Divide Two Integers Leetcode Solution

In this post, we are going to solve the Divide Two Integers Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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.

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

Now, let’s see the leetcode solution of Divide Two Integers Leetcode Solution.

Divide Two Integers Leetcode Solution in Python

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

Divide Two Integers Leetcode Solution in CPP

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;
  }
};

Divide Two Integers Leetcode Solution in Java

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;
  }
}

Note: This problem Divide Two Integers is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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