## Division without using *, / or % (Divide Two Integers)

Divide two integers without using multiplication, division and mod operator.

Thoughts:
The common way to do this is to count how many times that divisor adding up to dividend. Of course, we should take care of signs. That’s not hard though. However, LeetCode won’t let you pass for the reason of time exceed exception. I have to use nother trick: since $\log(\frac{a}{b}) = \log{a} - \log{b}$, we have $\frac{a}{b} = \exp (\log{a} - \log{b})$. The tricky part of this question does not lie on the algorithm though. It has something to do with overflows. For particular, if we use Math.abs to compute the absolute value of Integer.MIN(-2147483648), we get -2147483648 again. So we should manually make it equal to Integer.MAX(2147483647). Most of the cases this is fine, except for one case where you try to divide Integer.MIN by 2, i.e., -2147483648 / 2 = -1073741824. However, 2147483647 / 2 = 1073741823. I have to add one more edge condition that if the divisor is 2, we just do the bitwise operation: right shift. Another node is Integer.MAX / Integer.MIN = 0 (not -1).

Code (Java):

public class Solution {
public int divide(int dividend, int divisor) {
if(divisor == 0)
return 0;
if(divisor == 1)
return dividend;
if(dividend == divisor)
return 1;
if(divisor == 2)
return dividend >> 1;

boolean sign = false;
if( (dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0) )
sign = true;

if(dividend == Integer.MAX_VALUE && divisor == Integer.MIN_VALUE)
return 0;

dividend = dividend == Integer.MIN_VALUE ? Integer.MAX_VALUE : Math.abs(dividend);
divisor = divisor == Integer.MIN_VALUE ? Integer.MAX_VALUE : Math.abs(divisor);
int result = (int) Math.floor(Math.pow(Math.E, Math.log(dividend) - Math.log(divisor)));
return sign ? -result : result;
}
}

Code (C++):
class Solution {
public:
int divide(int dividend, int divisor) {
if(divisor == 0)
return 0;
if(divisor == 1)
return dividend;
if(dividend == divisor)
return 1;
if(divisor == 2)
return dividend >> 1;

bool sign = false;
if( (dividend > 0 && divisor < 0) ||
(dividend < 0 && divisor > 0))
sign = true;

if(dividend == numeric_limits<int>::max()
&& divisor == numeric_limits<int>::min())
return 0;

dividend = dividend == numeric_limits<int>::min() ?
numeric_limits<int>::max() : abs(dividend);
divisor = divisor == numeric_limits<int>::min() ?
numeric_limits<int>::max() : abs(divisor);

int result = (int) floor(exp(log(dividend) - log(divisor)));
return sign ? -result : result;
}
};

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