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LeetCode Solutions

67. Add Binary

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

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

			


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;
  }
};




			


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();
  }
}




			


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))