# Solving Leetcode Interviews in Seconds with AI: Bitwise AND of Numbers Range


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "201" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given two integers left and right that represent the range [left, right], return the bitwise AND of all numbers in this range, inclusive.   Example 1:  Input: left = 5, right = 7 Output: 4  Example 2:  Input: left = 0, right = 0 Output: 0  Example 3:  Input: left = 1, right = 2147483647 Output: 0    Constraints:  0 <= left <= right <= 231 - 1  

	# Explanation
	Here's the breakdown:

*   **Find the Common Prefix:** The bitwise AND of a range of numbers will result in a number that has a common prefix in the binary representation of all numbers within that range. This prefix is determined by the leftmost bits that are the same for both `left` and `right`.

*   **Right Shift and Count:** We can efficiently find this common prefix by repeatedly right-shifting both `left` and `right` until they become equal. The number of right shifts tells us how many bits we need to shift back left to get the result.

*   **Shift Back:** Finally, left-shift the now-equal number (which represents the common prefix) by the shift count.

*   **Complexity:** Time Complexity: O(log(n)), where n is the range size. Space Complexity: O(1).

	
	# Code
	```python
	def rangeBitwiseAnd(left: int, right: int) -> int:
    """
    Calculates the bitwise AND of all numbers in the range [left, right], inclusive.

    Args:
        left: The left endpoint of the range.
        right: The right endpoint of the range.

    Returns:
        The bitwise AND of all numbers in the range.
    """
    shift = 0
    while left != right:
        left >>= 1
        right >>= 1
        shift += 1
    return left << shift
	```