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The following code to find max Number from N+1 size array excluding number N from it.
After I run code and copy paste sample input of 3 testcases.... some of the testcases answer are already shown as seen in the attached picture. Why does that happen and how to avoid it???
package beginner;
import java.util.*;
import java.io.*;
public class FindMax {
public static void main(String[] args) throws IOException {
BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(sc.readLine());
while(t>0) {
String [] in = sc.readLine().split(" ");
int [] arr = new int[in.length];
for(int j=0; j<in.length; j++) {
arr[j] = Integer.parseInt(in[j]);
}
System.out.println(findMax(arr));
t--;
}
}
private static int findMax(int[] arr) {
int max = Integer.MIN_VALUE;
int n = arr.length-1;
for(int j=0; j<arr.length; j++) {
if(arr[j] == n) continue;
if(max < arr[j]) max = arr[j];
}
return max;
}
}
Since the text you're pasting to the console contains newline characters, it triggers the nextLine to return, and the result of findMax is printed after that.
Try the below code (It would print the min, max, avg , sum , omit the part you wont need) :
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
System.out.println("Enter how many numbers you want to enter : ");
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
int k = Integer.parseInt(str);
int arr[] = new int[k];
System.out.println("Enter the numbers : ");
for(int i=0;i<k;i++) {
BufferedReader br1= new BufferedReader(new InputStreamReader(System.in));
str=br1.readLine();
arr[i] = Integer.parseInt(str);
}
findMax(arr);
}
private static void findMax(int[] arr) {
int sum=0,i,count=0,max,min,n;
int len = arr.length;
float avg;
max = arr[0];
min = arr[0];
while(count < len) {
sum += arr[count];
if(min > arr[count])
min = arr[count];
if (max < arr[count])
max = arr[count];
count += 1;
}
avg = sum/len;
System.out.println("\nPrinting the results : \n");
System.out.println("Sum = " +sum);
System.out.println("Average = " +avg);
System.out.println("Maximum = " +max);
System.out.println("Minimum = " +min);
}
}
Output :
Enter how many numbers you want to enter :
3
Enter the numbers :
1
2
3
Printing the results :
Sum = 6
Average = 2.0
Maximum = 3
Minimum = 1
I want to read a number from the user and then sum the last seven digits of the entered number. What is the best way to do this? This is my code, but unfortunately it does not work:
class ersteAufgabe {
public static void main (String[] args)
{
Scanner s = new Scanner(System.in);
double [] a = new double[10];
for (int i = 0;i<10;i++)
{
a[i]=s.nextInt();
}
s.close();
System.out.println(a[0]);
}
}
I wanted only one number to be read and used as an array. Only now he expects 10 inputs from me.
public static int lastDigitsSum(int total) {
try (Scanner scan = new Scanner(System.in)) {
String str = scan.next();
int count = 0;
for (int i = str.length() - 1, j = 0; i >= 0 && j < total; i--, j++) {
if (Character.isDigit(str.charAt(i)))
count += str.charAt(i) - '0';
else
throw new RuntimeException("Input is not a number: " + str);
}
return count;
}
}
First you have to recognize if the entered value is a number and has at least 7 digits. Unless you have to output an error message. Convert the entered value to String and use the class Character.isDigit(); to check if the characters are numbers. Then you can use some methods from the String class like substring(..). At the end do a Unit-Test with erroneous/valid values to see if your code is robust. Close the BufferedReader and Resources when you are done by using finally { br.close() }. Push your code in methods and use an instance class erste-Aufgabe (first exercise).. When you are really really done use JFrame for a GUI-Application.
