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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, which means losing its fractional part. For example, truncate(8.345) = 8 and truncate(-2.7335) = -2.

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, assume that your function returns 231 − 1 when the division result overflows.

 

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = truncate(3.33333..) = 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = truncate(-2.33333..) = -2.

Example 3:

Input: dividend = 0, divisor = 1
Output: 0

Example 4:

Input: dividend = 1, divisor = 1
Output: 1

 

Constraints:

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

class Solution { public int divide(int dividend, int divisor) { if(dividend == Integer.MIN_VALUE && divisor == -1){ return Integer.MAX_VALUE; } int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1; long absDividend = Math.abs((long) dividend); long absDivisor = Math.abs((long) divisor); long ans = 0; while(absDividend >= absDivisor){ long temp = absDivisor; long mult = 1; while((temp<<1) <= absDividend){ temp <<= 1; mult <<= 1; } ans += mult; absDividend -= temp; } return sign * (int) ans; } }

This question requires the knowledge of Bit manipulation. If you know the concept that a left shift operation in bit manipulation represents multiplication by 2 then it is simple to code the same logic.
A detailed explanation is provided in the Video.

Time Complexity: O(32) // constant time

Space Complexity: O(1)