Adding binary numbers

Question

Does anyone know how to add 2 binary numbers, entered as binary, in Java?

For example, 1010 + 10 = 1100.

This may blow your mind. You can write C code in java. C is lower-level and you can do binary arithmetic with it. It's late, so I'm not writing out a sample, but you can look it up. — user817129, Oct 13 '13 at 10:01
This is more likely to be a leetcode question I guess. I came accross this question here: https://leetcode.com/problems/add-binary/ — Prometheus, Oct 21 '22 at 08:48

score 52 · Accepted Answer · edited Jun 28 '18 at 05:20

52

Use Integer.parseInt(String, int radix).

 public static String addBinary(){
 // The two input Strings, containing the binary representation of the two values:
    String input0 = "1010";
    String input1 = "10";

    // Use as radix 2 because it's binary    
    int number0 = Integer.parseInt(input0, 2);
    int number1 = Integer.parseInt(input1, 2);

    int sum = number0 + number1;
    return Integer.toBinaryString(sum); //returns the answer as a binary value;
}

edited Jun 28 '18 at 05:20

inspired_learner

142
1
11

answered Dec 17 '11 at 23:08

Martijn Courteaux

67,591
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198
287

wow. it worked! thanks bro. but i was wondering if you could explain why? – PulsePanda Dec 17 '11 at 23:13
`Integer` is a class, which contains a method to parse a string which represents an integer to its real integer value (`int`). You can take a look at the link I provided you. – Martijn Courteaux Dec 17 '11 at 23:14
Martin - it does not show the answer in binary. How do we do that? – Abhishek kumar Feb 16 '16 at 04:39
@Abhishekkumar just use `Integer.toBinaryString(sum);` – austin665 Mar 19 '16 at 18:31
11

This can throw java.lang.NumberFormatException: For input string: "10100000100100110110010000010101111011011001101110111111111101000000101111001110001111100001101" "110101001011101110001111100110001010100001101011101010000011011011001011101111001100000011011110011" – Tasneem Dec 08 '16 at 07:29
3

@Tasneem which makes sense since the max int value is `2,147,483,647` and the (first) binary value translates to `24,847,893,154,024,981,730,169,397,005` – Ivar Reukers May 17 '17 at 19:43

score 30 · Answer 2 · answered Mar 07 '14 at 18:16

To dive into fundamentals:

public static String binaryAddition(String s1, String s2) {
    if (s1 == null || s2 == null) return "";
    int first = s1.length() - 1;
    int second = s2.length() - 1;
    StringBuilder sb = new StringBuilder();
    int carry = 0;
    while (first >= 0 || second >= 0) {
        int sum = carry;
        if (first >= 0) {
            sum += s1.charAt(first) - '0';
            first--;
        }
        if (second >= 0) {
            sum += s2.charAt(second) - '0';
            second--;
        }
        carry = sum >> 1;
        sum = sum & 1;
        sb.append(sum == 0 ? '0' : '1');
    }
    if (carry > 0)
        sb.append('1');

    sb.reverse();
    return String.valueOf(sb);
}

@Serhanbaker Could you please explain, sum += s1.charAt(first) - '0'; bit. Thanks — kaila88, Feb 25 '17 at 17:36
this should be accepted answer, since the first answer doesn't work with large values — rahimli, Jul 07 '18 at 08:30

score 13 · Answer 3 · answered Dec 17 '11 at 23:51

13

Martijn is absolutely correct, to piggyback and complete the answer

Integer.toBinaryString(sum);

would give your output in binary as per the OP question.

answered Dec 17 '11 at 23:51

non sequitor

18,296
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haley · Answer 4 · 2015-06-08T15:14:17.690

9

You can just put 0b in front of the binary number to specify that it is binary.

