Minimum Reverse Operations - Video Solutions

Problem Overview: You start at position p in an array of length n. In one move, you can reverse any subarray of length k. Some indices are banned and cannot be visited. The task is to compute the minimum number of reverse operations needed to reach every index, or -1 if it is impossible.

Approach 1: Breadth-First Search with Ordered Set (O(n log n) time, O(n) space)

The state space is the index of the position after each reversal. Treat each index as a node in a graph and run Breadth-First Search from the starting position p. The challenge is efficiently finding all indices reachable by reversing a length-k subarray containing the current index. Reversing maps index i to l + r - i where [l, r] is the chosen window.

Instead of checking every possible window, compute the valid range of reachable indices mathematically. All reachable targets form a sequence with alternating parity. Maintain two ordered sets (even and odd indices) containing unvisited and non-banned positions. During BFS, query the ordered set for indices within the valid range and remove them once visited. This pruning prevents repeated scanning and keeps transitions efficient.

Each index is processed once and removed from the set once, while range queries cost O(log n). BFS guarantees the first time you visit an index is the minimum number of operations. This approach scales well even when n is large because it avoids enumerating all possible reversals explicitly.

Approach 2: Dynamic Programming Simulation (O(n * k) time, O(n) space)

A more direct strategy simulates transitions by evaluating all length-k subarrays that include the current index. For each valid window [l, r], compute the resulting index after reversal and update the minimum number of operations using dynamic programming or BFS-style relaxation. The DP array stores the best known move count for every position.

This approach is easier to reason about because it mirrors the problem statement: iterate over possible windows and apply the reversal mapping. However, each index may examine up to k candidate windows, which leads to O(n * k) time in the worst case. With large constraints, this becomes too slow compared to the ordered-set BFS optimization.

Recommended for interviews: The BFS with ordered sets is the expected solution. Interviewers want to see that you model the problem as a graph traversal, derive the reversal index formula, and optimize neighbor discovery using parity grouping and ordered sets. A brute-force or simulation approach demonstrates understanding of the mechanics, but the optimized BFS shows strong algorithmic reasoning with array transformations and efficient range queries.