29. Divide Two Integers

Description

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.

Constraints

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

• divisor != 0

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(n)
 * Space complexity : O(1)
 */

class Solution {
    public int divide(int dividend, int divisor) {
        if(dividend == 0 || divisor == 1) {
            return dividend;
        } else if(divisor == -1) {
            if(dividend == Integer.MIN_VALUE) {
                return Integer.MAX_VALUE;
            }
            return dividend * divisor;
        }

        int m = Math.abs(dividend);
        int n = Math.abs(divisor);
        int count = 0;
        
        while(m-n >= 0) {
            m -= n;
            count++;
        }
        
        return ((dividend >= 0 && divisor >=0) || 
                (dividend < 0 && divisor < 0))? count: count*-1;
    }
}

Solution 2

/**
 * Time complexity : O(logN)
 * Space complexity : O(1)
 */
 
 class Solution {
    public int divide(int dividend, int divisor) {
        if(dividend == 0 || divisor == 1) {
            return dividend;
        } else if(divisor == -1) {
            if(dividend == Integer.MIN_VALUE) {
                return Integer.MAX_VALUE;
            }
            return dividend * divisor;
        }
        
        long pDividend = Math.abs((long) dividend);
        long pDivisor = Math.abs((long) divisor);
        
        int count = 0;
        
        while(pDividend >= pDivisor) {
            int shift = 0;
            while(pDividend >= (pDivisor<<shift+1)) {
                shift++;
            }
            pDividend -= (pDivisor<<shift);
            count += (1<<shift);
        }
        return ((dividend >= 0 && divisor >= 1) || 
                    (dividend < 0 && divisor < 0))? count: count*-1;
    }
}

Follow up