Divide Two Integers LeetCode Solution

Problem – 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

Divide Two Integers LeetCode Solution in C++

    int divide(int A, int B) {
        if (A == INT_MIN && B == -1) return INT_MAX;
        int a = abs(A), b = abs(B), res = 0, x = 0;
        while (a - b >= 0) {
            for (x = 0; a - (b << x << 1) >= 0; x++);
            res += 1 << x;
            a -= b << x;
        }
        return (A > 0) == (B > 0) ? res : -res;
    }

Divide Two Integers LeetCode Solution in Java

    public int divide(int A, int B) {
        if (A == 1 << 31 && B == -1) return (1 << 31) - 1;
        int a = Math.abs(A), b = Math.abs(B), res = 0, x = 0;
        while (a - b >= 0) {
            for (x = 0; a - (b << x << 1) >= 0; x++);
            res += 1 << x;
            a -= b << x;
        }
        return (A > 0) == (B > 0) ? res : -res;
    }

Divide Two Integers LeetCode Solution in Python

    def divide(self, A, B):
        if (A == -2147483648 and B == -1): return 2147483647
        a, b, res = abs(A), abs(B), 0
        for x in range(32)[::-1]:
            if (a >> x) - b >= 0:
                res += 1 << x
                a -= b << x
        return res if (A > 0) == (B > 0) else -res
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