We Keep Coding

why does my code not giving me reverse number accurately [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 2 years ago.
Improve this question
Java Code Example: i am facing issue when i insert 010 it shows only 1 instead of whole number and when i insert 200 it gave me reverse number only 2 while it should show me exact 002 how can i get exact number in reverse:
class Reverse{
public static void main(String args[]){
int remander,sum=0;
int number=454;
while(n>0) // loop here
{
remander=n%10; // Getting remainder
sum=(sum*10)+remander;
number=number/10;
}
System.out.println("Reverse"+sum);
}
}

It is clear that you are new to Java, and new to programming. Welcome to StackOverflow. I'm going to add some extra advice that I don't normally offer, to help you get better faster.
Don't reuse a variable for more than one thing. Use it for only one thing. That means that if a variable holds you input, don't overwrite it with intermediate values. Variables are (relatively) cheap, create another. You can practice this by marking your variables final which will prevent you from changing their values after assigned.
Use meaningful variable names. "r" is not a meaningful name. "remainder" is a meaningful name. This helps others immediately, and will help you later, when the program isn't fresh on your mind and you don't remember what "r" means.
Once you master not reusing variables, you will have a lot of variables. Look for variables that are "set once and read once" and try to rearrange your code to remove them completely. It will take some time to understand what I mean, but if you have heard of "refactoring" the idea is to remove the variable name because it isn't used twice (so it probably isn't a key part of your problem.
Now, applying these rules
final int number = 454;
// get the number digits, smallest to largest
int remainder = number;
int accumulator = 0;
while (remainder > 0) {
final int digit = remainder % 10;
accumulator = accumulator * 10 + digit;
remainder = remainder / 10;
}
final int reversed = accumulator;
note that digit is never used twice
final int number = 454;
// get the number digits, smallest to largest
int remainder = number;
int accumulator = 0;
while (remainder > 0) {
accumulator = accumulator * 10 + remainder % 10;
remainder = remainder / 10;
}
final int reversed = accumulator;
Note that all that "logic" could have a really good name put to it, so we'll create a "method"
public int reverse(int value) {
int remainder = value;
int accumulator = 0;
while (remainder > 0) {
accumulator = accumulator * 10 + remainder % 10;
remainder = remainder / 10;
}
return accumulator;
}
which cleans up the other code to
final int number = 454;
final int reversed = reverse(number);
or maybe even
final int reversed = reverse(454);

Adding Values to a List and Converting to BigInteger - Java [closed]

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 2 years ago.
Improve this question
I'm writing a code that determines the smallest integer that is a sequence of sevens followed by some number of zeros (possibly none) that is divisible by int n. Since this number can be massive, the return value should be a BigInteger.
My code so far has an if-else ladder that covers the case that if any int n is not divisible by two or five is guaranteed to only contain sevens (no zeros). In the case where int n is not divisible by two or five, my thought process was to continue adding sevens to a LinkedList in a while loop, until that list (converted to a BigInteger) is divisible by int n. The same logic goes for the case where int n is divisible by two or five, except a two for-loops would add seven and zero to the list.
My code is getting a runtime error when converting the list to a string and then to a BigInteger, specifically on the line BigInteger numBig = new BigInteger(str);. The error is: "java.lang.NumberFormatException: Zero length BigInteger (in java.math.BigInteger)" Also, I'm not quite sure the logic is sound for the case where int n is divisible by two or five.

You don't need BigInteger for this task. The idea is the following:
First you determine the number of required zeros. Since number composed of only sevens cannot be divided by 2 or 5, the number of zeros is equal to the maximum power of 2 or 5 in number n.
Now we have a number n which is not divisible by 2 or 5. Suppose that a remainder of the division of a number composed of m sevens by n is equal to r:
777...m-times..777 mod n = r
Then number composed of (m+1) sevens will have a remainder 10*r + 7, because
777..(m+1)-times...777 = 777...m-times...7 * 10 + 7
So you can just recalculate the remainder until it becomes zero.
public static BigInteger method(int n) {
int two;
for (two = 0; n % 2 == 0; two++) n /= 2;
int five;
for (five = 0; n % 5 == 0; five++) n /= 5;
int zeros = Math.max(two, five);
int sevens = 1;
int r = 7 % n;
while (r != 0) {
r = (r * 10 + 7) % n;
sevens++;
}
// Now just make a number of 'sevens' sevens and 'zeros' zeros:
StringBuilder result = new StringBuilder();
for (int i = 0; i < sevens; i++) {
result.append("7");
}
for (int i = 0; i < zeros; i++) {
result.append("0");
}
return new BigInteger(result.toString());
}

