Given two integers a and b, return the sum of the two integers without using the operators + and -.

This problem tests "bit shifting" and the "half adder" logic. Simply put, the sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.

We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. We stop when the carry equals 0.

x holds the running answer, and y holds the carry at each iteration, after it's been left-shifted.

Reference 1: GeeksForGeeks

Reference 2: Back to Back SWE

class Solution {
 public:
  int getSum(int x, int y) {
    while (y != 0) {
      unsigned int carry = x & y;  // AND: calculate carry
      x = x ^ y;                   // XOR: actual addition for each current position
      y = carry << 1;              // left shift carry
    }
    return x;
  }
};