private static final int SUM_LAST_DIGITS = 7;
public void minimalSolution() {
String enteredValue = "";
showInfoMessage("Please enter your number with at least " + SUM_LAST_DIGITS + " digits!");
try (Scanner scan = new Scanner(System.in)) {
enteredValue = scan.next();
if (enteredValue.matches("^[0-9]{" + SUM_LAST_DIGITS + ",}$")) {
showInfoMessage(enteredValue, lastDigitsSum(enteredValue));
} else {
showErrorMessage(enteredValue);
}
} catch(Exception e) {
showErrorMessage(e.toString());
}
}
public int lastDigitsSum(String value) {
int count = 0;
for (int i = value.length() - 1, j = 0; i >= 0 && j < SUM_LAST_DIGITS; i--, j++)
count += value.charAt(i) - '0';
return count;
}
public void showInfoMessage(String parMessage) {
System.out.println(parMessage);
}
public void showInfoMessage(String parValue, int parSum) {
System.out.println("Your entered value: [" + parValue + "]");
System.out.println("The summed value of the last 7 digits are: [" + parSum + "]");
}
public void showErrorMessage(String parValue) {
System.err.println("Your entered value: [" + parValue + "] is not a valid number!");
}
I am trying to reverse the output of my code and it's working, the problem is when I input 12345678910 the "10" is not displayed.
import java.util.Scanner;
public class Test3{
public static void main(String args[]) {
Scanner in=new Scanner(System.in);
int num=0,rev=0,rev1=0,limit=0,num1=0;
System.out.print("Enter A Number: ");
limit=in.nextInt();
for (int i=0;i<limit;i++){
num=in.nextInt();
while (num >0) {
rev = rev*10 + num % 10;
num /= 10;
}
}
num1=rev;
while(num1>0){
rev1 = rev1 * 10 + num1 %10;
num1 /=10;
}
System.out.println(rev1);
}
}
12345678910 the "10" is not display
because this while loop while (num >0) is running twice for 10, as a result rev is overflowing and holding negative value (-539222978) since int can not hold more than 2^32-1 it wraps around to negative value (-539222978) . what is why you got revl initial value 0 got printed,
So try to use StringBuilder.
import java.util.Scanner;
import java.util.*;
public class Test3{
public static void main(String args[]) {
Scanner in=new Scanner(System.in);
StringBuilder sb = new StringBuilder();
String numStr;
System.out.print("Enter A Number: ");
int limit=in.nextInt();
for (int i=0;i<limit;i++){
numStr=in.next();
sb.append(numStr);
}
System.out.println(sb.reverse().toString());
}
}
Your code is a bit complicated:
public static String ReturnString()
{
long input = 12345678910L;
String inputAsString = Long.toString(input);
String rev = "";
for (int temp = inputAsString.length() - 1; temp > -1; temp--)
{
rev += inputAsString.charAt(temp);
}
return rev;
}
I am trying to convert decimal to binary numbers from the user's input using Java.
I'm getting errors.
package reversedBinary;
import java.util.Scanner;
public class ReversedBinary {
public static void main(String[] args) {
int number;
Scanner in = new Scanner(System.in);
System.out.println("Enter a positive integer");
number=in.nextInt();
if (number <0)
System.out.println("Error: Not a positive integer");
else {
System.out.print("Convert to binary is:");
System.out.print(binaryform(number));
}
}
private static Object binaryform(int number) {
int remainder;
if (number <=1) {
System.out.print(number);
}
remainder= number %2;
binaryform(number >>1);
System.out.print(remainder);
{
return null;
} } }
How do I convert Decimal to Binary in Java?
Integer.toBinaryString() is an in-built method and will do quite well.
Integer.toString(n,8) // decimal to octal
Integer.toString(n,2) // decimal to binary
Integer.toString(n,16) //decimal to Hex
where n = decimal number.
Your binaryForm method is getting caught in an infinite recursion, you need to return if number <= 1:
import java.util.Scanner;
public class ReversedBinary {
public static void main(String[] args) {
int number;
Scanner in = new Scanner(System.in);
System.out.println("Enter a positive integer");
number = in.nextInt();
if (number < 0) {
System.out.println("Error: Not a positive integer");
} else {
System.out.print("Convert to binary is:");
//System.out.print(binaryform(number));
printBinaryform(number);
}
}
private static void printBinaryform(int number) {
int remainder;
if (number <= 1) {
System.out.print(number);
return; // KICK OUT OF THE RECURSION
}
remainder = number % 2;
printBinaryform(number >> 1);
System.out.print(remainder);
}
}
I just want to add, for anyone who uses:
String x=Integer.toBinaryString()
to get a String of Binary numbers and wants to convert that string into an int. If you use
int y=Integer.parseInt(x)
you will get a NumberFormatException error.
