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[LeetCode] 029. Divide Two Integers

Problem (Medium)

029. Divide Two Integers

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

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

  • Input: dividend = 10, divisor = 3
  • Output: 3

Example 2:

  • Input: dividend = 7, divisor = -3
  • Output: -2

Note:

  • Both dividend and divisor will be 32-bit signed integers.
  • The divisor will never be 0.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 2^31 − 1 when the division result overflows.

Approach 1: Loop Subtraction (My Solution)

Idea

Since it is not allowed to use multiplication, divide and mod, the idea of incrementally increase the result by subtract divisor in loops is an option. (But for large numbers, it may exceed the time limit!)

Solution

class Solution1:
    def divide(self, dividend, divisor):
        """
        :type dividend: int
        :type divisor: int
        :rtype: int
        """
        isNeg = (dividend > 0) ^ (divisor > 0)
        dividend, divisor = abs(dividend), abs(divisor)

        res = 0
        while dividend >= divisor:
            res += 1
            dividend -= divisor
        if isNeg:
            res = -res
        return min(max(-2147483648, res), 2147483647)

Complexity

  • Time: $O(n)$
  • Space: $O(1)$

Approach 1 Improve: Bitwise Operation

Idea

Use bitwise operations to speed up the process in Approach 1.

Solution

class Solution2:
    def divide(self, dividend, divisor):
        positive = (dividend < 0) is (divisor < 0)
        dividend, divisor = abs(dividend), abs(divisor)
        res = 0
        while dividend >= divisor:
            temp, i = divisor, 1
            while dividend >= temp:
                dividend -= temp
                res += i
                i <<= 1
                temp <<= 1
        if not positive:
            res = -res
        return min(max(-2147483648, res), 2147483647)

Complexity

  • Time: $O(n)$
  • Space: $O(1)$



KF

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