How do you solve Minimum Reverse Operations on LeetCode?

Minimum Reverse Operations (LeetCode #2612) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Banned positions restrict movement, making some targets unreachable.

What is the brute force solution for Minimum Reverse Operations?

The brute force approach for Minimum Reverse Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will try to simulate every possible operation of reversing subarrays of size k and check if we can move the 1 to each position in the array. This method is straightforward but inefficient due to its repetitive nature. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Reverse Operations?

Minimum Reverse Operations uses the following concepts: Array, Hash Table, Breadth-First Search, Union-Find, Ordered Set. The recommended patterns to study are: Breadth-First Search, Graph Traversal.

What are the interview tips for Minimum Reverse Operations?

Always clarify edge cases, like what happens if k is larger than n. Discuss your thought process while coding to show your understanding. Test your solution with different inputs to ensure correctness.

What are common mistakes when solving Minimum Reverse Operations?

Not considering banned positions when reversing subarrays. Failing to mark positions as visited, leading to infinite loops.

#2612

Minimum Reverse Operations

Hard

ArrayHash TableBreadth-First SearchUnion-FindOrdered SetBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a breadth-first search (BFS) approach allows us to explore all possible positions of the 1 efficiently. We can treat each position as a node and the valid operations as edges, ensuring we find the shortest path to each position.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue for BFS and a visited set to track positions we've already processed.
2Step 2: Start from position p, mark it as visited, and enqueue it with 0 operations.
3Step 3: While the queue is not empty, dequeue a position and check all possible positions reachable by reversing subarrays of size k.
4Step 4: If a reachable position is not banned or visited, mark it, record the operation count, and enqueue it.

solution.py21 lines

1from collections import deque
2
3def minReverseOperations(n, p, banned, k):
4    banned_set = set(banned)
5    answer = [-1] * n
6    answer[p] = 0
7    queue = deque([p])
8    visited = set([p])
9
10    while queue:
11        current = queue.popleft()
12        for i in range(max(0, current - k + 1), min(n - k + 1, current + 1)):
13            if any(pos in banned_set for pos in range(i, i + k)):
14                continue
15            next_pos = i + k - 1 - (current - i)
16            if next_pos < 0 or next_pos >= n or next_pos in visited:
17                continue
18            visited.add(next_pos)
19            answer[next_pos] = answer[current] + 1
20            queue.append(next_pos)
21    return answer

ℹ

Complexity note: The BFS explores each position at most once, leading to linear time complexity. The space complexity is due to the queue and visited set.

1Banned positions restrict movement, making some targets unreachable.
2Reversing a subarray can significantly change the position of the 1.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.