# Solving Leetcode Interviews in Seconds with AI: Maximum OR


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2680" 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 a 0-indexed integer array nums of length n and an integer k. In an operation, you can choose an element and multiply it by 2. Return the maximum possible value of nums[0] | nums[1] | ... | nums[n - 1] that can be obtained after applying the operation on nums at most k times. Note that a | b denotes the bitwise or between two integers a and b.   Example 1:  Input: nums = [12,9], k = 1 Output: 30 Explanation: If we apply the operation to index 1, our new array nums will be equal to [12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.  Example 2:  Input: nums = [8,1,2], k = 2 Output: 35 Explanation: If we apply the operation twice on index 0, we yield a new array of [32,1,2]. Thus, we return 32|1|2 = 35.    Constraints:  1 <= nums.length <= 105 1 <= nums[i] <= 109 1 <= k <= 15  

	# Explanation
	*   **Maximize the most significant bits:** The bitwise OR operation is most influenced by the highest set bits. Therefore, the strategy is to apply the doubling operations to the element that will contribute the most high-order bits to the final result.
*   **Focus on the first element:** Because every element is combined using a bitwise OR, maximizing the first element has the most impact, as any bits set in the first element will definitely be in the final result. So we apply all k operations to `nums[0]`.
*   **Calculate and return the result:** After applying all the doubling operations to the first element, simply compute the bitwise OR of all elements in the modified array.

*   **Runtime Complexity:** O(n), where n is the length of the input array `nums`.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def solve():
    n, k = map(int, input().split())
    nums = list(map(int, input().split()))

    nums[0] *= (1 << k)

    result = 0
    for num in nums:
        result |= num

    print(result)

solve()
	```