Sum of Two Integers

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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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

✓ Bitwise Addition

⏱️ Time: O(1) Space: O(1)

Simulate binary addition using bitwise operations. XOR gives us the sum without carry, AND gives us carry bits, then shift carry left and repeat until no carry remains.

Loop-based Addition

⏱️ Time: O(b) Space: O(1)

Start with number a, then increment it one by one for b times. This simulates addition by repeated increment operations without using + or -.

Bitwise Addition — Algorithm Steps

XOR a and b to get sum without carry
AND a and b, shift left to get carry
Repeat until carry becomes 0
Return the sum

Visualization

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Step-by-Step Walkthrough

XOR Operation

a ^ b gives sum without carry

AND + Shift

(a & b) << 1 gives carry bits

Repeat

Continue until carry is 0

Code -

solution.c — C

#include <stdio.h>

int solution(int a, int b) {
    while (b != 0) {
        // XOR gives sum without carry
        int sumWithoutCarry = a ^ b;
        // AND and shift gives carry
        int carry = (a & b) << 1;
        
        a = sumWithoutCarry;
        b = carry;
    }
    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(1)

Fixed number of iterations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Bitwise Addition — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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