What is the time complexity of Set Matrix Zeroes?

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

What data structures are used to solve Set Matrix Zeroes?

This problem uses Array, Hash Table, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Set Matrix Zeroes?

Insight 1: The main challenge is achieving O(1) space complexity. We cannot use auxiliary arrays of size m or n to store rows and columns to be zeroed out. Insight 2: The first row and first column can be used as flags to indicate which rows and columns need to be zeroed out.

Which companies ask LeetCode 73. Set Matrix Zeroes in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Autodesk, Bloomberg and 15 more companies.

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Set Matrix Zeroes - LeetCode 73 Solution

Q: How to solve Set Matrix Zeroes?

The provided solution uses the first row and first column of the matrix to keep track of the rows and columns that need to be set to zero. It first checks if the first row and first column themselves need to be zeroed out. Then, it iterates through the rest of the matrix. If it finds a zero element, it sets the corresponding element in the first row and first column to zero. After this first pass, it iterates through the matrix again (excluding the first row and column) and sets elements to zero based on the values in the first row and column. Finally, it sets the first row and first column to zero if necessary, based on the initial flags.

Set Matrix Zeroes - Complete Solution Guide

Set Matrix Zeroes is LeetCode problem 73, 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 integer matrix matrix , if an element is 0 , set its entire row and column to 0 's. You must do it in place . Example 1: Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]] Example 2: Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]] Constraints: m == matrix.length n == matrix[0].length 1 <= m, n <= 200 -2 31 <= matrix[i][j] <= 2 31 - 1 Follow up: A straightforward solution using O(mn) space is probably a bad idea. A si

Detailed Explanation

The problem requires modifying a given m x n integer matrix in-place. If any element in the matrix is 0, we need to set its entire row and column to 0. The challenge lies in achieving this with constant space complexity, as using extra space proportional to the size of the matrix or even the sum of rows and columns is discouraged in the problem description.

Solution Approach

The provided solution uses the first row and first column of the matrix to keep track of the rows and columns that need to be set to zero. It first checks if the first row and first column themselves need to be zeroed out. Then, it iterates through the rest of the matrix. If it finds a zero element, it sets the corresponding element in the first row and first column to zero. After this first pass, it iterates through the matrix again (excluding the first row and column) and sets elements to zero based on the values in the first row and column. Finally, it sets the first row and first column to zero if necessary, based on the initial flags.

Step-by-Step Algorithm

Step 1: Determine if the first row and first column contain any zeros. Store these flags in `first_row_has_zero` and `first_col_has_zero`.
Step 2: Iterate through the matrix, starting from index (1, 1). If `matrix[i][j] == 0`, set `matrix[0][j] = 0` and `matrix[i][0] = 0`. This uses the first row and column as markers.
Step 3: Iterate through the matrix again, starting from index (1, 1). If `matrix[0][j] == 0` or `matrix[i][0] == 0`, set `matrix[i][j] = 0`. This zeroes out the appropriate elements based on the markers.
Step 4: If `first_row_has_zero` is true, set all elements in the first row to 0.
Step 5: If `first_col_has_zero` is true, set all elements in the first column to 0.

Key Insights

Insight 1: The main challenge is achieving O(1) space complexity. We cannot use auxiliary arrays of size m or n to store rows and columns to be zeroed out.
Insight 2: The first row and first column can be used as flags to indicate which rows and columns need to be zeroed out.
Insight 3: We need to handle the first row and first column separately because they are being used as flags. Otherwise, we might incorrectly zero out other rows and columns.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Matrix.

Companies

Asked at: Adobe, Amazon, Apple, Autodesk, Bloomberg, Expedia, Goldman Sachs, J.P. Morgan, Juspay, Meta, Microsoft, Nutanix, Nvidia, Oracle, ServiceNow, Sprinklr, Uber, Yahoo, eBay, tcs.

