Add divide two integers algorithm

bit_manipulation/divide_two_integers.py

@@ -0,0 +1,67 @@
+class Solution:
+    def divide(self, dividend: int, divisor: int) -> int:
+        """
+        Divide two integers without using *, /, or % operators.
+
+        LeetCode Problem: https://leetcode.com/problems/divide-two-integers/
+
+        Args:
+            dividend (int): The number to be divided.
+            divisor (int): The number to divide by.
+
+        Returns:
+            int: The quotient after truncating toward zero.
+
+        Examples:
+        >>> sol = Solution()
+        >>> sol.divide(43, 8)
+        5
+        >>> sol.divide(10, 3)
+        3
+        >>> sol.divide(-7, 2)
+        -3
+        >>> sol.divide(1, 1)
+        1
+        >>> sol.divide(-2147483648, -1)
+        2147483647
+        >>> sol.divide(24, 8)
+        3
+        >>> sol.divide(43, -8)
+        -5
+        """
+        # Edge case: if dividend equals divisor, result is 1
+        if dividend == divisor:
+            return 1
+
+        # Determine the sign of the result
+        sign = not ((dividend < 0) ^ (divisor < 0))  # True if same sign,
+
+        # Convert both numbers to positive
+        d = abs(dividend)  # remaining dividend
+        n = abs(divisor)  # divisor
+        quo = 0  # quotient
+
+        # Outer loop: subtract divisor multiples from dividend
+        while d >= n:
+            cnt = 0
+            # find largest power-of-two multiple of divisor that fits in dividend
+            while d >= (n << (cnt + 1)):
+                cnt += 1
+
+            # Add this power-of-two chunk to quotient
+            quo += 1 << cnt
+
+            # Subtract the chunk from remaining dividend
+            d -= n << cnt
+
+        # Handle overflow for 32-bit signed integers
+        if quo == (1 << 31):
+            return (2**31 - 1) if sign else -(2**31)
+
+        return quo if sign else -quo
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()