What is the time complexity of Minimum Reverse Operations?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Reverse Operations?

This problem uses Array, Breadth-First Search, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Reverse Operations?

Insight 1: The problem can be modeled as a graph problem where each index is a node and a valid reverse operation represents an edge. The goal is to find the shortest path from 'p' to every other node. Insight 2: Since we need to find the shortest path, Breadth-First Search (BFS) is a suitable algorithm. The queue in BFS will store the positions of '1' that can be reached with a certain number of reverse operations.

Which companies ask LeetCode 2612. Minimum Reverse Operations in interviews?

This problem is frequently asked at Infosys.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Minimum Reverse Operations - LeetCode 2612 Solution

Minimum Reverse Operations - Complete Solution Guide

Minimum Reverse Operations is LeetCode problem 2612, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i th result is the minimum number of operations needed to bring the single 1 to position

Detailed Explanation

The problem presents an array of size 'n' initialized with zeros, except for the element at index 'p', which is set to one. We are given a set of banned indices. The goal is to find the minimum number of reverse operations needed to move the '1' from its initial position 'p' to every other position in the array. A reverse operation involves reversing a subarray of length 'k'. Crucially, the '1' cannot be moved to any index present in the banned array during any reverse operation. The output is an array where the i-th element is the minimum operations required to move the '1' to index 'i', or -1 if it's impossible.

Solution Approach

The solution utilizes Breadth-First Search (BFS) to find the minimum number of operations required to move the '1' to each possible index. It uses Segment Trees to efficiently track available (unvisited and not banned) even and odd indices. The approach works because BFS explores the graph level by level, guaranteeing the shortest path. Segment Trees enables efficient management of available positions during the exploration.

Step-by-Step Algorithm

Step 1: Initialize an answer array of size 'n' with all elements set to -1. Set answer[p] = 0 because the '1' is initially at position 'p' and takes 0 operations to reach.
Step 2: Create a queue and add the initial position 'p' to it.
Step 3: Create sets 'avail_even_indices' and 'avail_odd_indices' containing the available even and odd indices that are not banned or equal to p.
Step 4: Construct Segment Trees for even and odd indices using the corresponding available sets.
Step 5: Perform BFS. While the queue is not empty:
Step 6: Dequeue an element 'curr' from the queue, representing the current position of '1'.
Step 7: Determine the minimum and maximum starting positions (j_min, j_max) for reverse operations of length 'k' that can be performed with 'curr' as one of the endpoints.
Step 8: Calculate the corresponding range of positions (i_prime_min, i_prime_max) where the '1' can be moved to after the reverse operation based on j_min, j_max and k. The target positions are called i_prime.
Step 9: Determine the parity of the target positions (even or odd). Select the corresponding segment tree and the mapping function that converts from the index of the segment tree to the actual array index.
Step 10: Use the Segment Tree to find and remove the available positions within the calculated range [st_L, st_R] in the Segment Tree. Add the corresponding indices to segment tree to a list.
Step 11: For each index found in the Segment Tree, calculate the actual next position in the array. If this 'next_pos' hasn't been visited (answer[next_pos] == -1), update the answer array with the number of operations (d + 1) and enqueue 'next_pos'.
Step 12: Return the 'answer' array.

Key Insights

Insight 1: The problem can be modeled as a graph problem where each index is a node and a valid reverse operation represents an edge. The goal is to find the shortest path from 'p' to every other node.
Insight 2: Since we need to find the shortest path, Breadth-First Search (BFS) is a suitable algorithm. The queue in BFS will store the positions of '1' that can be reached with a certain number of reverse operations.
Insight 3: The available positions after each reverse operation depend on the length 'k' of the subarray being reversed. We need to efficiently determine which positions are valid and haven't been visited before. The parity of the indices matters significantly for reachable positions after reverse operations.
Insight 4: A Segment Tree can efficiently determine the available and unvisited nodes for each parity(even or odd). It provides methods to search and remove indices, thus preventing revisits to the same node.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Breadth-First Search, Ordered Set.

Companies

Asked at: Infosys.

