Sum of Two Integers - Practice Coding Problems

Sum of Two Integers - Problem

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

This problem requires you to implement addition using only bitwise operations. You'll need to think about how addition works at the binary level and simulate the process of adding two numbers bit by bit.

Hint: Consider how XOR and AND operations can help simulate addition and carry operations.

Input & Output

Example 1 — Basic Addition

$ Input: a = 1, b = 2

› Output: 3

💡 Note: 1 + 2 = 3. Using bitwise: 1 (001) XOR 2 (010) = 3 (011), no carry needed.

Example 2 — With Carry

$ Input: a = 2, b = 3

› Output: 5

💡 Note: 2 + 3 = 5. Binary: 010 + 011. XOR gives 001, carry is (010 & 011) << 1 = 100. Then 001 + 100 = 101 (5).

Example 3 — Negative Numbers

$ Input: a = -1, b = 1

› Output: 0

💡 Note: -1 + 1 = 0. The bitwise operations handle two's complement representation correctly.

Constraints

-1000 ≤ a, b ≤ 1000

Visualization

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Understanding the Visualization

Input

Two integers a=1, b=2

Process

Use XOR and AND with bit shifting

Output

Sum = 3

Key Takeaway

🎯 Key Insight: XOR handles sum without carry, AND+shift handles carry propagation

Asked in

G Google 25 a Amazon 30 f Facebook 20 M Microsoft 15

The key insight is to use bitwise operations to simulate binary addition. XOR handles sum without carry, AND + left shift handles carry bits. Best approach is bitwise manipulation with Time: O(1), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Loop-based Addition	O(b)	O(1)	Increment one number by one, b times
Bitwise Addition	O(1)	O(1)	Use XOR for sum and AND+shift for carry

Loop-based Addition — Algorithm Steps

Start with result = a
Loop b times, incrementing result each time
Return the final result

Visualization

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Step-by-Step Walkthrough

Start

Begin with a = 1

Loop

Increment a for b=2 times

Result

Final result is 3

Code -

solution.c — C

#include <stdio.h>

int solution(int a, int b) {
    if (b == 0) return a;
    if (b > 0) {
        for (int i = 0; i < b; i++) {
            a = a ^ ((a ^ (a + 1)) & -(a ^ (a + 1)));
        }
        return a;
    } else {
        for (int i = 0; i < -b; i++) {
            a = a ^ ((a ^ (a - 1)) & -(a ^ (a - 1)));
        }
        return a;
    }
}

int main() {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", solution(a, b));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(b)

Loop runs b times, each increment is O(1)

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Loop-based Addition — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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