# Solving Leetcode Interviews in Seconds with AI: Minimum Impossible OR


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2568" 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. We say that an integer x is expressible from nums if there exist some integers 0 <= index1 < index2 < ... < indexk < nums.length for which nums[index1] | nums[index2] | ... | nums[indexk] = x. In other words, an integer is expressible if it can be written as the bitwise OR of some subsequence of nums. Return the minimum positive non-zero integer that is not expressible from nums.   Example 1:  Input: nums = [2,1] Output: 4 Explanation: 1 and 2 are already present in the array. We know that 3 is expressible, since nums[0] | nums[1] = 2 | 1 = 3. Since 4 is not expressible, we return 4.  Example 2:  Input: nums = [5,3,2] Output: 1 Explanation: We can show that 1 is the smallest number that is not expressible.    Constraints:  1 <= nums.length <= 105 1 <= nums[i] <= 109  

	# Explanation
	Here's an efficient solution to find the minimum positive non-zero integer not expressible from the bitwise OR of subsequences of a given array.

*   **Greedy Approach:**  Iteratively build expressible numbers from smallest to largest. If we can express all numbers from 1 to `k`, and `x` is the smallest number in `nums` that is greater than `k` and not yet used, then the next non-expressible number is `k+1`. If no such `x` exists, then `k+1` is the answer.

*   **Sorting:** Sort the input array in ascending order. This makes it easier to greedily determine the next smallest expressible number.

*   **Iterative Calculation:** Maintain the largest expressible number seen so far. Iterate through the sorted array. If an element is less than or equal to the current largest expressible number, then update the largest expressible number by taking the OR. Otherwise we've skipped a number.

*   **Complexity:**
    *   Runtime Complexity: O(N log N) due to sorting.
    *   Storage Complexity: O(1). (Sorting can be done in place by some implementations)

	
	# Code
	```python
	def smallest_non_expressible(nums):
    nums.sort()
    expressible = 0
    for num in nums:
        if num <= expressible + 1:
            expressible |= num
        else:
            break
    return expressible + 1
	```