#2612

Hard

auto_awesomeArray plus Breadth-First Search

LeetCode Problem Workspace

Minimum Reverse Operations

Find the minimum number of operations to move a 1 in an array from a given start position to other positions, considering restricted positions.

array breadth first search ordered set

Problem Statement

You are given an array of length n, initialized to all zeros, with a 1 placed at a given position p. You are also given an array banned that contains restricted positions where you cannot move the 1. Your task is to determine the minimum number of operations required to move the 1 from its initial position to each position in the array, while avoiding the banned positions. If it is impossible to reach a position, return -1 for that position.

The output should be an array of integers where the ith element represents the minimum number of operations to bring the 1 to position i. If a position is unreachable, the value for that position should be -1.

Examples

Example 1

Input: n = 4, p = 0, banned = [1,2], k = 4

Output: [0,-1,-1,1]

Example 2

Input: n = 5, p = 0, banned = [2,4], k = 3

Output: [0,-1,-1,-1,-1]

Example 3

Input: n = 4, p = 2, banned = [0,1,3], k = 1

Output: [-1,-1,0,-1]

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

Solution Approach

Breadth-First Search (BFS) for Efficient Exploration

Use BFS to explore all possible positions the 1 can reach. Start from the initial position and explore all adjacent, valid positions by performing operations. BFS ensures the minimum number of operations are used to reach each position because it explores all positions level by level.

Avoid Banned Positions

During BFS, skip positions listed in the banned array to ensure that the 1 does not move to restricted positions. These positions are blocked off and must be accounted for as obstacles in the search process.

Edge Case Handling

Consider edge cases where the 1 is already at a position or when all positions are blocked except the starting one. Additionally, if there is no valid path to a position due to restrictions, return -1.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the breadth-first search, which explores each valid position once. Therefore, the time complexity is O(n) where n is the size of the array. Space complexity is also O(n) due to the storage of the visited nodes and the queue used in BFS.

What Interviewers Usually Probe

Understanding BFS and how it guarantees minimum operations is key.
Watch for how the candidate handles blocked positions and restrictions.
Evaluate if the candidate can optimize the search process and handle edge cases efficiently.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle banned positions correctly, leading to invalid solutions.
Assuming the BFS will always find a valid path, without considering isolated or blocked positions.
Not handling edge cases, such as when the starting position is the only valid position.

Follow-up variants

Change the operation size k to affect how the 1 can move, adding more complexity to the problem.
Introduce multiple 1s in the array and calculate the minimum operations to bring all 1s to every position.
Consider dynamic constraints where the banned positions can change during the operation sequence.

FAQ

How do I use BFS to solve the Minimum Reverse Operations problem?

BFS is used to explore each position in the array from the starting position, ensuring the minimum number of operations are counted to reach each position, while avoiding banned ones.

What is the time complexity of the Minimum Reverse Operations problem?

The time complexity is O(n), where n is the size of the array, due to the BFS traversal of all valid positions.

Can the solution handle cases with multiple blocked positions?

Yes, the solution accounts for blocked positions using the banned array and ensures that BFS avoids these positions.

How do I handle the edge case where no valid moves are possible?

If no valid moves are possible to a particular position, return -1 for that position, indicating it is unreachable.

What is the best way to optimize the solution for larger arrays?

The current BFS solution is efficient with O(n) time complexity, but optimization could focus on pruning unnecessary checks or dynamically adjusting the BFS search depending on input constraints.

terminal

Solution

Solution 1: Ordered Set + BFS

We notice that for any index $i$ in the subarray interval $[l,..r]$, the flipped index $j = l + r - i$.

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class Solution:
    def minReverseOperations(
        self, n: int, p: int, banned: List[int], k: int
    ) -> List[int]:
        ans = [-1] * n
        ans[p] = 0
        ts = [SortedSet() for _ in range(2)]
        for i in range(n):
            ts[i % 2].add(i)
        ts[p % 2].remove(p)
        for i in banned:
            ts[i % 2].remove(i)
        ts[0].add(n)
        ts[1].add(n)
        q = deque([p])
        while q:
            i = q.popleft()
            mi = max(i - k + 1, k - i - 1)
            mx = min(i + k - 1, n * 2 - k - i - 1)
            s = ts[mi % 2]
            j = s.bisect_left(mi)
            while s[j] <= mx:
                q.append(s[j])
                ans[s[j]] = ans[i] + 1
                s.remove(s[j])
                j = s.bisect_left(mi)
        return ans

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