Develop Reference

Checking whether a number is an Unique number or not

I have written a program to check if a number is an Unique number.
[A Unique number is a number with no repeating digits and no leading zeros.]
I have written the following code:
Scanner sc=new Scanner(System.in)
System.out.println("Enter the number to be checked: ");
String num=sc.nextLine();
if(num.charAt(0)!='0')
{
Outer:
for(int i=0;i<num.length();i++)
{
for(int j=0;j<num.length();j++)
{
if(num.charAt(i)==num.charAt(j))
{
System.out.println("No, "+num+" is not a Unique number.");
break Outer;
}
}
if(i==num.length()-1)
{
System.out.println("Yes, "+num+" is a Unique number.");
}
}
}
else
System.out.println("No, "+num+" is not a Unique number as it has leading zeros.");
The problem is that is shows any number as NOT Unique, even 12345.
I would like to know where I have gone wrong.

Your code will always find "duplicate" characters when i == j.
You should change the indices of the loop in order not to compare a character to itself:
for(int i=0;i<num.length();i++) {
for(int j=i+1;j<num.length();j++) {
if(num.charAt(i)==num.charAt(j))
...
Besides, you should only output the "...is a Unique number." message after you are done with the outer loop.

Lets assume , length of input number to be 10 and "i" has reached the value of 5 in the for loop.
Now "j" will have the values 0 to 9.
So when "j" is equal to 5 , the if condition becomes true as you are comparing the digit at 5th position with itself (which is always true).
If you add i != j condition , it will fix the issue :-
if(num.charAt(i)==num.charAt(j) and i != j)
Alternatively, you can modify the loop for j to start from i + 1 so
that there are no overlaps.
for(int j=i+1;j<num.length();j++)
The second option is much better as it will reduce the number of comparisons from (n*n)
to (n * (n - 1))/2) , where n is the number of digits in the input number.

A possible solution is to use Stream to convert your String in a Set of char, then if the size of the set is the same as the length of your string, it is unique:
Scanner sc = new Scanner(System.in);
System.out.println("Enter the number to be checked: ");
String num = sc.nextLine();
boolean unique = Stream.of(num.split(""))
.map(s -> new String(s))
.collect(Collectors.toSet()).size() == num.length();
// With "1234" -> print true
// With "12342" -> print false
System.out.println(unique);

You can use below short and handy approach:
String a = "123452";
String[] split = a.split("");
List<String> list = Arrays.asList(a.split(""));
Set<String> set = new HashSet<>(list);
System.out.println("Unique: " + (list.size() == set.size()));

import java.util.*;
public class spnum
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter a number: ");
String num = sc.next();
int ctr = 0;
boolean isNumUnique = true;
for(int i = 0; i < num.length(); i++)
{
for(int j = 0; j < num.length(); j++)
{
if(num.charAt(i) == num.charAt(j))
{
ctr++;
}
}
if(ctr > 1)
{
isNumUnique = false;
}
ctr = 0;
}
if(isNumUnique == true)
{
System.out.println("Number is a unique number");
}
else
{
System.out.println("Number is not a unique number");
}
}
}
this code would give the right answer

program that reads integers between 1 and 100 and counts the occurrence of each

Write a program that reads integers between
1 and 100 and counts the occurrence of each (you should store the numbers in an array). Output should be in ascending order. Assume the input ends when the user enters a 0.
Hi guys, I know that this question has been posted before, perhaps a lot of times, but as I am a complete beginner at java, I don't completely understand the complexity of the codes that are posted. I just started taking Java classes, and would appreciate if you could help me figure out how to get my program to output the correct occurrences at the end. I'm so close to getting the answer but I can't figure it out for the life of me!! Thanks in advance!
import java.util.Scanner;
public class Problem1 {
public static void main(String[] args) {
//declarations
int [] myArray = new int [100];
int input = 5;
Scanner keyboard = new Scanner(System.in);
//input and processing
System.out.println("Please enter integers between 1 and 100 (enter 0 to stop): ");
while (input != 0)
{
input = keyboard.nextInt();
for (int i = 0; i < myArray.length; i++)
{
if (input == i)
{
myArray[i] = input;
}
}
}
//output (This is where I need help!!!!)
for (int k = 0; k < myArray.length; k++)
{
if (myArray[k] != 0)
{
System.out.print(k + " occurs " + myArray[k] + " time");
if (myArray[k] > 1)
{
System.out.println("s");
}
else
System.out.println("");
}
}
keyboard.close();
}
}

You are storing the number entered by the user in the array. Instead, you should store a counter in each position of the array for the corresponding integer. When the user inputs a number, you should increase the corresponding counter.
The second part of your code (output results) seems ok. It is the first one that needs fixing.

