What is the time complexity of Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

The optimal solution has a time complexity of O((m*n)*2^(m*n)).

What data structures are used to solve Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

This problem uses Array, Hash Table, Bit Manipulation, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

Insight 1: Since the matrix size is small, we can represent the entire matrix state as a single integer using bit manipulation. Each bit in the integer represents a cell in the matrix. This allows for efficient state representation and comparisons. Insight 2: The problem can be modeled as a graph search problem, where each state of the matrix is a node, and flipping a cell and its neighbors represents an edge connecting two states. Breadth-First Search (BFS) is suitable here to find the shortest path (minimum number of flips) to the zero matrix state.

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Minimum Number of Flips to Convert Binary Matrix to Zero Matrix - LeetCode 1284 Solution

Q: How to solve Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

The solution employs Breadth-First Search (BFS) to explore the possible states of the matrix. The initial matrix state is converted to an integer representation. The BFS algorithm starts from this initial state and explores neighboring states by flipping each cell and its neighbors. A queue is used to maintain the states to be explored, and a set is used to keep track of visited states. The algorithm continues until the zero matrix state is found or the queue is empty (meaning no solution exists).

Q: What is the space complexity of Minimum Number of Flips to Convert Binary Matrix to Zero Matrix?

The space complexity is O(2^(m*n)).

Q: Which companies ask LeetCode 1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix in interviews?

This problem is frequently asked at Airbnb.

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix - Complete Solution Guide

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix is LeetCode problem 1284, 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

Given a m x n binary matrix mat . In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1 ). A pair of cells are called neighbors if they share one edge. Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot. A binary matrix is a matrix with all cells equal to 0 or 1 only. A zero matrix is a matrix with all cells equal to 0 . Example 1: Input: mat = [[0,0],[0,1]] Output: 3 Explan

Detailed Explanation

The problem requires finding the minimum number of 'flips' needed to transform a given m x n binary matrix into a zero matrix (a matrix where all elements are 0). A 'flip' involves selecting a cell and inverting its value (0 becomes 1, and 1 becomes 0), as well as inverting the values of its immediate neighbors (up, down, left, right). The task is to return the fewest flips required, or -1 if it's impossible to achieve a zero matrix. The matrix dimensions are small (1 <= m, n <= 3).

Solution Approach

The solution employs Breadth-First Search (BFS) to explore the possible states of the matrix. The initial matrix state is converted to an integer representation. The BFS algorithm starts from this initial state and explores neighboring states by flipping each cell and its neighbors. A queue is used to maintain the states to be explored, and a set is used to keep track of visited states. The algorithm continues until the zero matrix state is found or the queue is empty (meaning no solution exists).

Step-by-Step Algorithm

Step 1: Convert the initial matrix 'mat' into an integer 'start_state'. Each cell's value is represented by a bit in the integer. For example, mat[r][c] contributes to the bit at position (r * n + c).
Step 2: Initialize a queue with the 'start_state' and its corresponding step count (0). Also, initialize a set 'visited' to keep track of visited states, adding the 'start_state' to it.
Step 3: Perform BFS: While the queue is not empty, dequeue a state and its step count.
Step 4: For each cell (r_flip, c_flip) in the matrix:
Step 5: Create a 'next_state' by flipping the cell (r_flip, c_flip) and its neighbors in the 'current_state'. Flipping is done using the XOR (^) operator on the corresponding bit position.
Step 6: If 'next_state' is the zero matrix (0), return the current 'steps' + 1.
Step 7: If 'next_state' has not been visited, add it to the 'visited' set and enqueue it with the incremented 'steps'.
Step 8: If the queue becomes empty without finding the zero matrix, return -1 (no solution).

Key Insights

Insight 1: Since the matrix size is small, we can represent the entire matrix state as a single integer using bit manipulation. Each bit in the integer represents a cell in the matrix. This allows for efficient state representation and comparisons.
Insight 2: The problem can be modeled as a graph search problem, where each state of the matrix is a node, and flipping a cell and its neighbors represents an edge connecting two states. Breadth-First Search (BFS) is suitable here to find the shortest path (minimum number of flips) to the zero matrix state.
Insight 3: We need to keep track of visited states to avoid cycles and redundant computations. A set (or hash table) is used to store the visited states efficiently.

Complexity Analysis

Time Complexity: O((m*n)*2^(m*n))

Space Complexity: O(2^(m*n))

Topics

This problem involves: Array, Hash Table, Bit Manipulation, Breadth-First Search, Matrix.

Companies

Asked at: Airbnb.

