LeetCode Meditations: Sum of Two Integers

The description given for Sum of Two Integers is very simple:

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

For example:

Input: a = 1, b = 2
Output: 3

Or:

Input: a = 2, b = 3
Output: 5

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In the very last problem of this series, we will close off with adding two integers using bit manipulation instead of our beloved plus operator.

Adding two bits, either of which can only be 1 or 0, doesn't have many varying results.
If we're adding two bits which are 1 and 0 (or, 0 and 1), the result will be 1. If we're adding two 0 bits, the result is 0. If, however, we're adding two 1 bits, we have a carry — which means we have to write 0 in the output, but also carry a 1.

For example, adding 2 and 3 will result in 5, and we will have a carry value during the operation:

Without thinking about the carry value, the output we need to have after adding two bits resembles a lot like what we would have after an XOR operation. If we have different bits (0 and 1, or, 1 and 0), the output will be 1, otherwise 0 (adding 0 and 0, or, 1 and 1).

So, an XOR operation can help us with the output.

What about the carry?

We have a carry value only when both of the bits are 1 — which looks like an AND operation.

So, an AND operation can help us with the carry.

Also note that the carry value is shifted to the left, for which we also have a handy left-shift operator.

So, our output and carry can look like this:

let output = a ^ b;
let carry = (a & b) << 1;

We can keep modifying the two values we have, and keep going until we don't have any carry values left. We can modify a to be the output, and b to be the carry, and return a, which holds the final output at the end.

Overall, the final solution might look like this in TypeScript:

function getSum(a: number, b: number): number {
  // while we still have carry
  while (b !== 0) {
    let output = a ^ b;
    let carry = (a & b) << 1;
    a = output;
    b = carry;
  }

  return a;
}

Time and space complexity

Both a and b are constant values, and we also don't need an additional data structure whose size will grow proportionately to the input, so both our time and space complexities will be constant, $O(1)$O(1) .

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And, that is the final problem of LeetCode Meditations series! We will wrap it up with a conclusion in the next post — until then, happy coding.

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