For this example, you can simply do:

Integer.toString(0b1010 + 0b10, 2);

This will add the two in binary, and Integer.toString() with 2 as the second parameter converts it back to binary.

edited Jun 08 '15 at 15:14

answered May 20 '14 at 17:11

haley

1,165
7
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Binary literals are allowed in java 7 too, see https://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1 – fabian Jun 07 '15 at 14:19
This should be the correct answer; no reason to convert to strings or roll your own solution. – Jason Jul 14 '16 at 18:16

score 5 · Answer 5 · answered Jul 25 '17 at 05:48

The original solution by Martijn will not work for large binary numbers. The below code can be used to overcome that.

public String addBinary(String s1, String s2) {
    StringBuilder sb = new StringBuilder();
    int i = s1.length() - 1, j = s2.length() -1, carry = 0;
    while (i >= 0 || j >= 0) {
        int sum = carry;
        if (j >= 0) sum += s2.charAt(j--) - '0';
        if (i >= 0) sum += s1.charAt(i--) - '0';
        sb.append(sum % 2);
        carry = sum / 2;
    }
    if (carry != 0) sb.append(carry);
    return sb.reverse().toString();
}

it works from me large binary numbers. – Santosh Kumar Jan 25 '22 at 03:33 — Santosh Kumar, Jan 25 '22 at 03:33

score 2 · Answer 6 · edited May 29 '13 at 21:32

public class BinaryArithmetic {

    /*-------------------------- add ------------------------------------------------------------*/
    static String add(double a, double b) {
        System.out.println(a + "first val :" + b);
        int a1 = (int) a;
        int b1 = (int) b;
        String s1 = Integer.toString(a1);
        String s2 = Integer.toString(b1);
        int number0 = Integer.parseInt(s1, 2);
        int number1 = Integer.parseInt(s2, 2);

        int sum = number0 + number1;
        String s3 = Integer.toBinaryString(sum);

        return s3;
    }
    /*-------------------------------multiply-------------------------------------------------------*/

    static String multiply(double a, double b) {
        System.out.println(a + "first val :" + b);
        int a1 = (int) a;
        int b1 = (int) b;
        String s1 = Integer.toString(a1);
        String s2 = Integer.toString(b1);
        int number0 = Integer.parseInt(s1, 2);
        int number1 = Integer.parseInt(s2, 2);

        int sum = number0 * number1;
        String s3 = Integer.toBinaryString(sum);

        return s3;
    }
    /*----------------------------------------substraction----------------------------------------------*/

    static String sub(double a, double b) {
        System.out.println(a + "first val :" + b);
        int a1 = (int) a;
        int b1 = (int) b;
        String s1 = Integer.toString(a1);
        String s2 = Integer.toString(b1);
        int number0 = Integer.parseInt(s1, 2);
        int number1 = Integer.parseInt(s2, 2);

        int sum = number0 - number1;
        String s3 = Integer.toBinaryString(sum);

        return s3;
    }

    /*--------------------------------------division------------------------------------------------*/
    static String div(double a, double b) {
        System.out.println(a + "first val :" + b);
        int a1 = (int) a;
        int b1 = (int) b;
        String s1 = Integer.toString(a1);
        String s2 = Integer.toString(b1);
        int number0 = Integer.parseInt(s1, 2);
        int number1 = Integer.parseInt(s2, 2);

        int sum = number0 / number1;
        String s3 = Integer.toBinaryString(sum);

        return s3;
    }
}

score 2 · Answer 7 · answered Feb 18 '14 at 14:29

2

Another interesting but long approach is to convert each of the two numbers to decimal, adding the decimal numbers and converting the answer obtained back to binary!

answered Feb 18 '14 at 14:29

programmer_360

21
7

score 2 · Answer 8 · answered Feb 19 '16 at 03:07

Java solution

static String addBinary(String a, String b) {

    int lenA = a.length();
    int lenB = b.length();
    int i = 0;

    StringBuilder sb = new StringBuilder();
    int rem = Math.abs(lenA-lenB);
    while(rem >0){
        sb.append('0');
        rem--;
    }
    if(lenA > lenB){
        sb.append(b);  
        b = sb.toString();
    }else{
        sb.append(a);
        a = sb.toString();
    }

    sb = new StringBuilder();
    char carry = '0';
    i = a.length();
    while(i > 0){
        if(a.charAt(i-1) == b.charAt(i-1)){
            sb.append(carry);
            if(a.charAt(i-1) == '1'){
                carry = '1';
            }else{
                carry = '0';
            }
        }else{
            if(carry == '1'){
                sb.append('0');
                carry = '1';
            }else{
                carry = '0';
                sb.append('1');
            }
        }
        i--;
    }

    if(carry == '1'){
        sb.append(carry);
    }

    sb.reverse();
    return sb.toString();

