371. Sum of Two Integers - Detailed Explanation

Problem Statement

Given two integers, a and b, return the sum of these two integers without using the addition (+) or subtraction (-) operators. The challenge is to compute the result using bit-level operations.

Example 1

Input:

a = 1, b = 2

Output:

Explanation:
The sum of 1 and 2 is 3.

Example 2

Input:

a = 2, b = 3

Output:

Explanation:
The sum of 2 and 3 is 5.

Constraints

The integers a and b can be positive, negative, or zero.
You cannot use the + or - operators in your solution.
The solution should handle typical 32-bit integer operations.

Hints

Bit Manipulation Insight:
Use bitwise XOR (^) to add two numbers without carrying. This operation simulates addition for each bit but does not account for the carry-over.
Carry Calculation:
Use bitwise AND (&) to identify which bit positions will generate a carry. Then, shift the carry left by one (using <<) to add it to the next higher bit.
Iterative or Recursive Approach:
The sum can be computed by iteratively updating the numbers until there is no carry left, or by using recursion that terminates when the carry becomes zero.

Approach: Bit Manipulation

Explanation

The key observation is that the sum of two integers can be broken into two parts:

Non-carry sum:
Using the XOR operation (a ^ b) adds the bits without considering any carry.
Carry bits:
Using the AND operation (a & b) finds all positions where both bits are 1, meaning a carry is generated. Shifting this result left by one ((a & b) << 1) gives the actual carry that must be added to the non-carry sum.

The algorithm then repeats the process:

Let sum = a ^ b (non-carry addition).
Let carry = (a & b) << 1 (carry bits).
Set a = sum and b = carry and repeat until carry becomes 0.

When there is no carry, the variable a contains the final sum.

Visual Example

Consider adding a = 2 and b = 3:

Binary representation:
- 2 → 10
- 3 → 11

Step 1:

Non-carry sum: a ^ b = 10 ^ 11 = 01 (which is 1 in decimal)
Carry: (a & b) << 1 = (10 & 11 = 10) << 1 = 100 (which is 4 in decimal)

Now, update:

a = 1
b = 4

Step 2:

Non-carry sum: a ^ b = 001 ^ 100 = 101 (which is 5 in decimal)
Carry: (a & b) << 1 = (001 & 100 = 000) << 1 = 000 (which is 0 in decimal)

Since the carry is 0, the final answer is a = 5.

Python Code

Python3

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Java Code

Java

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Complexity Analysis

Time Complexity:
In the worst case, the loop may run O(n) times where n is the number of bits (typically 32 for a standard integer). So, the time complexity is O(1) with respect to the size of the integers.
Space Complexity:
The solution uses O(1) extra space since only a fixed number of variables are used.

Step-by-Step Walkthrough

Initialization:
- Define a mask for 32-bit operations.
- Set a and b to the input integers.
Iteration:
- Calculate the non-carry sum using XOR.
- Calculate the carry using AND, then left-shift by one.
- Update a with the non-carry sum and b with the carry.
Termination:
- The loop ends when there is no carry left (i.e., b equals 0).
- If a is greater than the maximum 32-bit positive integer, convert it to a negative number using bitwise complement.
Return:
- a contains the final sum.

Common Mistakes and Edge Cases

Handling Negative Numbers:
Since Python uses unlimited precision for integers, you need to simulate 32-bit overflow using a mask. In Java, the integer type naturally uses 32 bits.
Infinite Loop:
Ensure the loop eventually terminates when the carry becomes zero.
Bit Masking:
Forgetting to mask intermediate results might lead to incorrect values in languages with fixed integer sizes.

Alternative Variations

Recursive Implementation:
Instead of iterating, you can implement the solution recursively until the carry is zero.
Subtraction Variant:
A similar technique can be used to implement subtraction using bitwise operations, though the logic is a bit more involved.