Divide Two Integers LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

Divide Two Integers LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++

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In this post, you will find the solution for the Divide Two Integers in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemDivide 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: [−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
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

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