I think the first for loop should be something like this:
for (int i = 0; i < myArray.length; i++)
{
if (input == i)
{
myArray[i] += 1;
}
}
}
This should store add 1 to the array everytime the numbers occurs.

hey this my source code that worked out or me.
package test2;
import java.util.Arrays;
import java.util.Scanner;
public class Test2 {
public static void main(String[] args) {
java.util.Scanner input = new java.util.Scanner(System.in);
// ask for user to input numbers
System.out.println("Enter some integers between 1 and 100 (and 0 when done): ");
int[] myArray = new int[1000];//create a new array for user inputs
int number;//variable for user inputs
int count = 0;
do
{
number = input.nextInt();
myArray[count] = number;
count++;
}
while (number != 0);
int[] mySort = new int [count - 1]; //create a new array with only the numbers
for(int i = 0; i< (count-1); i++) { //get the array until 0th number into new
mySort[i] = myArray[i];
}
java.util.Arrays.sort(mySort);// sort the array in ascending order
int n = 0;
for(int i = 0; i < mySort.length; i++) {//check if the number have checked before
int occurance = 0;
if(n >= mySort[i]) {
continue;
}
else {
n = mySort[i];//if a new number found do the calculation+
for (int j=0; j<mySort.length; j++)
if (n == mySort[j])
occurance++;
System.out.print(n + " occurs " + occurance );
{
if (occurance == 1) {
System.out.println(" time.");
}
else {
System.out.println(" times.");
}
}
}
}
}
}

Checking if String entered is a binary numer, getting incorrect output

I am trying to write a program that will check if the user-entered string is a binary number, and if it is, it will output the number of 1s in the number. I had this working fine with an integer value, but since an int can't hold more than 2 billion or whatever the max value is, I am trying to rewrite it to work with Strings.
As of right now, any number I enter will output "Number entered is not binary." and when I enter 0, I will get a StringIndexOutofBoundsException. I am a fairly novice programmer, so forgive any obvious errors I may have missed, I am just asking for a possible solution to my problem or a push in the right direction. Here is my code (after trying to make it work with Strings rather than integers):
import java.util.*;
public class BinaryHW {
public static void main(String[] args) {
Scanner kb = new Scanner(System.in);
System.out.println("Enter a binary number: ");
String bin = kb.nextLine();
//the method I used to check whether or not user entered a binary
//number requires changing the value of 'bin'.
//Therefore, 'origBin' set equal to 'bin' for later use.
String origBin = bin;
int count = 0;
boolean isBinary = true;
/* if bin = 0, then this loop will be skipped because
* 0 is a binary number and need not be checked.
*/
while (Integer.parseInt(bin) != 0) {
int lastDigit = bin.charAt(bin.length() - 1);
if (lastDigit > 1) {
System.out.println("Number entered is not binary.");
isBinary = false;
break;
} else {
bin = bin.substring(bin.length() - 2);
}
}
//Again, the method I used to count the 1s in the bin number
//requires changing the value of origBin, so origBin2 is introduced
String origBin2 = origBin;
for (int i = 0; i < origBin.length(); i++) {
if (origBin.charAt(origBin.length() - 1) == 1) {
count ++;
origBin2 = origBin.substring(origBin2.length() - 2);
} else {
origBin2 = origBin.substring(origBin2.length() - 2);
}
}
if (isBinary)
if (count == 1)
System.out.println("There is "
+ count + " 1 in the binary number entered.");
else
System.out.println("There are "
+ count + " 1s in the binary number entered.");
}
}

I think you are overcomplicating things... simply iterate through your binary string, and keep track of the number of 1's reached. If a number other than 0 or 1 is found, report that input is a non-binary number. Below is a snippet which accomplishes this:
public static void main(String[] args) {
Scanner kb = new Scanner(System.in);
System.out.println("Enter a binary number: ");
String bin = kb.nextLine();
int oneCount = 0;
boolean validBinaryNum = true;
for (int i = 0; i < bin.length() && validBinaryNum; i++) {
char currentChar = bin.charAt(i);
if (currentChar == '1') oneCount++;
else if (currentChar != '0') {
validBinaryNum = false;
}
}
if (validBinaryNum && bin.length() != 0) {
System.out.println("Number of 1's: " + oneCount);
} else {
System.out.println("Number is not binary");
}
}