What I did to convert String x to Integers, was first converted each individual Char in the String x to a single Char in a for loop.
char t = (x.charAt(z));
I then converted each Char back into an individual String,
String u=String.valueOf(t);
then Parsed each String into an Integer.
Id figure Id post this, because I took me a while to figure out how to get a binary such as 01010101 into Integer form.
/**
* #param no
* : Decimal no
* #return binary as integer array
*/
public int[] convertBinary(int no) {
int i = 0, temp[] = new int[7];
int binary[];
while (no > 0) {
temp[i++] = no % 2;
no /= 2;
}
binary = new int[i];
int k = 0;
for (int j = i - 1; j >= 0; j--) {
binary[k++] = temp[j];
}
return binary;
}
public static void main(String h[])
{
Scanner sc=new Scanner(System.in);
int decimal=sc.nextInt();
String binary="";
if(decimal<=0)
{
System.out.println("Please Enter greater than 0");
}
else
{
while(decimal>0)
{
binary=(decimal%2)+binary;
decimal=decimal/2;
}
System.out.println("binary is:"+binary);
}
}
The following converts decimal to Binary with Time Complexity : O(n) Linear Time and with out any java inbuilt function
private static int decimalToBinary(int N) {
StringBuilder builder = new StringBuilder();
int base = 2;
while (N != 0) {
int reminder = N % base;
builder.append(reminder);
N = N / base;
}
return Integer.parseInt(builder.reverse().toString());
}
In C# , but it's just the same as in Java :
public static void findOnes2(int num)
{
int count = 0; // count 1's
String snum = ""; // final binary representation
int rem = 0; // remainder
while (num != 0)
{
rem = num % 2; // grab remainder
snum += rem.ToString(); // build the binary rep
num = num / 2;
if (rem == 1) // check if we have a 1
count++; // if so add 1 to the count
}
char[] arr = snum.ToCharArray();
Array.Reverse(arr);
String snum2 = new string(arr);
Console.WriteLine("Reporting ...");
Console.WriteLine("The binary representation :" + snum2);
Console.WriteLine("The number of 1's is :" + count);
}
public static void Main()
{
findOnes2(10);
}
It might seem silly , but if u wanna try utility function
System.out.println(Integer.parseInt((Integer.toString(i,2))));
there must be some utility method to do it directly, I cant remember.
public static void main(String[] args)
{
Scanner in =new Scanner(System.in);
System.out.print("Put a number : ");
int a=in.nextInt();
StringBuffer b=new StringBuffer();
while(a>=1)
{
if(a%2!=0)
{
b.append(1);
}
else if(a%2==0)
{
b.append(0);
}
a /=2;
}
System.out.println(b.reverse());
}
Binary to Decimal without using Integer.ParseInt():
import java.util.Scanner;
//convert binary to decimal number in java without using Integer.parseInt() method.
public class BinaryToDecimalWithOutParseInt {
public static void main(String[] args) {
Scanner input = new Scanner( System.in );
System.out.println("Enter a binary number: ");
int binarynum =input.nextInt();
int binary=binarynum;
int decimal = 0;
int power = 0;
while(true){
if(binary == 0){
break;
} else {
int temp = binary%10;
decimal += temp*Math.pow(2, power);
binary = binary/10;
power++;
}
}
System.out.println("Binary="+binarynum+" Decimal="+decimal); ;
}
}
Output:
Enter a binary number:
1010
Binary=1010 Decimal=10
Binary to Decimal using Integer.parseInt():
import java.util.Scanner;
//convert binary to decimal number in java using Integer.parseInt() method.
public class BinaryToDecimalWithParseInt {
public static void main(String[] args) {
Scanner input = new Scanner( System.in );
System.out.println("Enter a binary number: ");
String binaryString =input.nextLine();
System.out.println("Result: "+Integer.parseInt(binaryString,2));
}
}
Output:
Enter a binary number:
1010
Result: 10
A rather simple than efficient program, yet it does the job.
