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public static int divide(int dividend, int divisor) {

        if(divisor == 0) throw new RuntimeException();

        int sign = 1;
        //figure out the sign of the result
        if((dividend < 0 && divisor > 0)
                || (dividend > 0 && divisor < 0)) {

            sign = -1;

        }

        if(dividend < 0) dividend = -dividend;
        if(divisor < 0) divisor = -divisor;

        int n = 1; //get the range of result [n, 2n], where n is doubled every round

        while(dividend > (n << 1) * divisor) {

            n <<= 1;

        }

        //now it's known that the result is between [n, 2n]
        //so in the dividend has a sum of more than n and less than 2n divisors in total
        //to figure out exactly how many divisors sum up to the dividend,
        // break down the problem to result = n + divide(dividend - n * divisor, divisor)
        // next round the dividend becomes (dividend - n * divisor) with n added to the result.
        // proceed to figure out the range [x, 2x] of result for the new dividend, where x can be n/2, n/4, n/8...
        // whenever the x is found, add x to the result and deduce x * divisor from the dividend
        // util n is 0 or dividend is 0

        //If written in math, the formula looks like dividend = ([T/F]* n + [T/F]* n/2 + [T/F]* n/ 4 + [T/F]* n/ 8...) * divisor
        //The quotient will be ([T/F]* n + [T/F]* n/2 + [T/F]* n/ 4 + [T/F]* n/ 8...), [T/F] depends on (dividend - n * divisor) >= 0
        int result = 0;

        while(n > 0 && dividend > 0) {

            if(dividend - n * divisor >= 0) {

                dividend -= n * divisor;
                result += n;

            }

            n >>= 1;

        }

        return result * sign;

    }

private static void Divide(int a, int b, out int quotient, out int reminder)
        {
            quotient = 0;
            reminder = 0;
            int aCopy = a;
            while (aCopy >= b)
            {
                quotient++;
                aCopy -= b;
            }

            reminder = aCopy;
        }

private static void Divide(int a, int b, out int quotient, out int reminder)
        {
            quotient = 0;
            reminder = 0;
            int aCopy = a;
            while (aCopy >= b)
            {
                quotient++;
                aCopy -= b;
            }

            reminder = aCopy;
        }

As the function returns an int we suppose we are trimming the decimals and we just only return the int part, e.g. 3/2 = 1.5, result = 1

now we have that a/b = c thus giving us a=c*b, then we need to find a value where c times b is equal to a, so we iterate over the possibles values of c

function divide(a,b) {
	i=0;
	while(b*i<=a){
            i++;
	}
	console.log(i-1);
}