Divide Two Integers
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# LeetCode #29: Divide Two Integers (C/C++).

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source: https://leetcode.com/problems/divide-two-integers/
C/C++ Solution to LeetCode problem 29. Divide Two Integers.

## Problem

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: [−231, 231 − 1]. For this problem, if the quotient is strictly greater than 231 - 1, then return 231 - 1, and if the quotient is strictly less than -231, then return -231.

## Examples

### 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

• -231 <= dividend, divisor <= 231 - 1
• divisor != 0

## Solution

• There are three cases that we can handle directly:
• Division by 1.
• Division of the minimum possible integer for a 32-bit signed integer.
• Divident and divisor are equal.
• Once handled those cases, we can proceed with the division. I can think of two possible solutions:
1. As long as the dividend >= divisor.
• Substract the divisor from the dividend.
• Count how many times has been substracted.
2. Using bitwise operations, we can count, with bigger steps, how many times we can substrac the divisor from the dividend. this is the optimal solution.
• As long as the dividen >= divisor
• As long as the divisor <= dividend
• Perform a left shift (same as multiplying by 2) with the divisor.
• We do a left shift to a counter that started at 1.
• Substract the final result of the divisor from the dividend.
• Add to our result the value of the counter we shifted too.
- Finally, we verify if the dividend < 0 XOR divisor < 0 to know if the result will be negative.

We need to prevent to go outside the range of a 32-bit integer, for that, we verify at every iteration if the INT_MAX - shifted_divisor <= shifted_divisor. Once this condition is true means that for next shifting the value will be equal or greather than the max value, so, we can prevent it.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 class Solution { public: int divide(int dividend, int divisor) { if (dividend == INT_MIN && divisor == 1) return INT_MIN; if (dividend == INT_MIN && divisor == -1) return INT_MAX; if (dividend == divisor) return (dividend < 0) != (divisor < 0) ? -1: 1; unsigned int a = abs(dividend); unsigned int b = abs(divisor); unsigned int tmp = 0; unsigned int quotient = 0; while (a >= b) { tmp = b; int i = 1; while(tmp << 1 <= a) { tmp <<= 1; i <<= 1; if (INT_MAX - tmp <= tmp) break; } a -= tmp; quotient += i; } return (dividend < 0) != (divisor < 0) ? -quotient : quotient; } };