Scanner sc = new Scanner(System.in);
System.out.println("Give me my binaries");
int str = sc.nextInt(2);
System.out.println(str);
All your problems can be solved with a one-liner!
To incorporate my solution into your project, simply remove your binaryform(int number) method, and replace System.out.print(binaryform(number)); with System.out.println(Integer.toBinaryString(number));.
/**
* converting decimal to binary
*
* #param n the number
*/
private static void toBinary(int n) {
if (n == 0) {
return; //end of recursion
} else {
toBinary(n / 2);
System.out.print(n % 2);
}
}
/**
* converting decimal to binary string
*
* #param n the number
* #return the binary string of n
*/
private static String toBinaryString(int n) {
Stack<Integer> bits = new Stack<>();
do {
bits.push(n % 2);
n /= 2;
} while (n != 0);
StringBuilder builder = new StringBuilder();
while (!bits.isEmpty()) {
builder.append(bits.pop());
}
return builder.toString();
}
Or you can use Integer.toString(int i, int radix)
e.g:(Convert 12 to binary)
Integer.toString(12, 2)
public static String convertToBinary(int dec)
{
String str = "";
while(dec!=0)
{
str += Integer.toString(dec%2);
dec /= 2;
}
return new StringBuffer(str).reverse().toString();
}
Practically you can write it as a recursive function. Each function call returns their results and add to the tail of the previous result. It is possible to write this method by using java as simple as you can find below:
public class Solution {
private static String convertDecimalToBinary(int n) {
String output = "";
if (n >= 1) {
output = convertDecimalToBinary(n >> 1) + (n % 2);
}
return output;
}
public static void main(String[] args) {
int num = 125;
String binaryStr = convertDecimalToBinary(num);
System.out.println(binaryStr);
}
}
Let us take a look how is the above recursion working:
After calling convertDecimalToBinary method once, it calls itself till the value of the number will be lesser than 1 and return all of the concatenated results to the place where it called first.
References:
Java - Bitwise and Bit Shift Operators https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html
If you want to reverse the calculated binary form , you can use the StringBuffer class and simply use the reverse() method . Here is a sample program that will explain its use and calculate the binary
public class Binary {
public StringBuffer calculateBinary(int number) {
StringBuffer sBuf = new StringBuffer();
int temp = 0;
while (number > 0) {
temp = number % 2;
sBuf.append(temp);
number = number / 2;
}
return sBuf.reverse();
}
}
public class Main {
public static void main(String[] args) throws IOException {
System.out.println("enter the number you want to convert");
BufferedReader bReader = new BufferedReader(newInputStreamReader(System.in));
int number = Integer.parseInt(bReader.readLine());
Binary binaryObject = new Binary();
StringBuffer result = binaryObject.calculateBinary(number);
System.out.println(result);
}
}
The better way of doing it:
public static void main(String [] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(bf.readLine().trim());
double ans = 0;
int i=0;
while(t!=0){
int digit = t & 1;
ans = ans + (digit*Math.pow(10,i));
i++;
t =t>>1;
}
System.out.println((int)ans);
}
I just solved this myself, and I wanted to share my answer because it includes the binary reversal and then conversion to decimal. I'm not a very experienced coder but hopefully this will be helpful to someone else.