}

score 2 · Answer 9 · answered Mar 27 '20 at 21:24

public String addBinary(String a, String b) { 
    int carry = 0;
    StringBuilder sb = new StringBuilder();
    for(int i = a.length() - 1, j = b.length() - 1;i >= 0 || j >= 0;i--,j--){
        int sum = carry + (i >= 0 ? a.charAt(i) - '0':0) + (j >= 0 ? b.charAt(j) - '0' : 0);
        sb.append(sum%2);
        carry =sum / 2;
    }
    if(carry > 0) sb.append(carry);
    sb.reverse();
    return sb.toString();
}

score 1 · Answer 10 · edited Oct 16 '15 at 14:11

I've actually managed to find a solution to this question without using the stringbuilder() function. Check this out:

public void BinaryAddition(String s1,String s2)
{
    int l1=s1.length();int c1=l1;
    int l2=s2.length();int c2=l2;
    int max=(int)Math.max(l1,l2);
    int arr1[]=new int[max];
    int arr2[]=new int[max];
    int sum[]=new int[max+1];
    for(int i=(arr1.length-1);i>=(max-l1);i--)
    {
        arr1[i]=(int)(s1.charAt(c1-1)-48);
        c1--;
    }
    for(int i=(arr2.length-1);i>=(max-l2);i--)
    {
        arr2[i]=(int)(s2.charAt(c2-1)-48);
        c2--;
    }
    for(int i=(sum.length-1);i>=1;i--)
    {
        sum[i]+=arr1[i-1]+arr2[i-1];
        if(sum[i]==2)
        {
            sum[i]=0;
            sum[i-1]=1;
        }
        else if(sum[i]==3)
        {
            sum[i]=1;
            sum[i-1]=1;
        }
    }
    int c=0;
    for(int i=0;i<sum.length;i++)
    {
        System.out.print(sum[i]);
    }
}

score 1 · Answer 11 · answered Aug 17 '19 at 06:43

The idea is same as discussed in few of the answers, but this one is a much shorter and easier to understand solution (steps are commented).

// Handles numbers which are way bigger.
public String addBinary(String a, String b) {
    StringBuilder sb = new StringBuilder();
    int i = a.length() - 1; 
    int j = b.length() -1;
    int carry = 0;
    while (i >= 0 || j >= 0) {
        int sum = carry;
        if (j >= 0) { sum += b.charAt(j--) - '0' };
        if (i >= 0) { sum += a.charAt(i--) - '0' };

        // Added number can be only 0 or 1
        sb.append(sum % 2);

        // Get the carry.
        carry = sum / 2;
    }

    if (carry != 0) { sb.append(carry); }

    // First reverse and then return.
    return sb.reverse().toString();
}

score 0 · Answer 12 · answered May 20 '14 at 15:54

i tried to make it simple this was sth i had to deal with with my cryptography prj its not efficient but i hope it

    public String binarysum(String a, String b){
    int carry=0;
    int maxim;
    int minim;
    maxim=Math.max(a.length(),b.length());
    minim=Math.min(a.length(),b.length());         
    char smin[]=new char[minim];
    char smax[]=new char[maxim];
    if(a.length()==minim){
     for(int i=0;i<smin.length;i++){
     smin[i]=a.charAt(i);
      }
      for(int i=0;i<smax.length;i++){
       smax[i]=b.charAt(i);
      }
      }
      else{
          for(int i=0;i<smin.length;i++){
          smin[i]=b.charAt(i);
             }
       for(int i=0;i<smax.length;i++){
       smax[i]=a.charAt(i);
      } 
      }
    char[]sum=new char[maxim];
    char[] st=new char[maxim];
    for(int i=0;i<st.length;i++){
    st[i]='0';
    }
    int k=st.length-1;
   for(int i=smin.length-1;i>-1;i--){
    st[k]=smin[i];
    k--;
    } 

