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** Link for the Problem** – Divide Two Integers– LeetCode Problem

Divide Two Integers– LeetCode Problem

**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: `[−2`

. For this problem, if the quotient is ^{31}, 2^{31} − 1]**strictly greater than** `2`

, then return ^{31} - 1`2`

, and if the quotient is ^{31} - 1**strictly less than** `-2`

, then return ^{31}`-2`

.^{31}

**Example 1:**

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

**Example 2:**

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

**Constraints:**

`-2`

^{31}<= dividend, divisor <= 2^{31}- 1`divisor != 0`

Divide Two Integers – LeetCode Solutions

class Solution { public: int divide(int dividend, int divisor) { // -2^{31} / -1 = 2^31 -> overflow so return 2^31 - 1 if (dividend == INT_MIN && divisor == -1) return INT_MAX; const int sign = dividend > 0 ^ divisor > 0 ? -1 : 1; long ans = 0; long dvd = labs(dividend); long dvs = labs(divisor); while (dvd >= dvs) { long k = 1; while (k * 2 * dvs <= dvd) k *= 2; dvd -= k * dvs; ans += k; } return sign * ans; } };

class Solution { public int divide(long dividend, long divisor) { // -2^{31} / -1 = 2^31 -> overflow so return 2^31 - 1 if (dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE; final int sign = dividend > 0 ^ divisor > 0 ? -1 : 1; long ans = 0; long dvd = Math.abs(dividend); long dvs = Math.abs(divisor); while (dvd >= dvs) { long k = 1; while (k * 2 * dvs <= dvd) k *= 2; dvd -= k * dvs; ans += k; } return sign * (int) ans; } }

class Solution: def divide(self, dividend: int, divisor: int) -> int: if dividend == -2**31 and divisor == -1: return 2**31 - 1 sign = -1 if (dividend > 0) ^ (divisor > 0) else 1 ans = 0 dvd = abs(dividend) dvs = abs(divisor) while dvd >= dvs: k = 1 while k * 2 * dvs <= dvd: k <<= 1 dvd -= k * dvs ans += k return sign * ans