What I did was push the binary data onto a stack as I was converting it, and then popped it off to reverse it and convert it back to decimal.
import java.util.Scanner;
import java.util.Stack;
public class ReversedBinary
{
private Stack<Integer> st;
public ReversedBinary()
{
st = new Stack<>();
}
private int decimaltoBinary(int dec)
{
if(dec == 0 || dec == 1)
{
st.push(dec % 2);
return dec;
}
st.push(dec % 2);
dec = decimaltoBinary(dec / 2);
return dec;
}
private int reversedtoDecimal()
{
int revDec = st.pop();
int i = 1;
while(!st.isEmpty())
{
revDec += st.pop() * Math.pow(2, i++);
}
return revDec;
}
public static void main(String[] args)
{
ReversedBinary rev = new ReversedBinary();
System.out.println("Please enter a positive integer:");
Scanner sc = new Scanner(System.in);
while(sc.hasNextLine())
{
int input = Integer.parseInt(sc.nextLine());
if(input < 1 || input > 1000000000)
{
System.out.println("Integer must be between 1 and 1000000000!");
}
else
{
rev.decimaltoBinary(input);
System.out.println("Binary to reversed, converted to decimal: " + rev.reversedtoDecimal());
}
}
}
}
import java.util.*;
public class BinaryNumber
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter the number");
int n = scan.nextInt();
int rem;
int num =n;
String str="";
while(num>0)
{
rem = num%2;
str = rem + str;
num=num/2;
}
System.out.println("the bunary number for "+n+" is : "+str);
}
}
This is a very basic procedure, I got this after putting a general procedure on paper.
import java.util.Scanner;
public class DecimalToBinary {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter a Number:");
int number = input.nextInt();
while(number!=0)
{
if(number%2==0)
{
number/=2;
System.out.print(0);//Example: 10/2 = 5 -> 0
}
else if(number%2==1)
{
number/=2;
System.out.print(1);// 5/2 = 2 -> 1
}
else if(number==2)
{
number/=2;
System.out.print(01);// 2/2 = 0 -> 01 ->0101
}
}
}
}
//converts decimal to binary string
String convertToBinary(int decimalNumber){
String binary="";
while(decimalNumber>0){
int remainder=decimalNumber%2;
//line below ensures the remainders are reversed
binary=remainder+binary;
decimalNumber=decimalNumber/2;
}
return binary;
}
One of the fastest solutions:
public static long getBinary(int n)
{
long res=0;
int t=0;
while(n>1)
{
t= (int) (Math.log(n)/Math.log(2));
res = res+(long)(Math.pow(10, t));
n-=Math.pow(2, t);
}
return res;
}
Even better with StringBuilder using insert() in front of the decimal string under construction, without calling reverse(),
static String toBinary(int n) {
if (n == 0) {
return "0";
}
StringBuilder bldr = new StringBuilder();
while (n > 0) {
bldr = bldr.insert(0, n % 2);
n = n / 2;
}
return bldr.toString();
}
No need of any java in-built functions. Simple recursion will do.
public class DecimaltoBinaryTest {
public static void main(String[] args) {
DecimaltoBinary decimaltoBinary = new DecimaltoBinary();
System.out.println("hello " + decimaltoBinary.convertToBinary(1000,0));
}
}
class DecimaltoBinary {
public DecimaltoBinary() {
}
public int convertToBinary(int num,int binary) {
if (num == 0 || num == 1) {
return num;
}
binary = convertToBinary(num / 2, binary);
binary = binary * 10 + (num % 2);
return binary;
}
}
int n = 13;
String binary = "";
//decimal to binary
while (n > 0) {
int d = n & 1;
binary = d + binary;
n = n >> 1;
}
System.out.println(binary);
//binary to decimal
int power = 1;
n = 0;
for (int i = binary.length() - 1; i >= 0; i--) {
n = n + Character.getNumericValue(binary.charAt(i)) * power;
power = power * 2;
}
System.out.println(n);
I'm having trouble in getting the binary. I do not know what's wrong. The binary number always ends up in gibberish. Also some parts like the new int[31] thing was from HW but I can't get around to make print the actual binary.