     //   *************************** sum begins here
   for(int i=maxim-1;i>-1;i--){
   char x= smax[i];
   char y= st[i];
    if(x==y && x=='0'){
         if(carry==0)
             sum[i]='0';
         else if(carry==1){
             sum[i]='1';
             carry=0;
    }
   }
    else if(x==y && x=='1'){
        if(carry==0){
            sum[i]='0';
            carry=1;
        }
        else if(carry==1){
          sum[i]='1';
          carry=1;
        }
     }
     else if(x!=y){
        if(carry==0){
            sum[i]='1';
            }
        else if(carry==1){
          sum[i]='0';
          carry=1;
        }
       }        }
      String s=new String(sum);
     return s;
      }

score 0 · Answer 13 · answered Jun 12 '16 at 07:17

class Sum{
    public int number;
    public int carry;

    Sum(int number, int carry){
        this.number = number; 
        this.carry = carry;
    }
}

public String addBinary(String a, String b) {

    int lengthOfA = a.length();
    int lengthOfB = b.length();

    if(lengthOfA > lengthOfB){
        for(int i=0; i<(lengthOfA - lengthOfB); i++){
            b="0"+b;
        }
    }
    else{
         for(int i=0; i<(lengthOfB - lengthOfA); i++){
            a="0"+a;
        }
    }

    String result = "";
    Sum s = new Sum(0,0);
    for(int i=a.length()-1; i>=0; i--){
        s = addNumber(Character.getNumericValue(a.charAt(i)), Character.getNumericValue(b.charAt(i)), s.carry);
        result = result + Integer.toString(s.number);
    }

    if(s.carry == 1) { result += s.carry ;}

    return new StringBuilder(result).reverse().toString();
}

Sum addNumber(int number1, int number2, int carry){
        Sum sum = new Sum(0,0);
        sum.number = number1 ^ number2 ^ carry;
        sum.carry = (number1 & number2) | (number2 & carry) | (number1 & carry);

        return sum;
}

score 0 · Answer 14 · edited Jul 14 '16 at 19:23

import java.util.*;
public class BitAddition {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int len = sc.nextInt();
        int[] arr1 = new int[len];
        int[] arr2 = new int[len];
        int[] sum = new int[len+1];
        Arrays.fill(sum, 0);
        for(int i=0;i<len;i++){
            arr1[i] =sc.nextInt();
        }
        for(int i=0;i<len;i++){
            arr2[i] =sc.nextInt();
        }
        for(int i=len-1;i>=0;i--){
            if(sum[i+1] == 0){
                if(arr1[i]!=arr2[i]){
                    sum[i+1] = 1;
                }
                else if(arr1[i] ==1 && arr2[i] == 1){
                    sum[i+1] =0 ;
                    sum[i] = 1;
                }
            }
            else{
                if((arr1[i]!=arr2[i])){
                    sum[i+1] = 0;
                    sum[i] = 1;
                }
                else if(arr1[i] == 1){
                    sum[i+1] = 1;
                    sum[i] = 1;
                }
            }
        }
        for(int i=0;i<=len;i++){

        System.out.print(sum[i]);
        }
    }

}

score 0 · Answer 15 · answered May 17 '17 at 19:37

One of the simple ways is as:

convert the two strings to char[] array and set carry=0.
set the smallest array length in for loop
start for loop from the last index and decrement it
check 4 conditions(0+0=0, 0+1=1, 1+0=1, 1+1=10(carry=1)) for binary addition for each element in both the arrays and reset the carry accordingly.
append the addition in stringbuffer
append rest of the elements from max size array to stringbuffer but check consider carry while appending
print stringbuffer in reverse order for the answer.

//The java code is as

static String binaryAdd(String a, String b){
    int len = 0;
    int size = 0;
    char[] c1 = a.toCharArray();
    char[] c2 = b.toCharArray();
    char[] max;


    if(c1.length > c2.length){
        len = c2.length;
        size = c1.length;
        max = c1;
    }       
    else
    {
        len = c1.length;
        size = c2.length;
        max = c2;
    }

