Divide Two Integers

                                            
class Solution {
public:
    int divide(int dividend, int divisor) {
        if(dividend == INT_MIN && divisor == -1) return INT_MAX;
        
        long dvd = labs(dividend);
        long dvs = labs(divisor);
        int sign = dividend > 0 ^ divisor > 0 ? -1 : 1;
        
        long result = 0;
        while(dvd >= dvs) {
            long temp = dvs, multiple = 1;
            while(dvd >= (temp << 1)) {
                temp <<= 1;
                multiple <<= 1;
            }
            dvd -= temp;
            result += multiple;
        }
        
        return sign * result;
    }
};

                                            
class Solution {
    public int divide(int dividend, int divisor) {
        if(dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }
        
        int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;
        
        long dvd = Math.abs((long)dividend);
        long dvs = Math.abs((long)divisor);
        
        int result = 0;
        while(dvd >= dvs) {
            long temp = dvs, multiple = 1;
            while(dvd >= (temp << 1)) {
                temp <<= 1;
                multiple <<= 1;
            }
            dvd -= temp;
            result += multiple;
        }
        
        return sign == 1 ? result : -result;
    }
}

                                            
class Solution(object):
    def divide(self, dividend, divisor):
        INT_MAX = 2**31 - 1
        INT_MIN = -2**31
        
        if dividend == 0:
            return 0
        
        if dividend == INT_MIN and divisor == -1:
            return INT_MAX
        
        negative = (dividend < 0) != (divisor < 0)
        
        dividend = abs(dividend)
        divisor = abs(divisor)
        
        quotient = 0
        
        while dividend >= divisor:
            temp, i = divisor, 1
            while dividend >= temp:
                dividend -= temp
                quotient += i
                i <<= 1
                temp <<= 1
        
        if negative:
            return -quotient
        else:
            return quotient

                                            
int divide(int dividend, int divisor) {
    int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1;
    long long dvd = labs(dividend);
    long long dvs = labs(divisor);
    long long quotient = 0;
    
    while (dvd >= dvs) {
        long long temp = dvs, multiple = 1;
        while (dvd >= (temp << 1)) {
            temp <<= 1;
            multiple <<= 1;
        }
        
        dvd -= temp;
        quotient += multiple;
    }
    
    quotient *= sign;
    if (quotient > INT_MAX) return INT_MAX;
    if (quotient < INT_MIN) return INT_MIN;
    
    return quotient;
}

                                            
function divide(dividend: number, divisor: number): number {
    if (dividend === 0) return 0;
    if (divisor === 1) return dividend;
    if (divisor === -1) return dividend === -Math.pow(2, 31) ? Math.pow(2, 31) - 1 : -dividend;

    const sign = (dividend < 0) !== (divisor < 0) ? -1 : 1;
    dividend = Math.abs(dividend);
    divisor = Math.abs(divisor);

    let result = 0;
    while (dividend >= divisor) {
        let currentDivisor = divisor;
        let currentResult = 1;

        while (dividend >= currentDivisor << 1) {
            currentDivisor <<= 1;
            currentResult <<= 1;
        }

        result += currentResult;
        dividend -= currentDivisor;
    }

    return sign === 1 ? Math.min(Math.pow(2, 31) - 1, result) : -result;
};

                                            
class Solution {
    /**
     * @param Integer $dividend
     * @param Integer $divisor
     * @return Integer
     */
    function divide($dividend, $divisor) {
        $sign = ($dividend < 0) ^ ($divisor < 0) ? -1 : 1;
        $dividend = abs($dividend);
        $divisor = abs($divisor);
        $result = 0;
        while ($dividend >= $divisor) {
            $temp = $divisor;
            $quotient = 1;
            while ($dividend >= ($temp << 1)) {
                $temp <<= 1;
                $quotient <<= 1;
            }
            $dividend -= $temp;
            $result += $quotient;
        }
        $result *= $sign;
        if ($result < -(pow(2, 31)) || $result > (pow(2, 31) - 1)) {
            return pow(2, 31) - 1;
        } else {
            return $result;
        }
    }
}

To solve this coding challenge, you'll need to develop an algorithm that divides two integers (dividend and divisor) without using the multiplication, division, or mod operators. The result should be the quotient with any fractional part truncated toward zero.

Explanation:

Handling Edge Cases :

If the dividend is 0, the quotient should be 0 since any number divided by non-zero results in 0.
If the dividend is the minimum possible 32-bit integer value (-2^31) and the divisor is -1, the result would overflow because the result 2^31 is beyond the maximum positive value for a 32-bit integer. Therefore, return the maximum 32-bit integer value (2^31 - 1).

Handling Signs :

Determine the sign of the result based on the signs of the dividend and divisor. If only one of them is negative, the result is negative. Otherwise, it's positive.

Convert to Positive :

Work with absolute values of the dividend and divisor to simplify the calculations.

Performing Division Using Subtraction :

Initialize a variable
quotient

to 0 to keep track of the result.
Use a loop to repeatedly subtract the divisor (multiplied by increasing powers of 2) from the dividend until the dividend is less than the divisor. This effectively simulates the division operation.

Adjusting the Quotient's Sign :

If the result should be negative, convert the quotient to negative. Otherwise, leave it as is.

Returning the Result :

Finally, return the computed quotient but ensure it is within the 32-bit signed integer range.

Step-by-Step Explanation/Detailed Steps in Pseudocode:

Check for Dividend and Divisor as Edge Cases :

Initialize constants for INT_MAX and INT_MIN.
Handle cases where the dividend is 0 or the problematic overflow case.

Determine the Sign of the Result :

Compare the signs of the dividend and divisor.

Convert to Positive Values :

Transform the dividend and divisor to their absolute values.

Binary Division via Subtraction :

Initialize a variable
quotient

to 0.
Use nested loops to perform the division using bitwise shifts to handle the subtraction in powers of 2.

Return Result :

Adjust the result if it's negative.
Ensure the result is within bounds and return it.

Pseudocode:

                                            
// Define 32-bit integer bounds
INT_MAX = 2^31 - 1
INT_MIN = -2^31

function divide(dividend, divisor):
    // Edge case: if dividend is 0, the result is 0
    if dividend == 0:
        return 0

    // Handle overflow case
    if dividend == INT_MIN and divisor == -1:
        return INT_MAX

    // Determine if the result is negative
    isNegative = (dividend < 0) != (divisor < 0)

    // Convert both dividend and divisor to positive values
    dividend = abs(dividend)
    divisor = abs(divisor)

    // Initialize quotient
    quotient = 0

    // Perform subtraction method
    while dividend >= divisor:
        temp_divisor = divisor
        current_quotient = 1

        // Subtract divisor multiplied by powers of 2 from dividend
        while dividend >= temp_divisor:
            dividend = dividend - temp_divisor
            quotient = quotient + current_quotient

            // Double the temp_divisor and current_quotient
            current_quotient = current_quotient << 1
            temp_divisor = temp_divisor << 1

    // Apply the sign to the result
    if isNegative:
        return -quotient
    else:
        return quotient

In essence, this approach ensures that we perform division through repeated subtraction and bitwise shifts, thereby adhering to the constraints of not using multiplication, division, or modulus operations. This method efficiently calculates the quotient up to the bounds of a 32-bit signed integer, ensuring correct handling and termination of the algorithm.