public class DectoBinary {
public static void main(String[]args) {
Scanner CONSOLE = new Scanner(System.in);
System.out.print("Please enter a nonnegative integer: ");
int value = CONSOLE.nextInt();
while (value < 0) {
System.out.print("number outside range.");
System.out.print
("Please enter a nonnegative interger more than 0: ");
value = CONSOLE.nextInt();
}
int[] intArray = new int[31];
decimalToBinary(value, intArray);
System.out.println(value + "" + intArray);
}
public static int[] decimalToBinary(int value, int[]intArray) {
int i = 0;
while (value != 0) {
if (value % 2 == 1)
intArray[i] = 1;
else
intArray[i] = 0;
value /= 2;
i++;
}
return intArray;
}
}
I think the error is on this line:
System.out.println(value + "" + intArray);
You cannot print an array of integers like this: you should either convert it to string, or write a loop that prints the array digit by digit:
for (int i : inrArray) {
System.out.print(intArray[i]);
}
System.out.println();
You do not need to pass in the output array as well: you can create it inside the function.
public static int[] decimalToBinary(int value) {
int count = 1;
int tmp = value;
while (tmp != 0) {
tmp /= 2;
count++;
}
int[] intArray = new int[count];
// Do the conversion here...
return intArray;
}
You can simply use Integer.toBinaryString(int).
Actually the is a very simple way to get binary numbers in java using BigInteger
public String dectoBin(int num){
String s = ""+num;
BigInteger bi = new BigInteger(s);
String bin = bi.toString(2);
return bin
}
BigInteger.toString(2) returns the number stored on the numerical base specified inside the parenthesis. Is a very easy way to get arround this problems.
System.out.println(value + "" + intArray);
the 'intArray' is a arrays's address, so, if you want to get actual binary you can use Arrays.toString(intArray)
As dasblinkenlight you need to print the array item by item. If you want a nice alternative, you can use a recursive printing of value mod 2 (modulo 2 gives you 1 or 0)
/** print directly*/
public static void decimalToBinary(int value) {
if(value > 1){
System.out.print(decimalToBinary(value/2) + "" + (value%2));
/**recursion with implicit cast to string*/
} else {
System.out.print( (value==0)?"":"1");
}
}
It works with any Base
Well actually to print the array, because all the slots in the array are initialized at 0 you need to detect where the first one begins, so.
you need to replace
System.out.println(value + "" + intArray);
with something like this;
System.out.println(vale + " ");
boolean sw = false;
for(int i=0;i<intArray.length;i++){
if(!sw)
sw = (intArray[i]==1);//This will detect when the first 1 appears
if(sw)
System.out.println(intArray[1]); //This will print when the sw changes to true everything that comes after
}
Here is a program to convert Decimal nos. into Binary.
import java.util.Scanner;
public class decimalToBinary {
static int howManyTerms (int n) {
int term = 0;
while (n != 0) {
term ++;
n /= 2;
}
return term;
}
static String revArrayofBin2Str (int[] Array) {
String ret = "";
for (int i = Array.length-1; i >= 0; i--)
ret += Integer.toString(Array[i]);
return ret;
}
public static void main (String[] args) {
Scanner sc=new Scanner (System.in);
System.out.print ("Enter any no.: ");
int num = sc.nextInt();
int[] bin = new int[howManyTerms (num)];
int dup = num, el = -1;
while (dup != 0) {
int rem = dup % 2;
bin [++el] = rem;
dup /= 2;
}
String d2b = revArrayofBin2Str(bin);
System.out.println("Binary of " + num + " is: " + d2b);
}
}
This is simple java code for decimal to binary using only primitive type int, hopefully it should help beginners.
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class DtoB {
public static void main(String[] args) {
try { // for Exception handling of taking input from user.
System.out.println("Please enter a number");
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String input = br.readLine();
int x = Integer.parseInt(input);
int bin = 0;
int p = 1;
while (x > 0) {
int r = x % 2;
bin = (r * p) + bin;
x = x / 2;
p *= 10;
}
System.out.println("Binary of " + input + " is = " + bin);
} catch (Exception e) {
System.out.println("Please enter a valid decimal number.");
System.exit(1);
e.printStackTrace();
}
}
}