LeetCode //C - 29. Divide Two Integers

29. Divide Two Integers

Given two integers dividend and divisor, divide two integers without using multiplication, division, and mod operator.

The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8, and -2.7335 would be truncated to -2.

Return the quotient after dividing dividend by divisor.

Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [ − 2 31 , 2 31 − 1 −2^{31}, 2^{31} − 1 −231,231−1]. For this problem, if the quotient is strictly greater than 2 31 − 1 2^{31} - 1 231−1, then return 2 31 − 1 2^{31} - 1 231−1, and if the quotient is strictly less than − 2 31 -2^{31} −231, then return − 2 31 -2^{31} −231.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Explanation: 10/3 = 3.33333... which is truncated to 3.

Example 2:

Input: dividend = 7, divisor = -3
Output: -2
Explanation: 7/-3 = -2.33333... which is truncated to -2.

Constraints:

• − 2 31 < = d i v i d e n d , d i v i s o r < = 2 31 − 1 -2^{31} <= dividend, divisor <= 2^{31} - 1 −231<=dividend,divisor<=231−1
• divisor != 0

From: LeetCode

Link: 29. Divide Two Integers

---

Solution:

Ideas:

1. Handle Signs: Compute the sign of the result based on the signs of dividend and divisor.
2. Use Absolute Values: Since we're working with possible negative numbers, we convert everything into positive values for ease of calculation.
3. Exponential Search: Use a method to subtract bigger chunks of the divisor from the dividend by shifting the divisor left (equivalent to multiplying the divisor by powers of 2) until we overshoot the dividend.
4. Subtraction and Accumulation: Once the optimal subtraction chunk is identified, subtract it from the dividend and accumulate the count of how many times this chunk fits into the dividend.
5. Overflow Handling: Make sure the result does not exceed the bounds defined by 32-bit integers.

Code:

int divide(int dividend, int divisor) {
    // Edge cases for overflow
    if (dividend == INT_MIN && divisor == -1) {
        return INT_MAX;
    }

    // Determine the sign of the result
    int sign = (dividend > 0) ^ (divisor > 0) ? -1 : 1;

    // Work with absolute values to avoid overflow issues
    long long ldividend = llabs((long long)dividend);
    long long ldivisor = llabs((long long)divisor);

    long long result = 0;

    // Perform the division using bit manipulation and subtraction
    while (ldividend >= ldivisor) {
        long long temp = ldivisor, multiple = 1;
        while (ldividend >= (temp << 1)) {
            temp <<= 1;
            multiple <<= 1;
        }
        ldividend -= temp;
        result += multiple;
    }

    // Apply the sign to the result
    result = sign * result;

    // Handle potential overflow
    if (result > INT_MAX) return INT_MAX;
    if (result < INT_MIN) return INT_MIN;

    return (int)result;
}