Divide Two Integers LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [๐Ÿ’ฏCorrect]

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Link for the ProblemDivide Two Integersโ€“ LeetCode Problem

Divide Two Integersโ€“ LeetCode 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: [โˆ’231, 231 โˆ’ 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.


  • -231 <= dividend, divisor <= 231 - 1
  • divisor != 0
Divide Two Integers
โ€“ LeetCode Solutions
class Solution {
  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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