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Sum of Two Integers

Medium
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Apple
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Topics:
Bit Manipulation

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

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

Can you implement a function to achieve this, considering potential edge cases and providing an explanation of the time and space complexity of your solution?

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What is the range of values for the integers a and b? Are they 32-bit integers, or could they be larger?
  2. Can a and b be negative, positive, or zero?
  3. Are there any specific error conditions I should handle, such as overflow?
  4. Should I optimize for a particular language or architecture when choosing my bit manipulation techniques?
  5. Are there any constraints on space complexity, or can I use auxiliary variables as needed?

Brute Force Solution

Approach

The brute force approach to finding the sum of two integers involves trying out different combinations. Think of it like manually testing every possible pair of numbers until you find the right one. We check each pair to see if it matches the desired total.

Here's how the algorithm would work step-by-step:

  1. Consider the first number as a potential part of the sum.
  2. Then, consider the second number as a potential part of the sum.
  3. Add these two numbers together.
  4. Check if the result of the addition equals the target sum.
  5. If it does, you've found the pair. If not, move on and try different numbers.

Code Implementation

def two_sum_brute_force(numbers, target):

    for first_number_index in range(len(numbers)):
        # Consider each number as the first addend in the potential sum

        for second_number_index in range(first_number_index + 1, len(numbers)):
            # Consider the next number as the second addend
            if numbers[first_number_index] + numbers[second_number_index] == target:

                # Check to see if the sum of these two is our target
                return [first_number_index, second_number_index]

    #If no such pair exists, return an empty list.
    return []

Big(O) Analysis

Time Complexity
O(n²)The provided 'brute force' strategy implies iterating through all possible pairs of numbers. If we have an input of size 'n' representing the number of potential integer values, we effectively consider each number in relation to every other number. For each of the 'n' numbers, we potentially compare it with 'n-1' other numbers. This nested-loop structure leads to roughly n * (n-1) comparisons, which simplifies to O(n²).
Space Complexity
O(1)The provided description involves checking pairs of numbers without using any additional data structures to store intermediate results or track visited pairs. No auxiliary arrays, hash maps, or significant variables are created. Therefore, the space used remains constant, irrespective of the input size N, representing the number of potential pairs to check. The algorithm operates in place, implying minimal memory overhead beyond the input itself, resulting in O(1) space complexity.

Optimal Solution

Approach

We'll use bitwise operations to mimic how addition works at the binary level. We will simulate the carry and sum operations separately, repeating until there's no carry left.

Here's how the algorithm would work step-by-step:

  1. Calculate the sum of the two numbers without considering the carry. This is like an XOR operation.
  2. Figure out the carry bits. The carry occurs when both corresponding bits in the two numbers are 1. This is similar to an AND operation, then shifted to the left.
  3. Repeat the process. The previous 'sum' becomes one of the numbers, and the previous 'carry' becomes the other number. Keep calculating sum and carry until the carry becomes zero, meaning there are no more carries to add.
  4. The final 'sum' is the result of adding the two original numbers.

Code Implementation

def get_sum(first_number, second_number):
    mask = 0xFFFFFFFF

    while (second_number & mask) > 0:
        # Calculate sum without carry
        sum_without_carry = (first_number ^ second_number) & mask

        # Calculate carry
        carry = ((first_number & second_number) << 1) & mask

        first_number = sum_without_carry
        second_number = carry

    # Handle overflow
    max_int = 0x7FFFFFFF
    return first_number if first_number <= max_int else ~(first_number ^ mask)

Big(O) Analysis

Time Complexity
O(1)The number of iterations in the while loop depends on the number of bits required to represent the integers a and b. Since integers are typically represented using a fixed number of bits (e.g., 32 or 64 bits), the loop will execute at most a constant number of times. The bitwise operations within the loop take constant time. Therefore, the time complexity is O(1), constant time, regardless of the magnitude of the input integers.
Space Complexity
O(1)The algorithm uses a fixed number of integer variables to store the intermediate sum and carry values. The number of variables does not depend on the magnitude of the input integers. Therefore, the auxiliary space required is constant, regardless of the input size N (where N is related to the number of bits needed to represent the input integers), resulting in a space complexity of O(1).

Edge Cases

CaseHow to Handle
Both a and b are zeroThe bitwise operations will correctly return 0.
One of a or b is zeroBitwise operations handle 0 correctly, simply returning the other number.
Both a and b are positiveThe bitwise operations accurately simulate addition in this common case.
Both a and b are negativeThe bitwise operations handle negative numbers represented in two's complement.
a is positive and b is negative (a > abs(b))Bitwise operations correctly handle positive and negative interaction and return the correct sum.
a is positive and b is negative (a < abs(b))The algorithm should compute the negative sum correctly using two's complement.
Integer overflow (sum exceeds MAX_INT or falls below MIN_INT)In languages like Java or Python, integer overflow wraps around, but in C/C++ it's undefined behavior; we should note this potential pitfall.
MAX_INT and 1 or MIN_INT and -1Check for overflow on these edge cases after bitwise operations return the intermediate result to avoid incorrect sums.