Minimum Reverse Operations | LeetCode 2612

Hard

Array Breadth-First Search Ordered Set

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] minReverseOperations(int n, int p, int[] banned, int k) {
        class SegmentTree {
            int size;
            int[] tree;
            Set<Integer> availablePosSet;

            public SegmentTree(int size, Set<Integer> availablePosSet) {
                this.size = size;
                this.tree = new int[4 * size];
                this.availablePosSet = availablePosSet;
                if (this.size > 0) {
                    build(1, 0, this.size - 1);
                }
            }

            private void build(int v, int tl, int tr) {
                if (tl == tr) {
                    if (availablePosSet.contains(tl)) {
                        tree[v] = 1;
                    }
                } else {
                    int tm = (tl + tr) / 2;
                    build(v * 2, tl, tm);
                    build(v * 2 + 1, tm + 1, tr);
                    tree[v] = tree[v * 2] + tree[v * 2 + 1];
                }
            }

            public List<Integer> findAndRemove(int v, int tl, int tr, int l, int r) {
                if (l > r || tree[v] == 0) {
                    return new ArrayList<>();
                }

                if (tl == tr) {
                    tree[v] = 0;
                    return Collections.singletonList(tl);
                }

                List<Integer> foundIndices = new ArrayList<>();
                int tm = (tl + tr) / 2;

                if (l <= tm && tree[v * 2] > 0) {
                    foundIndices.addAll(findAndRemove(v * 2, tl, tm, l, r));
                }
                if (r > tm && tree[v * 2 + 1] > 0) {
                    foundIndices.addAll(findAndRemove(v * 2 + 1, tm + 1, tr, l, r));
                }

                tree[v] = tree[v * 2] + tree[v * 2 + 1];
                return foundIndices;
            }
        }

        Set<Integer> bannedSet = new HashSet<>();
        for (int b : banned) {
            bannedSet.add(b);
        }

        int[] answer = new int[n];
        Arrays.fill(answer, -1);
        answer[p] = 0;

        Queue<Integer> q = new LinkedList<>();
        q.offer(p);

        Set<Integer> availEvenIndices = new HashSet<>();
        for (int i = 0; i < n; i += 2) {
            if (i != p && !bannedSet.contains(i)) {
                availEvenIndices.add(i / 2);
            }
        }

        Set<Integer> availOddIndices = new HashSet<>();
        for (int i = 1; i < n; i += 2) {
            if (i != p && !bannedSet.contains(i)) {
                availOddIndices.add((i - 1) / 2);
            }
        }

        int sizeEven = (n + 1) / 2;
        int sizeOdd = n / 2;

        SegmentTree stEven = new SegmentTree(sizeEven, availEvenIndices);
        SegmentTree stOdd = new SegmentTree(sizeOdd, availOddIndices);

        while (!q.isEmpty()) {
            int curr = q.poll();
            int d = answer[curr];

            int jMin = Math.max(0, curr - k + 1);
            int jMax = Math.min(n - k, curr);

            int iPrimeMin = 2 * jMin + k - 1 - curr;
            int iPrimeMax = 2 * jMax + k - 1 - curr;

            int targetParity = (curr + k - 1) % 2;

            SegmentTree st;
            int stSize;
            java.util.function.Function<Integer, Integer> mapFn;
            int stL;
            int stR;

            if (targetParity == 0) {
                st = stEven;
                stSize = sizeEven;
                mapFn = x -> 2 * x;
                stL = (iPrimeMin + 1) / 2;
                stR = iPrimeMax / 2;
            } else {
                st = stOdd;
                stSize = sizeOdd;
                mapFn = x -> 2 * x + 1;
                stL = iPrimeMin / 2;
                stR = (iPrimeMax - 1) / 2;
            }

            if (stSize > 0 && stL <= stR) {
                List<Integer> stIndices = st.findAndRemove(1, 0, stSize - 1, stL, stR);
                for (int stIdx : stIndices) {
                    int nextPos = mapFn.apply(stIdx);
                    if (answer[nextPos] == -1) {
                        answer[nextPos] = d + 1;
                        q.offer(nextPos);
                    }
                }
            }
        }

        return answer;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2612: Minimum Reverse Operations - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Ordered Set

