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Minimum Reverse Operations

Difficulty: Hard

Problem Description

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.

Key Insights

  • The starting position of the 1 is given by p, and it can only be moved if the reverse operation does not involve banned positions.
  • If the current position of the 1 is adjacent to a banned position, it may limit the possible moves.
  • The breadth-first search (BFS) algorithm can be applied to explore all possible positions the 1 can reach while keeping track of the number of operations needed.
  • Each position can be reached with a minimum number of moves or may be unreachable (resulting in -1).

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(n)

Solution

To solve the problem, we can utilize a breadth-first search (BFS) approach. We start from the initial position p and explore all possible positions we can move the 1 to by reversing subarrays of size k. We maintain a queue to track the current position and the number of moves taken. We also use a set to keep track of banned positions to ensure we do not attempt to move the 1 to those locations. The BFS will explore both left and right directions depending on the size of k, and we'll update the results as we go.

Code Solutions

from collections import deque

def minReverseOperations(n, p, banned, k):
    banned_set = set(banned)
    answer = [-1] * n
    answer[p] = 0
    queue = deque([p])
    
    while queue:
        current = queue.popleft()
        current_moves = answer[current]
        
        # Try to move left
        left = current - k + 1
        if left >= 0 and left not in banned_set and answer[left] == -1:
            answer[left] = current_moves + 1
            queue.append(left)
        
        # Try to move right
        right = current + k - 1
        if right < n and right not in banned_set and answer[right] == -1:
            answer[right] = current_moves + 1
            queue.append(right)

    return answer
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