What is the time complexity of 01 Matrix?

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

What is the space complexity of 01 Matrix?

The space complexity is O(m*n).

What data structures are used to solve 01 Matrix?

This problem uses Array, Dynamic Programming, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving 01 Matrix?

Insight 1: A Breadth-First Search (BFS) approach is suitable because we need to find the shortest path (distance) from each '1' to the nearest '0'. BFS explores outwards layer by layer. Insight 2: Start the BFS from all the '0' cells simultaneously. This ensures that the distances calculated for the '1' cells are the minimum distances.

Which companies ask LeetCode 542. 01 Matrix in interviews?

This problem is frequently asked at Accenture, DoorDash, Flipkart, Graviton.

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01 Matrix - LeetCode 542 Solution

Q: How to solve 01 Matrix?

The provided solution uses a Breadth-First Search (BFS) algorithm. It first identifies all the '0' cells in the matrix and adds them to a queue. Then, it iterates through the matrix, replacing all '1' values with -1, effectively marking them as unvisited. The BFS then proceeds level by level, starting from the '0' cells. For each cell visited, it checks its neighbors (up, down, left, right). If a neighbor is a '1' (marked as -1), it calculates the distance to the nearest '0' as the current cell's distance plus 1, updates the neighbor's value in the matrix, and enqueues the neighbor for further exploration. This process continues until the queue is empty, ensuring that every cell has been visited and assigned the shortest distance to the nearest '0'.

01 Matrix - Complete Solution Guide

01 Matrix is LeetCode problem 542, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an m x n binary matrix mat , return the distance of the nearest 0 for each cell . The distance between two cells sharing a common edge is 1 . Example 1: Input: mat = [[0,0,0],[0,1,0],[0,0,0]] Output: [[0,0,0],[0,1,0],[0,0,0]] Example 2: Input: mat = [[0,0,0],[0,1,0],[1,1,1]] Output: [[0,0,0],[0,1,0],[1,2,1]] Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 10 4 1 <= m * n <= 10 4 mat[i][j] is either 0 or 1 . There is at least one 0 in mat . Note: This question is the same as 17

Detailed Explanation

The problem asks us to find the distance to the nearest '0' for each cell in a given binary matrix (a matrix containing only 0s and 1s). The distance between two adjacent cells is defined as 1. The output should be a matrix of the same dimensions as the input, where each cell contains the distance to the nearest '0' in the original matrix. The constraints specify that the matrix dimensions are at most 100x100, and there is guaranteed to be at least one '0' in the input matrix.

Solution Approach

The provided solution uses a Breadth-First Search (BFS) algorithm. It first identifies all the '0' cells in the matrix and adds them to a queue. Then, it iterates through the matrix, replacing all '1' values with -1, effectively marking them as unvisited. The BFS then proceeds level by level, starting from the '0' cells. For each cell visited, it checks its neighbors (up, down, left, right). If a neighbor is a '1' (marked as -1), it calculates the distance to the nearest '0' as the current cell's distance plus 1, updates the neighbor's value in the matrix, and enqueues the neighbor for further exploration. This process continues until the queue is empty, ensuring that every cell has been visited and assigned the shortest distance to the nearest '0'.

Step-by-Step Algorithm

Step 1: Initialize a queue and iterate through the input matrix `mat`.
Step 2: For each cell, if its value is 0, add its coordinates to the queue. Otherwise, set its value to -1 to indicate it is unvisited.
Step 3: While the queue is not empty, dequeue a cell (r, c).
Step 4: For each of the four possible directions (up, down, left, right), calculate the coordinates of the neighboring cell (nr, nc).
Step 5: Check if the neighboring cell is within the bounds of the matrix, and if its value is -1 (meaning it's a '1' and hasn't been visited yet).
Step 6: If the neighboring cell satisfies the condition in step 5, update its value to `mat[r][c] + 1` (the distance from the current '0'), and enqueue its coordinates (nr, nc) for later processing.
Step 7: Repeat steps 3-6 until the queue is empty. At this point, the matrix `mat` will contain the distances to the nearest '0' for each cell.
Step 8: Return the modified matrix `mat`.

Key Insights

Insight 1: A Breadth-First Search (BFS) approach is suitable because we need to find the shortest path (distance) from each '1' to the nearest '0'. BFS explores outwards layer by layer.
Insight 2: Start the BFS from all the '0' cells simultaneously. This ensures that the distances calculated for the '1' cells are the minimum distances.
Insight 3: Mark unvisited '1' cells with a special value (-1) before the BFS to distinguish them from '0' cells and cells that have already been visited and assigned a distance. This prevents infinite loops.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Breadth-First Search, Matrix.