Set Matrix Zeroes | LeetCode 73

Medium

Array Hash Table Matrix

Time: O(m*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        
        boolean first_row_has_zero = false;
        boolean first_col_has_zero = false;
        
        for (int j = 0; j < n; j++) {
            if (matrix[0][j] == 0) {
                first_row_has_zero = true;
                break;
            }
        }
        
        for (int i = 0; i < m; i++) {
            if (matrix[i][0] == 0) {
                first_col_has_zero = true;
                break;
            }
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[0][j] == 0 || matrix[i][0] == 0) {
                    matrix[i][j] = 0;
                }
            }
        }
        
        if (first_row_has_zero) {
            for (int j = 0; j < n; j++) {
                matrix[0][j] = 0;
            }
        }
        
        if (first_col_has_zero) {
            for (int i = 0; i < m; i++) {
                matrix[i][0] = 0;
            }
        }
    }
}

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Problem Info

Description

Problem 73: Set Matrix Zeroes - Detailed explanation and solution approaches.

Topics

Array Hash Table Matrix

Companies

AdobeAmazonAppleAutodeskBloombergExpediaGoldman SachsJ.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:73

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

5 steps

1
Step 1: Determine if the first row and first column contain any zeros. Store these flags in `first_row_has_zero` and `first_col_has_zero`.
2
Step 2: Iterate through the matrix, starting from index (1, 1). If `matrix[i][j] == 0`, set `matrix[0][j] = 0` and `matrix[i][0] = 0`. This uses the first row and column as markers.
3
Step 3: Iterate through the matrix again, starting from index (1, 1). If `matrix[0][j] == 0` or `matrix[i][0] == 0`, set `matrix[i][j] = 0`. This zeroes out the appropriate elements based on the markers.
4
Step 4: If `first_row_has_zero` is true, set all elements in the first row to 0.
5
Step 5: If `first_col_has_zero` is true, set all elements in the first column to 0.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(m*n) because we iterate through the entire matrix multiple times. Specifically, we iterate through the matrix to check for zeros and then iterate again to set elements to zero based on the flags in the first row and column. The space complexity is O(1) because we are only using a few extra variables to store the flags for the first row and first column; we are modifying the matrix in place.

📖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

Using O(m+n) space

We can use two arrays, one of size m and another of size n, to store which rows and columns should be zeroed out. We iterate through the matrix once to mark the rows and columns that contain zeros. Then, we iterate through the matrix again and set the appropriate elements to zero.

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

Trade-offs: This approach is simpler to implement but uses O(m+n) space, which is not optimal if memory is a constraint. The in-place solution avoids using additional arrays but requires more careful handling of the first row and column.

✨

Key Insights

1Insight 1: The main challenge is achieving O(1) space complexity. We cannot use auxiliary arrays of size m or n to store rows and columns to be zeroed out.
2Insight 2: The first row and first column can be used as flags to indicate which rows and columns need to be zeroed out.
3Insight 3: We need to handle the first row and first column separately because they are being used as flags. Otherwise, we might incorrectly zero out other rows and columns.

📚

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: Matrix with dimensions 1x1. The solution correctly handles this case because the loops will simply not execute, and the flags will determine whether the single element should be zeroed.
Edge case 2: Matrix where the first row and/or column initially contains zeros. The solution explicitly checks for these cases at the beginning and handles them accordingly, preventing incorrect results.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the first row and first column separately. This leads to incorrect results because the flags in the first row and column might be affected by the zeros found in the rest of the matrix.

✗Common mistake 2: Forgetting to zero out the first row and/or first column at the end, if they initially contained zeros. This is an easy step to miss, leading to an incomplete solution.

✅

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. Representing images as matrices and zeroing out certain regions of an image based on specific conditions.

🏦Real-world scenario 2: Spreadsheet applications. Imagine a spreadsheet where certain calculations cause error values (represented as 0). This algorithm could efficiently mark all rows and columns related to the error for further processing or highlighting.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 13, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Set Matrix Zeroes"?

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

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