Leetcode: Divide Two Integers

Problem Statement:

Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator.

The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, and -2.7335 would be truncated to -2.

Return the quotient after dividing dividend by divisor.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For this problem, if the quotient is strictly greater than 231 - 1, then return 231 - 1, and if the quotient is strictly less than -231, then return -231.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = 3.33333.. which is truncated to 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = -2.33333.. which is truncated to -2.

Constraints:

• -231 <= dividend, divisor <= 231 - 1
• divisor != 0

Solution:

class Solution:
    def divide(self, dividend: int, divisor: int) -> int:
        # Save the original divisor for reference
        orig_divisor = divisor
        # Initialize the quotient
        n = 0

        # Handle the case where either dividend or divisor is zero
        if dividend == 0:
            return 0

        # Determine the sign of the result
        sign = (dividend > 0) == (divisor > 0)

        # Handle the case where divisor is 1 or -1 separately
        if divisor == 1:
            return dividend
        elif divisor == -1:
            # Handle division by -1, considering overflow cases
            dividend = dividend if sign else -dividend
            return min(max(dividend, -2**31), 2**31 - 1)

        # Take the absolute values for calculation
        dividend, divisor = abs(dividend), abs(divisor)

        # Main division logic
        while dividend >= divisor:
            dividend -= divisor
            n += 1

        # Apply the sign to the result
        result = n if sign else -n

        # Limit the result within the 32-bit signed integer range
        return min(max(result, -2**31), 2**31 - 1)

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