Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Mediumbit-manipulation

Sum of Two Integers

Problem (restated)

Return a+b without using + or, operators.

Intuition

a^b is sum without carry; (a&b)<<1 is carry. Iterate until carry is 0.

Approaches

XOR + carry loop

Unverified

Time O(1) 32-bitSpace O(1)

Idea. Full-adder simulation over bits. Python needs 32-bit mask for negatives.

Walkthrough. 1+2 → 3; -1+1 → 0.

Trade-offs. Built-in + is fine in production; problem is a bit-op exercise.

Solution

export function getSum(a: number, b: number): number {
  while (b !== 0) {
    const carry = (a & b) << 1;
    a = a ^ b;
    b = carry;
  }
  // JS bitwise is 32-bit signed already for |0
  return a | 0;
}

export function getSum(a: number, b: number): number {
  while (b !== 0) {
    const carry = (a & b) << 1;
    a = a ^ b;
    b = carry;
  }
  // JS bitwise is 32-bit signed already for |0
  return a | 0;
}

Template connection

Bitwise arithmetic simulation.

Reflection

XOR is the sum with no carry. a & b, shifted left by one, is the carry. Stop when the carry is 0.
Negatives are two’s complement. In Python, mask the carry to 32 bits or the loop does not settle.
a + 0 is a. (-1) + 1 is 0. At the int boundary the mask is what keeps the carry inside the word.