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Leetcode 371. Sum of Two Integers

Problem Statement

Leetcode 371: Sum of Two Integers

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

Clarifying Questions

  1. Q: What are the constraints on the integers a and b (e.g., size, range)?
    • A: The integers a and b will be within the range of a typical 32-bit signed integer: ([-2^{31}, 2^{31} - 1]).
  2. Q: Can I assume that a and b are always valid integers?
    • A: Yes, you can assume a and b are within the 32-bit signed integer range.
  3. Q: Do we need to handle any special cases like overflow?
    • A: No explicit overflow handling is needed as the problem is about implementing the addition using bitwise operations.

Strategy

To calculate the sum of two integers without using + or -, we can use bitwise operations:

  1. Sum Calculation:
    • The sum of two bits can be done with ^ (XOR) operator since 1 XOR 1 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, and 0 XOR 0 = 0.
  2. Carry Calculation:
    • The carry can be calculated using the & (AND) operator followed by a left shift <<, since carry = (a & b) << 1. This handles the carry operation in binary addition.
  3. Iterative Approach:
    • We use the iterative approach until there is no carry. In each iteration, we update a to be the sum (without carry) and b to be the carry. This process continues until b becomes zero.

Code

class Solution {
    public int getSum(int a, int b) {
        while (b != 0) {
            int carry = (a & b) << 1;  // Calculate carry
            a = a ^ b;                 // Calculate sum without carry
            b = carry;                 // Assign carry to b
        }
        return a;
    }
}

Time Complexity

This approach effectively computes the sum of two integers using bitwise operations while adhering to the constraints provided in the problem statement.

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