Companies

Asked at: Accenture, DoorDash, Flipkart, Graviton.

01 Matrix | LeetCode 542

Medium

Array Dynamic Programming Breadth-First Search Matrix

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Queue;

class Solution {
    public int[][] updateMatrix(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;
        Queue<int[]> q = new ArrayDeque<>();

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (mat[r][c] == 0) {
                    q.offer(new int[]{r, c});
                } else {
                    mat[r][c] = -1;
                }
            }
        }

        while (!q.isEmpty()) {
            int[] cell = q.poll();
            int r = cell[0];
            int c = cell[1];
            
            for (int[] dir : new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}) {
                int nr = r + dir[0];
                int nc = c + dir[1];

                if (0 <= nr && nr < m && 0 <= nc && nc < n && mat[nr][nc] == -1) {
                    mat[nr][nc] = mat[r][c] + 1;
                    q.offer(new int[]{nr, nc});
                }
            }
        }
        
        return mat;
    }
}

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Problem Info

Description

Problem 542: 01 Matrix - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Breadth-First Search Matrix

Companies

AccentureDoorDashFlipkartGraviton

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:542

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: Initialize a queue and iterate through the input matrix `mat`.
2
Step 2: For each cell, if its value is 0, add its coordinates to the queue. Otherwise, set its value to -1 to indicate it is unvisited.
3
Step 3: While the queue is not empty, dequeue a cell (r, c).
4
Step 4: For each of the four possible directions (up, down, left, right), calculate the coordinates of the neighboring cell (nr, nc).
5
Step 5: Check if the neighboring cell is within the bounds of the matrix, and if its value is -1 (meaning it's a '1' and hasn't been visited yet).
6
Step 6: If the neighboring cell satisfies the condition in step 5, update its value to `mat[r][c] + 1` (the distance from the current '0'), and enqueue its coordinates (nr, nc) for later processing.
7
Step 7: Repeat steps 3-6 until the queue is empty. At this point, the matrix `mat` will contain the distances to the nearest '0' for each cell.
8
Step 8: Return the modified matrix `mat`.

💡

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)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is determined by the BFS traversal, which visits each cell at most once. The space complexity is determined by the size of the queue, which in the worst case, could contain all the cells in the matrix.

📖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

Dynamic Programming

A dynamic programming approach can be used by iterating through the matrix in multiple passes. First from top-left to bottom-right, and then from bottom-right to top-left, updating the distance to the nearest '0' based on the distances of the adjacent cells.

Time: O(m*n)Space: O(1) if done in-place (modifying the original matrix), otherwise O(m*n) if using a separate distance matrix.

Trade-offs: Dynamic programming might be slightly faster in some cases because it avoids the overhead of the queue in BFS, but it may be harder to understand and implement correctly. The space complexity can be better (O(1) in-place) compared to the BFS's O(m*n) in the worst case where almost all cells are 0s.

✨

Key Insights

1Insight 1: A Breadth-First Search (BFS) approach is suitable because we need to find the shortest path (distance) from each '1' to the nearest '0'. BFS explores outwards layer by layer.
2Insight 2: Start the BFS from all the '0' cells simultaneously. This ensures that the distances calculated for the '1' cells are the minimum distances.
3Insight 3: Mark unvisited '1' cells with a special value (-1) before the BFS to distinguish them from '0' cells and cells that have already been visited and assigned a distance. This prevents infinite loops.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Matrix contains only 0s. The algorithm correctly assigns 0 to all cells because the initial population of the queue contains all cells and the while loop does not find any -1 neighbors.
Edge case 2: A single 0 surrounded by 1s. The algorithm correctly finds the shortest distance to that single 0 for each of the surrounding 1s by traversing from the single 0 outwards.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not marking the '1' cells as -1 initially. This can lead to infinite loops during BFS because the algorithm might revisit cells that have already been processed.

✗Common mistake 2: Performing BFS from each '1' cell individually. This approach has a time complexity of O(m*n*k), where k is the average distance from a '1' to the nearest '0', which is much less efficient than starting BFS from all '0' cells simultaneously.

✅

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: Image processing, where you might want to find the distance of each pixel to the nearest edge in a binary image (e.g., after applying an edge detection filter).

🏦Real-world scenario 2: Pathfinding in robotics, where you might want to find the distance of each location in a grid to the nearest obstacle (represented as '0'). This information can be used to plan a path for a robot to navigate around the obstacles.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 3, 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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "01 Matrix"?

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

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