Easy: Sum of Two Integers — solve it online. Python & JavaScript run locally; Java, Go & Rust need sign-in.

Sum of Two Integers – Solve Online

Longest Consecutive Sequence

Add two integers together without using the + or - operators.

Given two integers `a` and `b`, return their sum without using the `+` or `-` operators. Use bitwise operations (`AND`, `OR`, `XOR`, shifts) to simulate addition with carrying instead.

**Input / output**
- Input: `a: integer`, `b: integer`
- Output: `integer`, equal to `a + b`

**Examples**
1. `a = 1, b = 2` returns `3`.
2. `a = 2, b = 3` returns `5`.
3. `a = -2, b = 3` returns `1`.

**Constraints**
- `-1,000 <= a, b <= 1,000`

**Follow-up**
Can you explain why `a XOR b` gives the sum without carries, and `(a AND b) << 1` gives exactly the carries that must be added back in, repeating until there is no carry left?

- Compute `a + b` without using the `+` or `-` operators — you're only allowed bitwise operations.
- This is fundamentally about how binary addition works at the hardware level: each bit position produces a sum bit and a carry bit, and the carry ripples into the next position.
- `a XOR b` gives the sum of each bit ignoring carries; `a AND b` (shifted left by 1) gives exactly the carries that need to be added in.

Understanding the Problem

- `a ^ b` computes each bit's sum as if there were no carry (since `1^1=0`, `0^0=0`, `1^0=1` — matches binary addition without carry-out).
- `(a & b) << 1` computes the carry that must be added into the next higher bit position (a carry is produced exactly where both bits are `1`, and it affects the next bit up).
- Repeat: keep adding the running sum and the new carry the same way, until there's no carry left. This is the classic "add via repeated XOR/AND" bit trick, equivalent to how a full-adder circuit ripples carries.

XOR for Sum-Without-Carry, AND for Carry

- Loop: `carry = (a & b) << 1`, `a = a ^ b`, `b = carry`, until `b == 0`. At that point `a` holds the final sum.
- In languages without arbitrary-precision integers (Java/Go/Rust), this must be done with fixed-width (32-bit) arithmetic so negative numbers and overflow wrap around exactly like two's-complement hardware addition — Python needs an explicit mask + sign-fix step since its integers are unbounded.

Iterating to Convergence

`O(1)` time in practice — bounded by the fixed bit width (32 iterations max) — and `O(1)` space.

Final Solution (all languages)

- In Python, forgetting the 32-bit mask/sign-fix step: since Python ints are unbounded, the naive `while b: a, b = a ^ b, (a & b) << 1` loop never terminates correctly for negative inputs without explicit masking.
- In JS, forgetting that `<<` and `^` already operate on 32-bit signed integers, so no extra masking is needed there (unlike Python) — but using them on values outside the 32-bit range gives wrong results.
- Follow-ups: how would you implement subtraction the same way (hint: `a - b == a + (~b + 1)`, i.e. add the two's-complement negation)? How would this generalize to 64-bit integers?