# Solving Leetcode Interviews in Seconds with AI: Minimum Reverse Operations


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2612" 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 an integer n and an integer p representing an array arr of length n where all elements are set to 0's, except position p which is set to 1. You are also given an integer array banned containing restricted positions. Perform the following operation on arr:  Reverse a subarray with size k if the single 1 is not set to a position in banned.  Return an integer array answer with n results where the ith result is the minimum number of operations needed to bring the single 1 to position i in arr, or -1 if it is impossible.   Example 1:  Input: n = 4, p = 0, banned = [1,2], k = 4 Output: [0,-1,-1,1] Explanation:  Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0. We can never place 1 on the banned positions, so the answer for positions 1 and 2 is -1. Perform the operation of size 4 to reverse the whole array. After a single operation 1 is at position 3 so the answer for position 3 is 1.   Example 2:  Input: n = 5, p = 0, banned = [2,4], k = 3 Output: [0,-1,-1,-1,-1] Explanation:  Initially 1 is placed at position 0 so the number of operations we need for position 0 is 0. We cannot perform the operation on the subarray positions [0, 2] because position 2 is in banned. Because 1 cannot be set at position 2, it is impossible to set 1 at other positions in more operations.   Example 3:  Input: n = 4, p = 2, banned = [0,1,3], k = 1 Output: [-1,-1,0,-1] Explanation: Perform operations of size 1 and 1 never changes its position.    Constraints:  1 <= n <= 105 0 <= p <= n - 1 0 <= banned.length <= n - 1 0 <= banned[i] <= n - 1 1 <= k <= n  banned[i] != p all values in banned are unique   

	# Explanation
	Here's a breakdown of the solution:

*   **BFS Approach:** Use Breadth-First Search (BFS) to explore possible positions of the '1' after each subarray reversal. The starting position is 'p'.
*   **Banned Set:** Maintain a set of banned positions for quick lookup to avoid placing the '1' at those locations.
*   **Visited Set:** Track visited positions to prevent cycles in the BFS and ensure we find the *minimum* number of operations.

*   **Time Complexity: O(n), Space Complexity: O(n)** (due to BFS queue and visited/banned sets).

	
	# Code
	```python
	from collections import deque

def minOperations(n: int, p: int, banned: list[int], k: int) -> list[int]:
    """
    Calculates the minimum operations to move a '1' in an array to each position.

    Args:
        n: The size of the array.
        p: The initial position of the '1'.
        banned: A list of banned positions.
        k: The size of the subarray to reverse.

    Returns:
        A list of minimum operations for each position, or -1 if impossible.
    """

    banned_set = set(banned)
    visited = set()
    distances = [-1] * n
    distances[p] = 0

    queue = deque([(p, 0)])  # (position, operations)
    visited.add(p)

    while queue:
        curr_pos, curr_ops = queue.popleft()

        # Iterate through possible start positions of the subarray to reverse
        for start in range(max(0, curr_pos - (k -1)), min(n - k + 1, curr_pos+1)):
            end = start + k - 1
            
            new_pos = start + end - curr_pos

            if 0 <= new_pos < n and new_pos not in banned_set and new_pos not in visited:
                distances[new_pos] = curr_ops + 1
                visited.add(new_pos)
                queue.append((new_pos, curr_ops + 1))

    return distances
	```