Divide Two Integers

Problem

https://leetcode.com/problems/divide-two-integers/

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: [−2:sup:`31`, 2:sup:`31`− 1]. For this problem, if the quotient is strictly greater than 2:sup:`31`- 1, then return 2:sup:`31`- 1, and if the quotient is strictly less than -2:sup:`31`, then return -2:sup:`31`.

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:

• -2:sup:`31`<= dividend, divisor <= 2:sup:`31`- 1

• divisor != 0

Pattern

Math, Bit Manipulation

Solution

We can use repeated subtraction of the dividend by the divisor to find the quotient without using multiplication, division, or modulus. However, loops in Python are slow so we will time out for large dividends and small divisors. Instead use len(range(0, stop, step)) to find the quotient. Since range(0, stop, step) yields 1 element even if stop < step, put stop = abs(dividend) - abs(divisor) + 1 and stop = abs(divisor). Adjust the sign of the quotient when \((dividend < 0) \\oplus (divisor < 0)\). Clamp the quotient to be in \([-2^{31}, 2^{31} - 1]\).

Code

def divide(dividend: int, divisor: int) -> int:
    quotient = len(range(0, abs(dividend) - abs(divisor) + 1, abs(divisor)))

    if (dividend < 0) ^ (divisor < 0):
        quotient = -quotient

    return max(min(quotient, 2**31 - 1), -2**31)

Test

>>> from divide_two_integers__approach_1 import divide
>>> divide(10, 3)
3
>>> divide(7, -3)
-2

Complexity

\(q\) is the quotient

Time: \(O(|q|)\) — calculating the quotient by counting the number of

steps in the range | Auxiliary Space: \(O(1)\) — constant extra space

divide_two_integers__approach_1.divide(dividend: int, divisor: int) → int