# Leetcode Add Binary problem solution

In this Leetcode Add Binary problem solution we have Given two binary strings a and b, return their sum as a binary string.

## Problem solution in Python.

```class Solution:
def addBinary(self, a: str, b: str) -> str:
carry = 0
result = ''

a = list(a)
b = list(b)

while a or b or carry:
if a:
carry += int(a.pop())
if b:
carry += int(b.pop())

result += str(carry %2)
carry //= 2

return result[::-1]
```

## Problem solution in Java.

```class Solution {
public String addBinary(String a, String b) {

int i = a.length() - 1;
int j = b.length() - 1;

StringBuilder sb = new StringBuilder();
int sum = 0;

while(i >= 0 || j >= 0){

sum /= 2;
if(i >= 0) sum += a.charAt(i) - '0';
if(j >= 0) sum += b.charAt(j) - '0';

sb.append(sum % 2);
i--;
j--;

}

if(sum / 2 != 0) sb.append(1);
return sb.reverse().toString();
}
}
```

## Problem solution in C++.

```class Solution {
public:
string addBinary(string a, string b) {
string result;
int carry = 0;
for(int idxA = a.size()-1, idxB = b.size()-1; idxA>=0 || idxB>=0; idxA--, idxB--) {
int aNum = idxA >=0 ? a[idxA]-'0' : 0;
int bNum = idxB >=0 ? b[idxB]-'0' : 0;
int curPosNum = aNum + bNum + carry;
int tempRe = curPosNum%2;
carry = curPosNum/2;
result.insert(result.begin(), tempRe+'0');
}
if(carry > 0) result.insert(result.begin(), carry+'0');
return result;
}
};
```

## Problem solution in C.

```char* addBinary(char* a, char* b) {
int i = strlen(a) - 1, j = strlen(b) - 1;
if (i < j) {
char* t = a;
a = b, b = t;
int k = i;
i = j, j = k;
}
const int aLen = i+1;;

while (i > 0 && j >= 0) {
a[i] += b[j] - '0';
a[i - 1] += (a[i] - '0') >> 1;
a[i] = (a[i] & 1) + '0';
j--;
i--;
}
if (j == 0)
a[0] += b[0] - '0';
else
while (i > 0 && a[i] > '1') {
a[i - 1] += (a[i] - '0') >> 1;
a[i] = (a[i] & 1) + '0';
i--;
}

if (a[0] > '1') {
char *c = malloc(aLen+2);
memcpy(c+1, a, aLen);
c[0] = '0' + ((c[1] - '0') >> 1);
c[1] = (c[1]&1) + '0';
c[aLen+1] = 0;
a = c;
}
return a;
}
```