### Leetcode

• Time:O(\max(|\texttt{a}|, |\texttt{b}|))
• Space:O(\max(|\texttt{a}|, |\texttt{b}|))

## C++

``````class Solution {
public:
string addBinary(string a, string b) {
string ans;
int carry = 0;
int i = a.length() - 1;
int j = b.length() - 1;

while (i >= 0 || j >= 0 || carry) {
if (i >= 0)
carry += a[i--] - '0';
if (j >= 0)
carry += b[j--] - '0';
ans += carry % 2 + '0';
carry /= 2;
}

reverse(begin(ans), end(ans));
return ans;
}
};
``````

## JAVA

``````class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int carry = 0;
int i = a.length() - 1;
int j = b.length() - 1;

while (i >= 0 || j >= 0 || carry == 1) {
if (i >= 0)
carry += a.charAt(i--) - '0';
if (j >= 0)
carry += b.charAt(j--) - '0';
sb.append(carry % 2);
carry /= 2;
}

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

## Python

``````class Solution:
def addBinary(self, a: str, b: str) -> str:
s = []
carry = 0
i = len(a) - 1
j = len(b) - 1

while i >= 0 or j >= 0 or carry:
if i >= 0:
carry += int(a[i])
i -= 1
if j >= 0:
carry += int(b[j])
j -= 1
s.append(str(carry % 2))
carry //= 2

return ''.join(reversed(s))
``````