Solution Explanation for Divide Two Integers

This problem challenges you to implement integer division without using the multiplication, division, or modulo operators. The solution leverages bit manipulation and a clever approach to simulate division through repeated subtraction. However, naive repeated subtraction is inefficient. The optimal solution uses a technique similar to binary exponentiation to speed up the process significantly.

Core Idea:

The core idea is to repeatedly subtract the divisor from the dividend, but not just one divisor at a time. Instead, we try to subtract multiples of the divisor (powers of 2) to accelerate the process.

Algorithm:

1. Handle Special Cases:

   • Check for divisor == 0 (throws exception or returns error code).
   • Check for dividend == INT_MIN and divisor == -1 (handles potential overflow).

2. Determine Sign:

   • Store the sign of the result (positive if dividend and divisor have the same sign, otherwise negative). We work with absolute values to avoid overflow issues.

3. Convert to Negative:

   • Convert both dividend and divisor to their negative counterparts. This is crucial to avoid overflow when handling INT_MIN.

4. Iterative Subtraction with Bit Shifting:

   • This is the heart of the algorithm. We iteratively try to subtract multiples of the divisor from the dividend using bit shifts.
   • Initialize quotient = 0.
   • While dividend <= divisor:
     • Start with temp_divisor = divisor and multiplier = 1.
     • Repeatedly double temp_divisor and multiplier (temp_divisor <<= 1; multiplier <<= 1) until doubling temp_divisor would make it greater than the dividend.
     • Subtract temp_divisor from dividend (dividend -= temp_divisor).
     • Add multiplier to the quotient (quotient += multiplier).

5. Adjust Sign:

   • If the original sign was negative, negate the quotient.

6. Handle Overflow:

   • Return quotient after ensuring it's within the 32-bit signed integer range (INT_MIN to INT_MAX).

Time Complexity Analysis:

The time complexity is O(log n), where n is the absolute value of the dividend. This is because each iteration of the while loop effectively halves the remaining dividend (in the best case) due to the doubling of temp_divisor. The number of times you can halve a number before reaching 1 is approximately log₂(n).

Space Complexity Analysis:

The space complexity is O(1) because we only use a constant number of variables regardless of the input size.

Code Examples (Python):

def divide(dividend: int, divisor: int) -> int:
    if divisor == 0:
        raise ZeroDivisionError("Cannot divide by zero")
    if dividend == -2**31 and divisor == -1:
        return 2**31 - 1  # Handle potential overflow
 
    sign = (dividend < 0) ^ (divisor < 0)  # XOR for sign determination
    dividend = abs(dividend)
    divisor = abs(divisor)
 
    quotient = 0
    while dividend >= divisor:
        temp_divisor = divisor
        multiplier = 1
        while dividend >= temp_divisor << 1:  # Efficient doubling with bit shift
            temp_divisor <<= 1
            multiplier <<= 1
        dividend -= temp_divisor
        quotient += multiplier
 
    return quotient if not sign else -quotient
 

The other languages (Java, C++, Go, TypeScript, C#, PHP) would implement the same logic using their respective syntax and data types. The core bit-shifting optimization and iterative subtraction strategy remain the same across all languages.