Inserting dashes between two odd numbers

folks. In my program I take a user input of numbers of a String type and put dashes between two odd numbers. For example:
Input = 99946 Output = 9-9-946
Input = 56730 Output = 567-30
But in my code, if I, for example, write 9933444 then the ouput that I'm getting is: 9-9-9-3-3-3-344444. It correctly separates the odd numbers by dashes but also adds extra numbers. What could be causing this bug ?
import java.util.Arrays;
import java.util.Scanner;
public class DashInsert {
public static void main(String[] args)
{
Scanner kbd = new Scanner(System.in);
System.out.println("Enter the numbers: ");
String myString = kbd.nextLine();
char[] numbers = myString.toCharArray();
String result = "";
for(int i = 1; i < numbers.length; i++)
{
int value1 = Character.getNumericValue(numbers[i]);
int value2 = Character.getNumericValue(numbers[i-1]);
if(value1 % 2 != 0 && value2 % 2 != 0)
{
result += numbers[i-1] + "-" + numbers[i] + "-";
}
else
result += numbers[i-1] + "" + numbers[i];
}
System.out.println(result);
}
}

There is a trivial one-line solution:
str = str.replaceAll("(?<=[13579])(?=[13579])", "-");
This works by matching between odd numbers and replacing the (zero-width) match with a dash. The regex is a look behind and a look ahead.
It can be done without look arounds by capturing the odd digits and putting them back using a back reference:
str = str.replaceAll("([13579])([13579])", "$1-$2");
Both solutions achieve the same result.

The code can be simplified a bit (as well as solve the "double char" bug):
String str = "9933444";
char[] numbers = str.toCharArray();
String result = "";
for(int i = 1; i < numbers.length; i++)
{
int value1 = Character.getNumericValue(numbers[i-1]);
int value2 = Character.getNumericValue(numbers[i]);
result += value1;
if(value1 % 2 != 0 && value2 % 2 != 0) {
result += "-";
}
}
result += numbers[numbers.length - 1];
System.out.println(result);
OUTPUT
9-9-3-3444
The reason for the "double char" bug is that every loop prints both the items on the places i-1 and i. which means that i will be printed again on the next loop (where it will become i-1).
In case you're using Java 8 - you can use a Stream do something that looks more like what you were originally trying to do:
public static void main(String[] args){
String str = "9933444";
List<String> lst = Arrays.asList(str.split(""));
String res = lst.stream().reduce((a,b) -> {
if (isOdd(a) && isOdd(b)) {
return a + "-" + b;
}
else {
return a + b;
}
}).get();
System.out.println(res);
}
// grep the last digit from the string and check if it's odd/even
public static boolean isOdd(String x) {
if (x.length() > 1) {
if (x.substring(x.length()-1).equals("-")) {
x = x.substring(x.length()-3, x.length()-2);
}
else {
x = x.substring(x.length() - 1);
}
}
return Integer.parseInt(x) % 2 == 1;
}
OUTPUT
9-9-3-3444

The bug is caused by the fact that, even though you are looping through your list of numbers one at a time, you write out two numbers with each loop iteration. Logically, this design will always yield repeated numbers.
Either change your loop to iterate by twos, or print a single number in each loop iteration.

Don't bother concatenating two odd numbers with the "-" in between them, during the evaluation, just add the "-" after the number that you're checking in each iteration.
public static void main(String[] args) throws Exception {
Scanner kbd = new Scanner(System.in);
System.out.println("Enter the numbers: ");
String myString = kbd.nextLine();
char[] numbers = myString.toCharArray();
String result = "";
for(int i = 0; i < numbers.length; i++) {
int value1 = Character.getNumericValue(numbers[i]);
int value2 = i + 1 < numbers.length
? Character.getNumericValue(numbers[i + 1])
: 0;
if(value1 % 2 != 0 && value2 % 2 != 0) {
result += numbers[i] + "-";
} else {
result += numbers[i];
}
}
System.out.println(result);
}
Results:
Input: 99946 Output: 9-9-946
Input: 56730 Output: 567-30
Input: 9933444 Output: 9-9-3-3444

