Is Minimum Reverse Operations easy or hard?

Minimum Reverse Operations is classified as a Hard problem with a low acceptance rate around 16%. The difficulty comes from deriving the index transformation after reversal and optimizing neighbor discovery so the BFS does not scan every possible subarray.

Minimum Reverse Operations Python/Java solution

Python implementations commonly use BFS with sorted containers or bisect-based structures, while Java solutions often rely on TreeSet for ordered index retrieval. Both follow the same idea: compute the reachable index range, query the ordered set for candidates, and push them into the BFS queue with distance +1.

How to solve Minimum Reverse Operations in O(n)?

A near-linear solution comes from modeling the problem as BFS and deriving the range of indices reachable after reversing a window containing the current position. By grouping indices by parity and removing them from ordered sets once visited, you avoid revisiting states and keep the total operations close to linear, typically implemented as O(n log n).

What is the best approach for Minimum Reverse Operations?

The most efficient approach uses Breadth-First Search combined with ordered sets. BFS guarantees minimum operations, while ordered sets allow you to quickly retrieve unvisited indices within the reachable range created by reversing a k-length subarray. By separating indices into even and odd sets, you avoid scanning the entire array and achieve about O(n log n) time complexity.

Is Minimum Reverse Operations asked at Google/Amazon/Meta?

Hard BFS and graph-transformation problems like this frequently appear in interviews at companies such as Google, Meta, and Amazon. The problem tests graph modeling, mathematical reasoning about array reversals, and efficient data structures like ordered sets or balanced trees.

What data structure is used in Minimum Reverse Operations?

The optimal solution uses a queue for BFS and ordered sets (often implemented with TreeSet, SortedSet, or balanced BST structures). Two sets are typically maintained for even and odd indices to efficiently fetch valid unvisited positions within a computed range.

What is the time complexity of Minimum Reverse Operations?

The optimized solution runs in O(n log n) time and O(n) space. Each index is inserted and removed from an ordered set at most once, and BFS processes each node a single time. Range queries on the ordered set add the logarithmic factor.

Description Hints 3 Solution Video 24:16

2612. Minimum Reverse Operations

Hard17.5% acceptance1 company

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 i^th 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

Asked at 1 company

Infosys

Topics

Hints open one at a time

Hint 1

Can we use a breadth-first search to find the minimum number of operations?

Hint 2

Find the beginning and end indices of the subarray of size k that can be reversed to bring 1 to a particular position.

Hint 3

Can we visit every index or do we need to consider the parity of k?

24:16

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

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1class Solution:
2    def minReverseOperations(self, n: int, p: int, banned: List[int], k: int) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

4
0
[1,2]
4

Input

5
0
[2,4]
3

Input

4
2
[0,1,3]
1

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