Add Binary LeetCode Solution

Problem – Add Binary

Given two binary strings a and b, return their sum as a binary string.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Constraints:

  • 1 <= a.length, b.length <= 104
  • a and b consist only of '0' or '1' characters.
  • Each string does not contain leading zeros except for the zero itself.

Add Binary LeetCode Solution in Java

public class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        int i = a.length() - 1, j = b.length() -1, 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';
            sb.append(sum % 2);
            carry = sum / 2;
        }
        if (carry != 0) sb.append(carry);
        return sb.reverse().toString();
    }
}

Add Binary LeetCode Solution in C++

class Solution
{
public:
    string addBinary(string a, string b)
    {
        string s = "";
        
        int c = 0, i = a.size() - 1, j = b.size() - 1;
        while(i >= 0 || j >= 0 || c == 1)
        {
            c += i >= 0 ? a[i --] - '0' : 0;
            c += j >= 0 ? b[j --] - '0' : 0;
            s = char(c % 2 + '0') + s;
            c /= 2;
        }
        
        return s;
    }
};

Add Binary LeetCode 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]
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