# 67 Add Binary – Easy

### Problem:

Given two binary strings, return their sum (also a binary string).

For example, a = "11" b = "1" Return "100".

### Thoughts:

Very straight forward solution.

### Solutions:

Improve logic. This is a version of solution that is more into clean code. If looking for a more efficient way, need to avoid unnecessary calculation, e.g. when you have a lot of 0 in the end for both number, you don’t want to keep calculate 0 + 0 + 0.

``````public class Solution {
public String addBinary(String a, String b) {
String result = "";
int i = a.length() - 1, j = b.length() - 1;
int carry = 0;
while (i >=0 || j >=0) {
int tmp = (i >=0?(a.charAt(i)- '0'):0 ) + (j >=0?(b.charAt(j) - '0'):0) + carry;
carry = tmp / 2;
int digit = tmp % 2;
result = digit + result;
i --;
j --;
}
if (carry > 0) {
result = carry + result;
}
return result;
}
}
``````

An example to avoid unnecessary calculation for special case.

``````public class Solution {
public String addBinary(String a, String b) {
String result = "";
int i = a.length() - 1, j = b.length() - 1;
int carry = 0;
while (i >=0 && j >=0 && a.charAt(i) == '0' && b.charAt(j) == '0') {
result = '0' + result;
i --;
j --;
}
while (i >=0 || j >=0) {
int tmp = (i >=0?(a.charAt(i)- '0'):0 ) + (j >=0?(b.charAt(j) - '0'):0) + carry;
carry = tmp / 2;
int digit = tmp % 2;
result = digit + result;
i --;
j --;
}
if (carry > 0) {
result = carry + result;
}
return result;
}
}
``````