"Zero length BigInteger" means you're trying to create a BigInteger from something that has length of 0. The stack trace would tell you on which line exactly.
I would guess the bug is in your convert method. If you pass in an empty list, it tries to convert an empty string into BigInteger with new BigInteger("")
I don't know what your algorithm is supposed to do in this case. If for example you want to convert an empty list into the number zero, you can do:
if (res.isEmpty()) return BigInteger.ZERO;

Reverse the digits of an integer [duplicate]

This question already has answers here:
Java reverse an int value without using array
(33 answers)
Closed 3 years ago.
I'm a Java beginner so please pardon me if the question seems silly but I already searched the forums but it seems like no one has my problem.
I need to reverse the digits of an integer, and my class hasn't covered while or if loops yet, so I can't use those. All answers I can find on stackoverflow use those, so I can't use those.
the input I am given is below 10000 and above 0 and the code I have written has no problem reversing the integer if the input is 4 digits (e.g. 1000 - 9999) but once the input is between 1 - 999 it creates zeroes on the right hand side but according to the answer sheets its wrong.
For example: 1534 gets turned into 4351, but
403 becomes 3040 instead of the 304 it should be, and 4 becomes 4000 instead of 4.
I've tried different things in the code but it seems to just keep giving the same answer. Or maybe I'm just missing some key mathematics, I'm not sure.
Scanner scan = new Scanner(System.in);
System.out.println ("Enter an integer:");
int value = scan.nextInt();
int digit = (value % 10);
value = (value / 10);
int digit2 = (value % 10);
value = (value / 10);
int digit3 = (value % 10);
value = (value / 10);
int digit4 = (value % 10);
String reversednum = ("" + digit + digit2 + digit3 + digit4);
System.out.println ( reversednum);
and
Scanner scan = new Scanner(System.in);
System.out.println ("Enter an integer:");
int value = scan.nextInt();
int digit = (value % 10);
int reversednum = (digit);
value = (value /10);
digit = (value % 10);
reversednum = (reversednum * 10 + digit);
value = (value / 10);
digit = (value % 10);
reversednum = (reversednum * 10 + digit);
value = (value / 10);
digit = (value);
reversednum = (reversednum * 10 + digit);
System.out.println (reversednum);
What am I doing wrong?

You can convert from int to String -> reverse String -> convert again in int.
This is a code example.
public int getReverseInt(int value) {
String revertedStr = new StringBuilder(value).reverse().toString();
return Integer.parseInt(revertedStr);
}

Your code assumes that the number can be divided by 1000, which is clearly not the case for numbers below 1000. So add some if statements:
public int reverseNumber(int n) {
// step one: we find the factors using integer maths
int s = n;
int thousands = s / 1000; // this will be 0 if the number is <1000
s = s - thousands*1000;
int hundreds = s / 100; // this will be 0 if the number is <100
s = s - hundreds*100;
int tens = s / 10; // etc.
s = s - tens*10;
int ones = s;
// then: let's start reversing. single digit?
if (n<10) return n;
// two digits?
if (n<100) {
return ones*10 + tens;
}
// etc.
if (n<1000) {
return ones*100 + tens*10 + hundreds;
}
if (n<10000) {
return ones*1000 + tens*100 + hundreds*10 + thousands;
}
// if we get here, we have no idea what to do with this number.
return n;
}

Without spoon-feeding you code (leaving the value of writing your own homework code intact)...
Although you've said you can't use a loop, I don't think there's a sane approach that doesn't use one. Your basic problem is you have hard-coded a solution that works when the number happens to have 4 digits, rather than using code that adapts to a variable length. ie, are not using a loop.
All is not lost with your code however. You have figured out the essence of the solution. You just need to convert it to work processing one digit at a time. Consider using recursion, that divides the number by 10 each time and continues until the number is zero. Of course, you’ll have to capture the end digit before it’s lost by division.
Pseudo code may look like:
pass in the number and the current result
if the number is 0 return result
multiply result by 10 and add remainder of number divided by 10
return the result of calling self with number divided by 10 and result
then call this passing number and zero

Using modulus and division:
int nbr = 123; // reverse to 321 or 3*10*10 + 2*10 + 1
int rev = 0;
while(nbr > 0) {
rev *= 10; // shift left 1 digit
int temp = nbr % 10; // get LO digit
rev += temp; // add in next digit
nbr /= 10; // move to next digit
}
Or a recursive method:
public static int reverseInt(int number, int value) {
switch(number) { // is this conditional statement allowed???
case 0:
return value;
}
value *= 10;
int lod = number % 10;
value += lod;
number /= 10;
return reverseInt(number, value);
}

(Java)Why does my decimal to binary conversion method only work some of the time when counting consecutive integers in a character array?

