29. Divide Two Integers LeetCode Solution

In this guide we will provide 29. Divide Two Integers LeetCode Solution with best time and space complexity. The solution to Divide Two Integers problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Problem Statement

Complexity Analysis

Divide Two Integers solution in C++
Divide Two Integers soution in Java
Divide Two Integers solution Python
Additional Resources

29. Divide Two Integers LeetCode Solution image

Problem Statement of 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: [−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

Complexity Analysis

Time Complexity: O(\log^2 n)
Space Complexity: O(1)

29. Divide Two Integers LeetCode Solution in C++

class Solution {
 public:
  int divide(int dividend, int divisor) {
    // -2^{31} / -1 = 2^31 will 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;
  }
};
/* code provided by PROGIEZ */

29. Divide Two Integers LeetCode Solution in Java

class Solution {
  public int divide(long dividend, long divisor) {
    // -2^{31} / -1 = 2^31 will 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;
  }
}
// code provided by PROGIEZ

29. Divide Two Integers LeetCode Solution in Python

class Solution:
  def divide(self, dividend: int, divisor: int) -> int:
    # -2^{31} / -1 = 2^31 will overflow, so return 2^31 - 1.
    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
 #code by PROGIEZ