Add Binary

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

For example,
a = “11”
b = “1”
Return “100”.

/**
 * @param {string} a
 * @param {string} b
 * @return {string}
 */

var addBinary = function(a, b) {
  a = a.split('').reverse();
  b = b.split('').reverse();

  var c = [];
  var add = 0;

  for(var i = 0, len = Math.max(a.length, b.length);i < len; i++) {
    var sum = (a[i] === undefined ? 0 : Number(a[i])) + (b[i] === undefined ? 0: Number(b[i])) + add;
    c[i] = sum & 1;
    if (sum >= 2) 
      add = 1;
    else 
      add = 0;
  }

  if (add)
    c[len] = 1;

  return c.reverse().join('').toString();
};

Add Digit

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
  while(num.toString().length !== 1) {
    var sum = 0;
    while (num) {
      sum += num % 10;
      num = ~~(num / 10);
    }
    num = sum;
  }

  return num;
};

Add Strings

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

1. The length of both num1 and num2 is < 5100.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var addStrings = function(num1, num2) {
  let [i, j] = [num1.length, num2.length];
  let ans = '';
  let add = 0;

  i -= 1, j -= 1;
  for ( ; i >= 0 || j >= 0; i--, j--) {
    let a = i >= 0 ? +num1[i] : 0;
    let b = j >= 0 ? +num2[j] : 0;
    let sum = a + b + add;
    ans = sum % 10 + ans;
    add = ~~(sum / 10);
  }

  add && (ans = add + ans);
  return ans;
};