Minimum Number of Flips to Convert Binary Matrix to Zero Matrix | LeetCode 1284

Hard

Array Hash Table Bit Manipulation Breadth-First Search Matrix

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int minFlips(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;

        int startState = 0;
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (mat[r][c] == 1) {
                    startState |= (1 << (r * n + c));
                }
            }
        }

        if (startState == 0) {
            return 0;
        }

        Deque<int[]> queue = new ArrayDeque<>();
        queue.offer(new int[]{startState, 0});
        Set<Integer> visited = new HashSet<>();
        visited.add(startState);

        while (!queue.isEmpty()) {
            int[] current = queue.poll();
            int currentState = current[0];
            int steps = current[1];

            for (int rFlip = 0; rFlip < m; rFlip++) {
                for (int cFlip = 0; cFlip < n; cFlip++) {

                    int nextState = currentState;

                    int[][] directions = {{0, 0}, {1, 0}, {-1, 0}, {0, 1}, {0, -1}};
                    for (int[] dir : directions) {
                        int nr = rFlip + dir[0];
                        int nc = cFlip + dir[1];

                        if (0 <= nr && nr < m && 0 <= nc && nc < n) {
                            int pos = nr * n + nc;
                            nextState ^= (1 << pos);
                        }
                    }

                    if (nextState == 0) {
                        return steps + 1;
                    }

                    if (!visited.contains(nextState)) {
                        visited.add(nextState);
                        queue.offer(new int[]{nextState, steps + 1});
                    }
                }
            }
        }

        return -1;
    }
}

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Problem Info

Description

Problem 1284: Minimum Number of Flips to Convert Binary Matrix to Zero Matrix - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Breadth-First Search Matrix

Companies

Airbnb

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1284

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

8 steps

1
Step 1: Convert the initial matrix 'mat' into an integer 'start_state'. Each cell's value is represented by a bit in the integer. For example, mat[r][c] contributes to the bit at position (r * n + c).
2
Step 2: Initialize a queue with the 'start_state' and its corresponding step count (0). Also, initialize a set 'visited' to keep track of visited states, adding the 'start_state' to it.
3
Step 3: Perform BFS: While the queue is not empty, dequeue a state and its step count.
4
Step 4: For each cell (r_flip, c_flip) in the matrix:
5
Step 5: Create a 'next_state' by flipping the cell (r_flip, c_flip) and its neighbors in the 'current_state'. Flipping is done using the XOR (^) operator on the corresponding bit position.
6
Step 6: If 'next_state' is the zero matrix (0), return the current 'steps' + 1.
7
Step 7: If 'next_state' has not been visited, add it to the 'visited' set and enqueue it with the incremented 'steps'.
8
Step 8: If the queue becomes empty without finding the zero matrix, return -1 (no solution).

💡

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((m*n)*2^(m*n))

💾

Space Complexity

O(2^(m*n))

📝

Why?

Time Complexity: There are at most 2^(m*n) possible states for the matrix. For each state, we iterate through all the cells (m*n) to consider flipping them. Hence, the time complexity is O((m*n)*2^(m*n)). Space Complexity: We store at most 2^(m*n) states in the 'visited' set and the queue. Therefore, the space complexity is O(2^(m*n)).

📖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

A* Search

Instead of BFS, A* search could be employed to potentially reduce the number of explored states. A heuristic function could be used to estimate the distance to the zero matrix, guiding the search towards more promising states.

Time: Potentially better than BFS in many cases, but worst-case remains O((m*n)*2^(m*n)). Depends heavily on the heuristic.Space: O(2^(m*n))

Trade-offs: A* search might be faster than BFS in some cases if a good heuristic is available. However, a poorly designed heuristic could lead to worse performance than BFS. The implementation would be more complex.

✨

Key Insights

1Insight 1: Since the matrix size is small, we can represent the entire matrix state as a single integer using bit manipulation. Each bit in the integer represents a cell in the matrix. This allows for efficient state representation and comparisons.
2Insight 2: The problem can be modeled as a graph search problem, where each state of the matrix is a node, and flipping a cell and its neighbors represents an edge connecting two states. Breadth-First Search (BFS) is suitable here to find the shortest path (minimum number of flips) to the zero matrix state.
3Insight 3: We need to keep track of visited states to avoid cycles and redundant computations. A set (or hash table) is used to store the visited states efficiently.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: The input matrix is already a zero matrix. The solution correctly returns 0 in this case because the initial check `if (start_state == 0)` handles this.
Edge case 2: The matrix cannot be transformed into a zero matrix. The solution correctly returns -1 in this case because the BFS algorithm will exhaust all possible states without finding the zero matrix, and the final `return -1` handles this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a visited set/hash table. This would lead to infinite loops and TLE (Time Limit Exceeded) errors as the BFS algorithm would revisit previously explored states repeatedly.

✗Common mistake 2: Incorrectly calculating the next state after a flip. It's important to use XOR (^) to correctly invert the bits representing the flipped cell and its neighbors.

✗Common mistake 3: Not handling the edge case where the input matrix is already a zero matrix. Missing this simple check can lead to 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: Game development. This algorithm could be used in puzzle games where the objective is to reach a target state through a series of moves, and finding the minimum number of moves is important.

🏦Real-world scenario 2: Network configuration. Imagine a small network of devices where each device can be in one of two states (e.g., on/off). The algorithm can be used to determine the minimum number of configurations to achieve a desired network state, considering the impact of changing the state of one device on its neighbors.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 25, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 Number of Flips to Convert Binary Matrix to Zero Matrix"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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