    StringBuilder sb = new StringBuilder();
    int carry = 0;
    int p = c1.length - 1;
    int q = c2.length - 1;

    for(int i=len-1; i>=0; i--){
        if(c1[p] == '0' && c2[q] == '0'){
            if(carry == 0){
                sb.append(0);
                carry = 0;
            }   
            else{
                sb.append(1);
                carry = 0;
            }   
        }
        if((c1[p] == '0' && c2[q] == '1') || (c1[p] == '1' && c2[q] == '0')){
            if(carry == 0){
                sb.append(1);
                carry = 0;
            }   
            else{
                sb.append(0);
                carry = 1;
            }                   
        }
        if((c1[p] == '1' && c2[q] == '1')){
            if(carry == 0){
                sb.append(0);
                carry = 1;
            }   
            else{
                sb.append(1);
                carry = 1;
            }
        }
        p--;
        q--;
    }

    for(int j = size-len-1; j>=0; j--){
        if(max[j] == '0'){ 
            if(carry == 0){     
                sb.append(0);
                carry = 0;
            }   
            else{
                sb.append(1);
                carry = 0;
            }   
        }
        if(max[j] == '1'){
            if(carry == 0){     
                sb.append(1);
                carry = 0;
            }   
            else{
                sb.append(0);
                carry = 1;
            }   
        }           
    }
    if(carry == 1)
        sb.append(1);   
    return sb.reverse().toString();
}

score 0 · Answer 16 · edited Sep 16 '17 at 05:40

0

import java.io.; 
import java.util.; 
public class adtbin {
  static Scanner sc=new Scanner(System.in); 
   public void fun(int n1) {
      int i=0; 
      int sum[]=new int[20]; 
      while(n1>0) { 
        sum[i]=n1%2; n1=n1/2; i++; 
      } 
      for(int a=i-1;a>=0;a--) { 
          System.out.print(sum[a]); 
      }  
   } 
   public static void main() { 
     int m,n,add; 
     adtbin ob=new adtbin(); 
     System.out.println("enter the value of m and n"); 
     m=sc.nextInt(); 
     n=sc.nextInt(); 
     add=m+n; 
     ob.fun(add); 
   } 
}

edited Sep 16 '17 at 05:40

Kiran

20,167
11
67
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answered Sep 16 '17 at 05:02

james

1

a bit of explanation is welcome with code. code only answers might be flagged as "low quality" and removed, even if they are technically valid. – Gilles Gouaillardet Sep 16 '17 at 05:19

score 0 · Answer 17 · answered Nov 03 '17 at 07:50

you can write your own One.

long a =100011111111L;
long b =1000001111L;

int carry = 0 ;
long result = 0;

long multiplicity = 1;

while(a!=0 || b!=0 || carry ==1){
    if(a%10==1){
        if(b%10==1){
            result+= (carry*multiplicity);
            carry = 1;
        }else if(carry == 1){
            carry = 1;  
        }else{
            result += multiplicity;
        }
    }else if (b%10 == 1){
        if(carry == 1){
            carry = 1;
        }else {
            result += multiplicity; 
        }
    }else {
        result += (carry*multiplicity);
        carry = 0;
    }

    a/=10;
    b/=10;
    multiplicity *= 10;

}


System.out.print(result);

it works just by numbers , no need string , no need SubString and ...

score 0 · Answer 18 · edited Jan 05 '18 at 15:05

package Assignment19thDec;

import java.util.Scanner;

public class addTwoBinaryNumbers {

    private static Scanner sc;

    public static void main(String[] args) {

        sc = new Scanner(System.in);
        System.out.println("Enter 1st Binary Number");
        int number1=sc.nextInt();
        int reminder1=0;
        int number2=sc.nextInt();
        int reminder2=0;
        int carry=0;
        double sumResult=0 ;int add = 0
        ;
        int n;
        int power=0;
        while (number1>0 || number2>0) {

            /*System.out.println(number1 + " " +number2);*/

            reminder1=number1%10;
            number1=number1/10;
            reminder2=number2%10;
            number2=number2/10;
            /*System.out.println(reminder1 +"  "+ reminder2);*/


            if(reminder1>1 || reminder2>1 ) {
                System.out.println("not a binary number");
                System.exit(0);
            }

            n=reminder1+reminder2+carry;
            switch(n) {

            case 0: 
                    add=0; carry=0;                                 
                    break;                  
            case 1: add=1; carry=0;
                    break;
            case 2: add=0; carry=1;
                    break;
            case 3: add=1;carry=1;
                    break;
            default: System.out.println("not a binary number ");

            }

            sumResult=add*(Math.pow(10, power))+sumResult;
            power++;


        }

        sumResult=carry*(Math.pow(10, power))+sumResult;
        System.out.println("\n"+(int)sumResult);