I'm a novice Java coder working on a problem dealing with counting consecutive integers in the binary forms of numbers.
The numbers are read from the input, and converted to binary using the method called conversion. The binary form is then sent to a character array where the for loop checks for consecutive characters(specifically the number 1) and prints the maximum count as the final answer.
I've managed to get the code to a state where I feel it should be working, but I've only had success with about half of the test cases. The larger number conversions like 262,141 tend to produce incorrect answers. Can anyone tell me where I've gone wrong?
I have a suspicion that it's something to do with the character array, but after several hours of research I haven't been able to find a solution to my particular problem.
import java.io.*;
import java.util.*;
public class Solution {
public static int conversion(int decimal){//this will take the decimal from the input and convert it to binary
int result = 0;//the result from each step of the conversion
int base = 1;//used to multiply the remainder by 1, 10, 100 etc
while(decimal > 0){
int remainder = decimal % 2;//takes the remainder of the iteration
decimal = decimal / 2;//halves the decimal number
result = result + (remainder * base);//pseudo concatenation of the binary
base = base * 10;//increases the base multiplier to continue filling out the binary leftward
}
return result;//returns result after loop has finished
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();//scan the input to obtain the decimal number
int binaryForm = conversion(n);//convert the decimal to binary and assign to binaryForm variable
String stringForm = Integer.toString(binaryForm);//convert binaryForm to a String
int counter = 1;
int max = 1;
char testArray[] = stringForm.toCharArray();//send stringForm to fill out testArray
for(int i = 0; i < testArray.length - 1; i++){//loops through testArray to test stringForm values
if(testArray[i] == testArray[i + 1] && testArray[i] == '1'){//if consecutive values equal char 1, increase counter
counter += 1;
if(counter > max){
max = counter;//if counter is higher than current maxCounter, increase maxCounter
}
}
else {//if consecutive values do not equal 1, reset counter
counter = 1;
}
}
System.out.print(max);//print the maximum consecutive values for the decimal input when converted to binary
}
}

You are trying to create a binary representation of a decimal number using an integer. This will work for smaller numbers but it doesn't take long for you to reach an overflow. You should use a string representation of the binary number like so
String numBin = "";
while(num > 0)
{
numBin = num % 2 + numBin;
num = num / 2;
}
System.out.println("Binary Representation: " + numBin);
Then take that string and loop through it calculating the consecutive counts of 1's
int consecutiveCount = 0;
for(int i = 0; i < numBin.length() - 1; i++)
{
if(numBin.charAt(i) == '1' && numBin.charAt(i + 1) == '1')
{
consecutiveCount++;
}
}
System.out.println("Consecutive Count: " + consecutiveCount);
Output
Number: 261141
Binary Representation: 111111110000010101
Consecutive Count: 7
Number: 3
Binary Representation: 11
Consecutive Count: 1
Number: 18
Binary Representation: 10010
Consecutive Count: 0
Number: 1111111
Binary Representation: 100001111010001000111
Consecutive Count: 5

Retrieving the first digit of a number [duplicate]

This question already has answers here:
Return first digit of an integer
(25 answers)
Closed 5 years ago.
I am just learning Java and am trying to get my program to retrieve the first digit of a number - for example 543 should return 5, etc. I thought to convert to a string, but I am not sure how I can convert it back? Thanks for any help.
int number = 534;
String numberString = Integer.toString(number);
char firstLetterChar = numberString.charAt(0);
int firstDigit = ????

Almost certainly more efficient than using Strings:
int firstDigit(int x) {
while (x > 9) {
x /= 10;
}
return x;
}
(Works only for nonnegative integers.)

int number = 534;
int firstDigit = Integer.parseInt(Integer.toString(number).substring(0, 1));

firstDigit = number/((int)(pow(10,(int)log(number))));
This should get your first digit using math instead of strings.
In your example log(543) = 2.73 which casted to an int is 2.
pow(10, 2) = 100
543/100 = 5.43 but since it's an int it gets truncated to 5

int firstDigit = Integer.parseInt(Character.toString(firstLetterChar));

int number = 534;
String numberString = "" + number;
char firstLetterchar = numberString.charAt(0);
int firstDigit = Integer.parseInt("" + firstLetterChar);

Integer.parseInt will take a string and return a int.

This example works for any double, not just positive integers and takes into account negative numbers or those less than one. For example, 0.000053 would return 5.
private static int getMostSignificantDigit(double value) {
value = Math.abs(value);
if (value == 0) return 0;
while (value < 1) value *= 10;
char firstChar = String.valueOf(value).charAt(0);
return Integer.parseInt(firstChar + "");
}
To get the first digit, this sticks with String manipulation as it is far easier to read.

int number = 534;
int firstDigit = number/100;
( / ) operator in java divide the numbers without considering the reminder so when we divide 534 by 100 , it gives us (5) .
but if you want to get the last number , you can use (%) operator
int lastDigit = number%10;
which gives us the reminder of the division , so 534%10 , will yield the number 4 .

This way might makes more sense if you don't want to use str methods
int first = 1;
for (int i = 10; i < number; i *= 10) {
first = number / i;
}