LeetCode #2612 — HARD

Minimum Reverse Operations

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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.

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 
Patterns Used
🔗Union-Find→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Union-Find · Segment Tree

Example 1

4
0
[1,2]
4

Example 2

5
0
[2,4]
3

Example 3

4
2
[0,1,3]
1
Step 02

Core Insight

What unlocks the optimal approach

  • Can we use a breadth-first search to find the minimum number of operations?
  • Find the beginning and end indices of the subarray of size k that can be reversed to bring 1 to a particular position.
  • Can we visit every index or do we need to consider the parity of k?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2612: Minimum Reverse Operations
class Solution {
    public int[] minReverseOperations(int n, int p, int[] banned, int k) {
        int[] ans = new int[n];
        TreeSet<Integer>[] ts = new TreeSet[] {new TreeSet<>(), new TreeSet<>()};
        for (int i = 0; i < n; ++i) {
            ts[i % 2].add(i);
            ans[i] = i == p ? 0 : -1;
        }
        ts[p % 2].remove(p);
        for (int i : banned) {
            ts[i % 2].remove(i);
        }
        ts[0].add(n);
        ts[1].add(n);
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(p);
        while (!q.isEmpty()) {
            int i = q.poll();
            int mi = Math.max(i - k + 1, k - i - 1);
            int mx = Math.min(i + k - 1, n * 2 - k - i - 1);
            var s = ts[mi % 2];
            for (int j = s.ceiling(mi); j <= mx; j = s.ceiling(mi)) {
                q.offer(j);
                ans[j] = ans[i] + 1;
                s.remove(j);
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND
O(α(n)) time
O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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