int countZeros(int x){
String a = Integer.toString(x);
int numOfZeros = 0;
for(int i = a.length() - 1; i >= 0; i--)
if (a.charAt(i) != '0') break;
else numOfZeros ++;
return numOfZeros;
}
Testing with :
System.out.println(countZeros(25000)); will print 3
System.out.println(countZeros(25)); will print 0
Hope this helps.

Not tried this code but this should work.
int counterForZeros=0;
for(long i=10;true;)
{
if(num%i==0)
{
counterForZeros++;
i*=10;
}
else
{
break;
}
}
System.out.println("Number of zeros in "+num+" is "+counterForZeros);

Well, if this is a contest to see who can do it in the fewest lines:
trailingZeroes = String.valueOf(num).length() - String.valueOf(num).replaceAll("0*$","").length();

algorithm to create a char array from permutations list in java

I'm sorry for the slightly confusing title, I'm unsure of how to phrase it.
I need to create a char array allowing for every possible permutation of a character set.
If I were to give you:
char[] charSet = {"a", "b", "c"};
BigInteger value = n; //where n is a number >= 0
char[] charArray = createCharArray(value, charSet);
How can I create charArray from value and charSet such that if I ran:
createCharArray(new BigInteger("6"), {"a", "b", "c"});
it would return {"a", "c"}
because
a=1
b=2
c=3
aa=4
ab=5
ac=6
Here's what I have so far:
private char[] createCharArray(BigInteger value, char[] charSet){
List<Character> charArray = new ArrayList<Character>();
if (value.compareTo(this.max) == 0)
System.out.println("");
BigInteger csSize = new BigInteger(String.valueOf(charSet.length));
if(this.powers.isEmpty())
this.powers.add(0, csSize.pow(0));
if(this.sumPowers.isEmpty())
this.sumPowers.add(0, csSize.pow(0));
BigInteger curPow;
int i = 1;
while((curPow = csSize.pow(i)).compareTo(value) <= -1){
if(this.powers.size() <= i)
this.powers.add(i, curPow);
if(this.sumPowers.size() <= i)
this.sumPowers.add(i, this.sumPowers.get(i-1).add(curPow));
i += 1;
}
i -= 1;
while (i >= 0 && value.compareTo(BigInteger.ZERO) >= 0){
if (i <= 1){
int charNum = value.divide(this.sumPowers.get(0)).intValue() - 1;
charArray.add(charSet[charNum]);
}
else{
int charNum = value.divide(this.sumPowers.get(i-1).subtract(BigInteger.ONE)).intValue() - 1;
charArray.add(charSet[charNum]);
}
value = value.subtract(this.powers.get(i));
i -= 1;
}
char[] returnArray = new char[charArray.size()];
int j = 0;
while(j<charArray.size()){
returnArray[j] = charArray.get(j);
j += 1;
}
return returnArray;
}
It certainly could use some help, as a value of 0 fails, values of 1 and 2 succeed, 3-8 fail, 9, 10 succeed, etc.
EDIT: To be clear, the value parameter must be able to be ANY number n > 0. This is why I've chosen BigInteger

Make a class that has two fields:
private char letter;
private int value;
public <classname>(char letter){
this.letter = letter;
value = 0;
}
//Setters and getters
Then in your main when initializing an array (via for loop) set the value to i + 1 (getting rid of 0)
for(int i = 0; i < <yourarray>.length; i ++){
//Assuming you initialized your objects before
<yourarray>[i].<setterforvalue>(i + 1);
}
And then to calculate them together:
for(int i = 0; i < <yourarray>.length; i ++){
for(int j = 0; j < <yourarray>.length; j ++){
if(<yourarray>[i] + <yourarray>[j] == <needednumber>){
//Do what you need to do with the value
}
}
}

Okay after much thinking and eventually breaking it down to using numbers 0-9 instead of characters. Here's the breakdown: Think about how regular base 10 numerals are created.
The number 194 is made up of a 4 in the ones column, a 9 in the tens, and a 1 in the hundreds. The difference between ones, tens, and hundreds is multiplication/division by 10, which is the base.
So I figured that I could mod the 194 by the base (10) to get 4, the ones. then divide by 10 to remove the ones column. mod again to get 9, then divide by 10, mod again to get 1, divide by 10. once the division creates a number that is exactly 0, we are done. This is because we cannot make a number 000194.
For my function, My base is the length of the character set, and the value is like 194 in the example above.
private static void createCharArray(BigInteger value, char[] charSet){
List<Character> charArray = new ArrayList<Character>();
BigInteger csSize = BigInteger.valueOf(charSet.length);
if (value.compareTo(BigInteger.ZERO) == 0)
charArray.add(0, charSet [0]);
else{
BigInteger modded = value.mod(csSize);
BigInteger digit = value.divide(csSize);
while (modded.compareTo(BigInteger.ZERO) != 0 || digit.compareTo(BigInteger.ZERO) != 0){
if(modded.compareTo(BigInteger.ZERO) == 0){
charArray.add(0, charSet[csSize.subtract(BigInteger.ONE).intValue()]);
value = value.subtract(BigInteger.ONE);
}
else
charArray.add(0, charSet[modded.subtract(BigInteger.ONE).intValue()]);
value = value.divide(csSize);
modded = value.mod(csSize);
digit = value.divide(csSize);
}
}
for(char c : charArray)
System.out.print(c);
System.out.println();
}
public static void main(String[] args) {
long start = System.nanoTime();
String characters = "";
characters += "0123456789";
characters += "abcdefghijklmnopqrstuvwxyz";
characters += "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
characters += " !\"#$%&'()*+,-./:;<=>?#[\\]^_`{|}~";
char[] cs = characters.toCharArray();
Arrays.sort(cs);
createCharArray(new BigInteger("1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890"), cs);
long total = System.nanoTime() - start;
System.out.println("Completed in: " + total + " billionths of a second");
System.out.println("Completed in: " + total/1000000 + " thousandth(s) of a second");
}
If you run this, note that that BigInteger 4 lines from the bottom is 100 characters long. On my machine, it takes only 1/1000 th of a second (1 Millisecond).