Naive Solution

A simple approach is to repeatedly subtract the divisor from the dividend until the dividend becomes smaller than the divisor. The number of subtractions will be the quotient.

Edge Cases:

• Divisor is 0: This should throw an error as division by zero is undefined.
• Dividend is 0: The quotient is 0.
• Signs: Handle the signs of the dividend and divisor to determine the sign of the quotient.
• Overflow: Check for potential overflow issues if the quotient is too large or too small.

Code (Python):

def divide_naive(dividend, divisor):
    if divisor == 0:
        raise ZeroDivisionError("Cannot divide by zero")

    sign = -1 if (dividend < 0) ^ (divisor < 0) else 1
    dividend = abs(dividend)
    divisor = abs(divisor)

    quotient = 0
    while dividend >= divisor:
        dividend -= divisor
        quotient += 1

    quotient *= sign

    # Handle overflow
    INT_MAX = 2**31 - 1
    INT_MIN = -2**31
    if quotient > INT_MAX:
        return INT_MAX
    elif quotient < INT_MIN:
        return INT_MIN
    else:
        return quotient

Time Complexity: O(dividend / divisor). In the worst case, if the dividend is very large and the divisor is 1, this can be very slow.

Space Complexity: O(1). Constant space is used.

Optimal Solution (Bit Manipulation)

A much more efficient approach involves using bit manipulation. The idea is to use repeated subtraction, but instead of subtracting the divisor one at a time, we subtract multiples of the divisor (powers of 2 times the divisor).

1. Determine the sign of the result.
2. Take the absolute values of the dividend and divisor.
3. Repeatedly subtract left-shifted versions of the divisor from the dividend.
4. Keep track of how many times each shifted divisor can be subtracted.
5. Handle overflow.

Edge Cases:

• Divisor is 0: Throw an error.
• Dividend is 0: Return 0.
• Signs: Handle signs correctly.
• Overflow: Properly handle integer overflow by clamping the result to the maximum or minimum integer value.

Code (Python):

def divide(dividend, divisor):
    if divisor == 0:
        raise ZeroDivisionError("Cannot divide by zero")

    sign = -1 if (dividend < 0) ^ (divisor < 0) else 1
    dividend = abs(dividend)
    divisor = abs(divisor)

    quotient = 0
    while dividend >= divisor:
        temp = divisor
        multiple = 1
        while dividend >= (temp << 1):
            if temp > 2**30: # avoid left shift overflow
                break
            temp <<= 1
            multiple <<= 1
        dividend -= temp
        quotient += multiple

    quotient *= sign

    INT_MAX = 2**31 - 1
    INT_MIN = -2**31
    if quotient > INT_MAX:
        return INT_MAX
    elif quotient < INT_MIN:
        return INT_MIN
    else:
        return quotient

Time Complexity: O(log(dividend)). This is because, in each iteration, we are essentially halving the dividend.

Space Complexity: O(1). Constant space is used.