Question

link

Stats

Frequency	3
Diffficulty	4
Adjusted Difficulty	4
Time to use	----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This question might be more difficult than you thought. So do not overlook it because of its seemingly simple description. First of all, please refer to [Fundamental] Java Bit Operation for information on bit operators.

And remember… overflow can happen, especially when you dealing with Integer.MAX_VALUE and Integer.MIN_VALUE.

Solution

This solution is a while loop that keeps subtracting (divisor * (2 ^ n)) from dividend. You can find a good solution from this blog.

code

public class Solution {
    public int divide(int dividend, int divisor) {
        int sign = 1;
        long x = dividend;
        long y = divisor;
        if (x < 0) {
            x = x * -1;
            sign *= -1;
        }
        if (y < 0) {
            y = y * -1;
            sign *= -1;
        }
        return divide(sign, x, y);
    }

    private int divide(int sign, long x, long y) {
        // both x and y are positive numbers
        if (x < y) {
            return 0;
        }
        long quotient = 0;
        long partialQuotient;
        long partialSubtract;
        while (x >= y) {
            // the idea is to subtract a certain times of x from y
            // and save the remainder value back to x
            partialQuotient = 1;
            partialSubtract = y;
            while ((partialSubtract << 1) <= x) {
                partialQuotient <<= 1; // doubles quotient
                partialSubtract <<= 1; // doubles subtraction
            }
            x -= partialSubtract;
            quotient += partialQuotient;
        }
        long finalQuo = sign * quotient;
        if (finalQuo < Integer.MIN_VALUE || finalQuo > Integer.MAX_VALUE) {
            return Integer.MAX_VALUE;
        } else {
            return (int) finalQuo;
        }
    }
}