Problem Statement

Given two integers a and b, return the sum of the two integers without using the operators + and -. https://leetcode.com/problems/sum-of-two-integers/

Understanding the Problem

At first glance, the problem seems counterintuitive. How can we possibly add two numbers without using the basic arithmetic operators? The answer lies in the realm of bitwise operations. By leveraging the properties of binary arithmetic, we can simulate the addition of two numbers.

Bitwise Operations: The Key to the Solution

Three primary bitwise operations form the backbone of our solution:

  1. Bitwise XOR (^): This operation helps us perform addition without considering any carry.
  2. Bitwise AND (&): This operation helps us determine where the carry occurs.
  3. Left Shift (<<): This operation helps us position the carry correctly for the next round of addition.

Algorithmic Approach

  1. Calculate the Sum without Carry: Use the XOR operation to add the two numbers without considering the carry.
  2. Determine the Carry: Use the AND operation followed by a left shift to determine and position the carry.
  3. Recursive Addition: If there is a carry, add the result from the XOR operation and the carry using the same approach. Continue this process until there is no carry left.

Time and Space Complexity

The time complexity of the algorithm depends on the number of bits needed to represent the numbers. In the worst-case scenario, the recursive approach might need to iterate for each bit in the numbers. Thus, the time complexity can be approximated as (O(m)), where (m) is the number of bits.

The space complexity is (O(m)) because of the recursive call stack, where each call can be related to each bit operation.

Example

Let’s walk through the example of adding 5 and 3 using their binary representations.

Binary Representations:

  • 5 in binary is 101
  • 3 in binary is 011

Walkthrough with Binary Representations:

  1. Example: getSum(5, 3)

    Initial values:

    • a = 5 = 101 (in binary)
    • b = 3 = 011 (in binary)

    First Iteration:

    • a^b = 5^3 = 6 = 110 (in binary)
    • (a&b)<<1 = (5&3)<<1 = 2 = 010 (in binary when shifted left by 1)

    Recursive call with getSum(6, 2)

    New values:

    • a = 6 = 110 (in binary)
    • b = 2 = 010 (in binary)

    Second Iteration:

    • a^b = 6^2 = 4 = 100 (in binary)
    • (a&b)<<1 = (6&2)<<1 = 4 = 100 (in binary when shifted left by 1)

    Recursive call with getSum(4, 4)

    New values:

    • a = 4 = 100 (in binary)
    • b = 4 = 100 (in binary)

    Third Iteration:

    • a^b = 4^4 = 0 = 000 (in binary)
    • (a&b)<<1 = (4&4)<<1 = 8 = 1000 (in binary when shifted left by 1)

    Recursive call with getSum(0, 8)

    New values:

    • a = 0 = 000 (in binary)
    • b = 8 = 1000 (in binary)

    Since b is 0 in decimal (and 000 in binary), return a which is 8 in decimal and 1000 in binary.

    Final result: 8 = 1000 (in binary)

Code Implementation

/**
 * @param {number} a
 * @param {number} b
 * @return {number}
 */
var getSum = function(a, b) {
    if (b === 0) {
        return a;
    }
    return getSum(a^b, (a&b)<<1);
};

Conclusion

The “Sum of Two Integers” problem provides a unique opportunity to delve deep into the intricacies of bitwise operations. By understanding the underlying principles of binary arithmetic and utilizing bitwise operations effectively, we can tackle challenges that initially seem insurmountable. This solution not only offers an efficient way to solve the problem but also broadens our understanding of how basic arithmetic can be simulated at the binary level, all while being efficient in terms of both time and space.

<
Blog Archive
Archive of all previous blog posts
>
Next Post
Finding Factors of a Number Efficiently in Node.js