    }

}

I think a little explanation will help too – Valerica Jan 05 '18 at 14:52 — Valerica, Jan 05 '18 at 14:52

balusu48 · Answer 19 · 2019-10-03T14:43:48.850

Try this, tested with binary and decimal and its self explanatory

public String add(String s1, String s2, int radix){
        int s1Length = s1.length();
        int s2Length = s2.length();
        int reminder = 0;
        int carry = 0;
        StringBuilder result = new StringBuilder();
        int i = s1Length -1;
        int j = s2Length -1;
        while (i >=0 && j>=0) {
            int operand1 = Integer.valueOf(s1.charAt(i)+"");
            int operand2 = Integer.valueOf(s2.charAt(j)+"");
            reminder = (operand1+operand2+carry) % radix;
            carry = (operand1+operand2+carry) / radix;
            result.append(reminder);
            i--;j--;
        }
        while(i>=0){
            int operand1 = Integer.valueOf(s1.charAt(i)+"");
            reminder = (operand1+carry) % radix;
            carry = (operand1+carry) / radix;
            result.append(reminder);
            i--;
        }
        while(j>=0){
            int operand1 = Integer.valueOf(s2.charAt(j)+"");
            reminder = (operand1+carry) % radix;
            carry = (operand1+carry) / radix;
            result.append(reminder);
            j--;
        }
        return result.reverse().toString();
    }
}

score 0 · Answer 20 · answered Dec 09 '22 at 01:57

static int addBinaryNumbers(String a, String b) {
    int firstToDecimal = 0;
    int secondToDecimal = 0;
    for (int i = a.length() - 1, count = 0; i >= 0; i--, count++) {
        firstToDecimal += (Math.pow(2, count) * Integer.parseInt(String.valueOf(a.toCharArray()[i])));

    }
    for (int i = b.length() - 1, count = 0; i >= 0; i--, count++) {
        secondToDecimal += (Math.pow(2, count) * Integer.parseInt(String.valueOf(b.toCharArray()[i])));

    }
    return firstToDecimal + secondToDecimal;
}

public static void main(String[] args) {
    System.out.println(addBinaryNumbers("101", "110"));

}

score -1 · Answer 21 · answered Nov 16 '14 at 04:35

here's a python version that

def binAdd(s1, s2):
    if not s1 or not s2:
        return ''

    maxlen = max(len(s1), len(s2))


    s1 = s1.zfill(maxlen)
    s2 = s2.zfill(maxlen)

    result  = ''
    carry   = 0

    i = maxlen - 1
    while(i >= 0):
        s = int(s1[i]) + int(s2[i])
        if s == 2: #1+1
            if carry == 0:
                carry = 1
                result = "%s%s" % (result, '0')
            else:
                result = "%s%s" % (result, '1')
        elif s == 1: # 1+0
            if carry == 1:
                result = "%s%s" % (result, '0')
            else:
                result = "%s%s" % (result, '1')
        else: # 0+0
            if carry == 1:
                result = "%s%s" % (result, '1')
                carry = 0   
            else:
                result = "%s%s" % (result, '0') 

        i = i - 1;

    if carry>0:
        result = "%s%s" % (result, '1')
    return result[::-1]

Question specifically says `Java`. – Jared Burrows Apr 19 '15 at 23:19 — Jared Burrows, Apr 19 '15 at 23:19

Vivek Kamble · Answer 22 · 2019-06-07T19:15:03.873

-2

import java.util.Scanner;

{
    public static void main(String[] args) 
    {
        String b1,b2;
        Scanner sc= new Scanner(System.in);
        System.out.println("Enter 1st binary no. : ") ;
        b1=sc.next();
        System.out.println("Enter 2nd binary no. : ") ;
        b2=sc.next();
        int num1=Integer.parseInt(b1,2);
        int num2=Integer.parseInt(b2,2);
        int sum=num1+num2;
        System.out.println("Additon is : "+Integer.toBinaryString(sum));
    }

}

edited Jun 07 '19 at 19:15

answered Jun 07 '19 at 19:06

Vivek Kamble

1
1

please add details to your answer – Muhammad Jun 07 '19 at 19:29

Adding binary numbers

22 Answers22

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