# Solving Leetcode Interviews in Seconds with AI: Minimum Array End


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3133" 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
	> You are given two integers n and x. You have to construct an array of positive integers nums of size n where for every 0 <= i < n - 1, nums[i + 1] is greater than nums[i], and the result of the bitwise AND operation between all elements of nums is x. Return the minimum possible value of nums[n - 1].   Example 1:  Input: n = 3, x = 4 Output: 6 Explanation: nums can be [4,5,6] and its last element is 6.  Example 2:  Input: n = 2, x = 7 Output: 15 Explanation: nums can be [7,15] and its last element is 15.    Constraints:  1 <= n, x <= 108  

	# Explanation
	Here's the breakdown of the solution:

*   **Understanding the AND constraint:** All numbers in the array must have `x` as a common bitwise AND result. This means `x` must be a subset of the bits present in *every* number in `nums`. Therefore, every number in `nums` must have all the bits set in `x`.
*   **Minimizing the last element:** To minimize `nums[n-1]`, we must minimize all the preceding elements. We start with `x` as the first element, which is the smallest possible value that satisfies the bitwise AND condition. Then, to make each subsequent element greater than the previous and still keep `x` as their bitwise AND, we simply increment by 1. This results in the sequence `x, x+1, x+2, ..., x+n-1`. Thus, the last element would be `x + n - 1`.

*   **Runtime & Storage Complexity:** O(1) runtime, O(1) storage.

	
	# Code
	```python
	def solve():
    n, x = map(int, input().split())
    print(x + n - 1)

solve()
	```