Divide-Two-Integers

Solve a problem on leetcode.com website divide two integers. Its detail of problem can be found in the Description (link of the problem) website leetcode.com.

Description of Problem

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

Solve The Problem

I solve this problem C# language.

• Step 1:

First of all we need to check if divisor equal to 0 the inputs is wrong and we return Exception error in program because we can't have divide by zero in order to slove this we write code below:

if (divisor == 0)
{
       throw new DivideByZeroException();
}  

• Step 2:

And then we need to calculate the divide function and name it result

double result = (double)dividend/(double)divisor; 

• You can also do it with the long variable like this

long result = (long)dividend/(long)divisor; 

You need to know we can't return result with long or double because the method is int so we need to convert it to int

• Step 3:

But before return the result we need to check the result which isn't out of the range of problem as I said before the problem have range: 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. to solve this we need to write code below:

if (result<int.MinValue || result > int.MaxValue)
{
       return (int)Math.Pow(2,31)-1;
} 

• (int) for converting the result value to int, and the answer will be truncate like example 1 and 2.

• Last Step:

For last step we need to return the result to the method as I said before we need to convert the variable result it dosn't matter you do it with long double we need to convert it write code below like did before:

return (int)result;

Final Code

public class Solution {
    
    public int Divide(int dividend, int divisor) {
        
        if (divisor == 0)
        {
            throw new DivideByZeroException();
        }
        
        long result = (long)dividend/(long)divisor; 
        
	    if(result<int.MinValue || result > int.MaxValue)
        {
		    return (int)Math.Pow(2,31)-1;
	    }
        
	    return (int)result;
    }
}

Unit Tests Listing

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