Adding binary numbers

Question

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

For example, 1010 + 10 = 1100.

• squiguy: @twiddles No more homework tagging! :)
• user817129: 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.
• Prometheus: This is more likely to be a leetcode question I guess. I came accross this question here: leetcode.com/problems/add-binary

Answer 1

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;
}

• PulsePanda: wow. it worked! thanks bro. but i was wondering if you could explain why?
• Martijn Courteaux: 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.
• Abhishek kumar: Martin - it does not show the answer in binary. How do we do that?
• Tasneem: This can throw java.lang.NumberFormatException: For input string: "10100000100100110110010000010101111011011001101110111111111101000000101111001110001111100001101" "110101001011101110001111100110001010100001101011101010000011011011001011101111001100000011011110011"
• Ivafo: @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

Answer 2

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);
}

• kaila88: @Serhanbaker Could you please explain, sum += s1.charAt(first) - '0'; bit. Thanks
• rahimli: this should be accepted answer, since the first answer doesn't work with large values
• Diego Borba: @kaila88 check this out: Java: Subtract '0' from char to get an int... why does this work?.

Answer 3

Martijn is absolutely correct, to piggyback and complete the answer

Integer.toBinaryString(sum);

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

Answer 4

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.

• fabian: Binary literals are allowed in java 7 too, see docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.1
• Jason: This should be the correct answer; no reason to convert to strings or roll your own solution.

Answer 5

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();
}

• Santosh Kumar: it works from me large binary numbers.

Answer 6

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;
 }
}

Answer 7

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!

Answer 8

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();
}

Answer 9

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();
}

• NIKHIL CHAURASIA: Explanation please?

Answer 10

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]);
 }
}

Answer 11

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();
}

Answer 12

• For future leetcoders looking to see if any builtin methods exist to overcome the hassle of converting it, you're at the right place.
• And we'll have to import & instantiate before using it, since I think leetcode doesn't allow the usage of static classes directly.

 // Use BigInteger to convert the string to binary.
 // BigInteger is used when long with range [-2^63, 2^63 - 1] 
 // couldn't handle the given number
 BigInteger(String s, int radix);
 // Where s is the input string & radix in this case would be 2 for binary

BigInteger Java docs - link here

Leetcode Challenge: https://leetcode.com/problems/add-binary/

Answer 13

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;
 }

Answer 14

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;
}

Answer 15

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]);
 }
 }
}

Answer 16

One of the simple ways is as:

1. convert the two strings to char[] array and set carry=0.
2. set the smallest array length in for loop
3. start for loop from the last index and decrement it
4. 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.
5. append the addition in stringbuffer
6. append rest of the elements from max size array to stringbuffer but check consider carry while appending
7. 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();
}

Answer 17

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); 
 } 
}

• Gilles Gouaillardet: 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.

Answer 18

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 ...

Answer 19

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);
 }
}

• Valerica: I think a little explanation will help too

Answer 20

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();
 }
}

Answer 21

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"));
}

Answer 22

import java.util.Scanner;
public class Test 
{
 public static void main(String args[]) 
 {
 Scanner sc=new Scanner(System.in);
 int b1,b2;
 b1=sc.nextInt();
 b2=sc.nextInt();
 sc.close();
 System.out.println(add(b1,b2));
 }
 static int add(int b1, int b2)
 {
 
 int l1,l2,c=0;
 StringBuilder str=new StringBuilder();
 while(b1>0 || b2>0)
 {
 l1=b1%2;
 l2=b2%2;
 if(l1+l2+c==3)
 {
 str.append(1);
 c=1;
 }
 else if(l1+l2+c==2)
 {
 str.append(0);
 c=1;
 
 }
 else if(l1+l2+c==1)
 {
 str.append(1);
 c=0;
 }
 else if(l1+l2+c==0)
 {
 str.append(0);
 c=0;
 }
 b1/=10;
 b2/=10;
 }
 str.append(c);
 str= new StringBuilder(str.toString()).reverse();
 return Integer.parseInt(str.toString());
 }
}

Legend:

c-> carry

b1,b2 -> binary numbers

l1,l2 -> last digits

Constraints : It can't add binary numbers which are greater than 1023, cause of the max size of int is 2^31.

Answer 23

To handle large numbers use BigInteger -

public String addBinary(String a, String b) {
 java.math.BigInteger fNum = new java.math.BigInteger(a, 2);
 java.math.BigInteger sNum = new java.math.BigInteger(b, 2);
 return fNum.add(sNum).toString(2);
 }

Answer 24

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]

• Jared Burrows: Question specifically says Java.

Answer 25

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));
 }
}

• Muhammad: please add details to your answer