binary to base10 in java with main method and TWO classes/method (boolean and

I am a beginner programmer. This is what I have so far. The directions for the question are kind of difficult. Here is what I am trying to accomplish..
You will write a program that converts binary numbers to base 10 numbers. This program will ask the user to enter a binary number. You will have to verify that what is entered by the user only has 0s and 1s. In the case the user entered letters, you have to keep asking the user for another input. When the input is valid, you will convert the binary number to a base 10 number. Please use the Question1.java file provided in the A2.zip file.
Valid input - In order to check if the input is valid your program should call the CheckInputCorrect method, which takes a String as an input parameter and returns a boolean value. An input string is considered valid if it corresponds to a number in binary representation.
More specifically, in the CheckInputCorrect method, you should scan this string to make sure it only contains ‘0’ or ‘1’ characters. As soon as you find a character that is not ‘0’ or ‘1’, the method should returns false. If the method reaches the end of the input string (i.e. all characters are ‘0’ or ‘1’) the method should return true.
Converter - At this point we assume that the input string is valid (checked with the CheckInputCorrect method). To convert the input from binary to base 10, you must implement the BinaryToNumber method. The BinaryToNumber method should take as parameter a String and return an integer, which corresponds to converted value in base 10.
The binary conversion should work as follows. For each digit in the binary number, if the digit is ‘1’ you should add the corresponding decimal value ( 20 for the rightmost digit, 21 for the next digits to the left, 22 for the next one, and so on) to a variable that will hold the final result to be returned. This can be easily accomplished by using a loop.
1) Am I on the right path?
2) I don't exactly know what I am doing and need you to help me figure that out....
Update1:
When I run this vvvv: It says "Please enter a binary number for me to convert: and then a place for me to type in my answer but whatever i put it just returns another box for me to type in but stops and doesn't evaluated anything.
import java.util.Scanner;
public class Question1
{
public static void main(String[] args)
{
System.out.println("Please enter a binary number for me to convert to decimal: ");
Scanner inputKeyboard = new Scanner(System.in);
String inputUser = inputKeyboard.nextLine();
boolean BinaryNumber = false;
String inputString = "";
while (!BinaryNumber){
inputString = inputKeyboard.next();
BinaryNumber = CheckInputCorrect(inputString);
System.out.println("You have given me a " + BinaryNumber + "string of binary numbers.");
}
int finalNumber = BinaryToNumber(inputString);
System.out.println("Congratulations, your binary number is " + finalNumber + ".");
}
public static boolean CheckInputCorrect(String input)
{
for (int i = 0; i < input.length(); i++)
{
while (i < input.length());
if (input.charAt(i) != '0' && input.charAt(i) != '1')
{return false;}
i++;
}
return true;
}
public static int BinaryToNumber(String numberInput)
{
int total = 0;
for (int i = 0; i < numberInput.length(); i++){
if (numberInput.charAt(i)=='1')
{
total += (int)Math.pow(2,numberInput.length() - 1 - i);
}
}
return total;
}
}
Original:
import java.util.Scanner;
public class Question1
{
public static void main(String[] args)
{
int binarynumber;
int arraySize = {0,1};
int[] binaryinput = new int[arraySize];
Scanner input = new Scanner(System.in);
System.out.println("Please enter a binary number");
binarynumber = in.nextInt();
if (binarynumber <0)
{
System.out.println("Error: Not a positive integer");
}
if (CheckInputCorrect) = true;
{
System.out.print(CheckInputCorrect);
}
public static boolean CheckInputCorrect(String input);
{
boolean b = true;
int x, y;
x = 0
y = 1
b = x || y
while (b >= 0 {
System.out.print("Please enter a binary number")
for (int i = 0; i < binarynumber.length; i++)
{
binaryinput[i] = in.nextInt();
if (binaryinput[i] = b.nextInt();
System.out.printl("Binary number is:" + binaryinput);
break outer;
if (binarynumber != b)
{
System.out.println("Error: Not a binary number")
}
return true;
}
}
public static int BinaryToNumber(String numberInput)
{
int remainder;
if (binarynumber <= 1) {
System.out.print(number);
return; // KICK OUT OF THE RECURSION
}
remainder = number %2;
printBinaryform(number >> 1);
System.out.print(remainder);
return 0;
}
}
}

As mentioned in my comments, your updated code contains two errors
while (i < input.length()); is an infinite loop, because it has no body. Therefore i cannot be increased and will stay lower than input.length().
inputString = inputKeyboard.next(); request another input after the first one and the first input will be ignored.
This is a fixed and commented version of your updated code:
public class Question1 {
public static void main(String[] args) {
System.out.println("Please enter a binary number for me to convert to decimal: ");
Scanner inputKeyboard = new Scanner(System.in);
String inputUser = inputKeyboard.nextLine();
//boolean BinaryNumber = false; // not needed anymore
//String inputString = ""; // not needed too
while (!checkInputCorrect(inputUser)) { // there is no reason for using a boolean var here .. directly use the returned value of this method
System.out.println("You have given me an invalid input. Please enter a binary number: "); // changed this message a little bit
inputUser = inputKeyboard.nextLine(); // re-use the "old" variable inputUser here
}
int finalNumber = binaryToNumber(inputUser);
System.out.println("Congratulations, your decimal number is " + finalNumber + ".");
}
public static boolean checkInputCorrect(String input) { // method names should start with a lower case letter
for (int i = 0; i < input.length(); i++) {
//while (i < input.length()); // this loop is deadly. Please think about why
if (input.charAt(i) != '0' && input.charAt(i) != '1') {
return false;
}
//i++; // what is the purpose of this? The "for" loop will increment "i" for you
}
return true;
}
public static int binaryToNumber(String numberInput) { //method name ... lower case letter ;)
int total = 0;
for (int i = 0; i < numberInput.length(); i++) {
if (numberInput.charAt(i) == '1') {
total += (int) Math.pow(2, numberInput.length() - 1 - i);
}
}
return total;
}
}