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2612

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Initialize an answer array of size 'n' with all elements set to -1. Set answer[p] = 0 because the '1' is initially at position 'p' and takes 0 operations to reach.
2
Step 2: Create a queue and add the initial position 'p' to it.
3
Step 3: Create sets 'avail_even_indices' and 'avail_odd_indices' containing the available even and odd indices that are not banned or equal to p.
4
Step 4: Construct Segment Trees for even and odd indices using the corresponding available sets.
5
Step 5: Perform BFS. While the queue is not empty:
6
Step 6: Dequeue an element 'curr' from the queue, representing the current position of '1'.
7
Step 7: Determine the minimum and maximum starting positions (j_min, j_max) for reverse operations of length 'k' that can be performed with 'curr' as one of the endpoints.
8
Step 8: Calculate the corresponding range of positions (i_prime_min, i_prime_max) where the '1' can be moved to after the reverse operation based on j_min, j_max and k. The target positions are called i_prime.
9
Step 9: Determine the parity of the target positions (even or odd). Select the corresponding segment tree and the mapping function that converts from the index of the segment tree to the actual array index.
10
Step 10: Use the Segment Tree to find and remove the available positions within the calculated range [st_L, st_R] in the Segment Tree. Add the corresponding indices to segment tree to a list.
11
Step 11: For each index found in the Segment Tree, calculate the actual next position in the array. If this 'next_pos' hasn't been visited (answer[next_pos] == -1), update the answer array with the number of operations (d + 1) and enqueue 'next_pos'.
12
Step 12: Return the 'answer' array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The BFS visits each node at most once, contributing O(n). The Segment Tree operations (find and remove) take O(log n) time each, and we might perform them for each node in the array in the worst case. Therefore, the dominant factor is O(n log n). The space complexity is O(n) due to the answer array, the banned set, and the Segment Tree.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force BFS

Instead of using Segment Trees to find the next available positions, we can iterate through all possible positions and check if they are valid (not banned, not visited).

Time: O(n^2)Space: O(n)

Trade-offs: This approach has a time complexity of O(n^2) because for each position we visit, we might iterate through all other positions to find the next available positions. This approach is less efficient than the Segment Tree approach and would likely result in a Time Limit Exceeded (TLE) error for larger input sizes.

Using Sorted Sets without Segment Tree

Instead of a segment tree, sorted sets (e.g., TreeSet in Java, set in C++) can be used to store the available even and odd positions. We can find a range within a sorted set in O(log n) and remove elements also in O(log n).

Time: O(n log n)Space: O(n)

Trade-offs: Using sorted sets is a more intuitive approach compared to segment trees, even though the time complexity remains the same. The overhead of segment tree might make sorted sets slightly faster in practice.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph problem where each index is a node and a valid reverse operation represents an edge. The goal is to find the shortest path from 'p' to every other node.
2Insight 2: Since we need to find the shortest path, Breadth-First Search (BFS) is a suitable algorithm. The queue in BFS will store the positions of '1' that can be reached with a certain number of reverse operations.
3Insight 3: The available positions after each reverse operation depend on the length 'k' of the subarray being reversed. We need to efficiently determine which positions are valid and haven't been visited before. The parity of the indices matters significantly for reachable positions after reverse operations.
4Insight 4: A Segment Tree can efficiently determine the available and unvisited nodes for each parity(even or odd). It provides methods to search and remove indices, thus preventing revisits to the same node.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: n = 1, The array has only one element which is '1' itself. The algorithm should return [0].
Edge case 2: k = n, It's possible to reverse the whole array. The algorithm should consider this reverse operation to find possible next positions.
Edge case 3: banned array contains all positions except 'p'. Then, no reverse operations are possible and the output should be [-1, -1, ..., 0, ...,-1].
Edge case 4: k = 1. Each reverse operation keeps the '1' at its original position. If 'p' is not banned the output is [-1,-1, ..., 0, ..., -1], otherwise, all elements are -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the banned positions. Any reverse operation should avoid placing '1' on a banned position.

✗Common mistake 2: Not considering the parity (even or odd) of indices, which is critical in deciding valid reverse operations.

✗Common mistake 3: Incorrectly calculating the starting and ending positions (j_min, j_max) for reverse operations and the corresponding positions to which '1' can be moved.

✗Common mistake 4: Not checking if the next position has already been visited before adding it to the queue. This would lead to an infinite loop or incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in a maze with certain blocked cells. The maze can be represented as a grid, and the goal is to find the shortest path from a starting cell to an ending cell, avoiding the blocked cells. The reverse operations are analogous to movements in specific directions.

🏦Real-world scenario 2: Optimizing data rearrangement. Suppose we have a database and certain records need to be rearranged to optimize performance. The reverse operation can be seen as a specific data rearrangement procedure, and the goal is to find the minimum number of rearrangements needed to achieve the desired data layout.

🎮Real-world scenario 3: Robot Navigation: A robot navigating in a constrained environment where it can only perform specific rotation maneuvers (reverse operations). The problem is to find the minimum number of maneuvers to reach a target location.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Reverse Operations"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?