What is the time complexity of Modify the Matrix?

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

What is the space complexity of Modify the Matrix?

The space complexity is O(m*n).

What data structures are used to solve Modify the Matrix?

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

What are the key insights for solving Modify the Matrix?

The problem requires iterating through the matrix twice: once to find the maximum value in each column and another to perform the replacement. Using an auxiliary array to store column maximum values optimizes the second iteration, avoiding repeated maximum calculations.

Which companies ask LeetCode 3033. Modify the Matrix in interviews?

This problem is frequently asked at Fidelity.

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Modify the Matrix - LeetCode 3033 Solution

Modify the Matrix - Complete Solution Guide

Modify the Matrix is LeetCode problem 3033, a Easy 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 0-indexed m x n integer matrix matrix , create a new 0-indexed matrix called answer . Make answer equal to matrix , then replace each element with the value -1 with the maximum element in its respective column. Return the matrix answer . Example 1: Input: matrix = [[1,2,-1],[4,-1,6],[7,8,9]] Output: [[1,2,9],[4,8,6],[7,8,9]] Explanation: The diagram above shows the elements that are changed (in blue). - We replace the value in the cell [1][1] with the maximum value in the column 1, that

Detailed Explanation

The problem asks you to modify a given matrix. The input is a 0-indexed `m x n` integer matrix called `matrix`. The task is to create a new matrix, `answer`, that's initially a copy of `matrix`. Then, you need to find all instances of the value -1 in `answer`. For each -1, you must replace it with the maximum value found in its corresponding column. The function should return the modified matrix `answer`.

Solution Approach

The provided solutions utilize a two-pass approach. The first pass iterates through each column of the matrix to find the maximum value within that column. These maximum values are stored in a separate array (or implicitly in the original matrix in the C++ solution). The second pass then iterates through the matrix again. If a cell contains -1, it's replaced with the corresponding column's maximum value stored from the first pass. The modified matrix is then returned.

Step-by-Step Algorithm

Initialize an `answer` matrix as a deep copy of the input `matrix` (except the C++ solution modifies the input matrix directly).
Iterate through each column (outer loop).
For each column, find the maximum value among its elements (inner loop).
Store the maximum value for each column (in an array or implicitly).
Iterate through the `answer` matrix again (outer loop).
If an element is -1, replace it with the corresponding column's maximum value found in the previous step (inner loop).
Return the modified `answer` matrix.

Key Insights

The problem requires iterating through the matrix twice: once to find the maximum value in each column and another to perform the replacement.
Using an auxiliary array to store column maximum values optimizes the second iteration, avoiding repeated maximum calculations.
Handling edge cases like empty columns or matrices where no -1 exists needs consideration. The problem statement guarantees at least one non-negative integer per column, simplifying this aspect.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix.

Companies

Asked at: Fidelity.

Modify the Matrix | LeetCode 3033

Easy

Array Matrix

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

Solution Code

Available in:

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

        int[] maxValues = new int[n];
        for (int j = 0; j < n; j++) {
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < m; i++) {
                if (matrix[i][j] > max) {
                    max = matrix[i][j];
                }
            }
            maxValues[j] = max;
        }

        int[][] answer = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == -1) {
                    answer[i][j] = maxValues[j];
                } else {
                    answer[i][j] = matrix[i][j];
                }
            }
        }

        return answer;
    }
}

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Problem Info

Description

Problem 3033: Modify the Matrix - Detailed explanation and solution approaches.

Topics

Array Matrix

Companies

Fidelity

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3033

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

7 steps

1
Initialize an `answer` matrix as a deep copy of the input `matrix` (except the C++ solution modifies the input matrix directly).
2
Iterate through each column (outer loop).
3
For each column, find the maximum value among its elements (inner loop).
4
Store the maximum value for each column (in an array or implicitly).
5
Iterate through the `answer` matrix again (outer loop).
6
If an element is -1, replace it with the corresponding column's maximum value found in the previous step (inner loop).
7
Return the modified `answer` matrix.

💡

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 algorithm iterates through the matrix twice (O(m*n) for each pass). Finding the column maximums takes O(m) time per column, resulting in a total of O(m*n) time for finding the column maximums. The replacement step takes O(m*n). Therefore, the overall time complexity is dominated by O(m*n). The space complexity depends on the method: The Python and Java solutions use an auxiliary array of size n to store column maximums; the C solution uses O(m*n) due to explicit memory allocation for answer; and the C++ solution uses O(1) because it directly modifies the input 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

In-place Modification

Modify the input matrix directly instead of creating a copy. This reduces space complexity.

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

Trade-offs: This approach modifies the input matrix, which might not always be desired. The C++ solution demonstrates this approach. It is preferred when memory is a major concern and modifying the input is acceptable.

✨

Key Insights

1The problem requires iterating through the matrix twice: once to find the maximum value in each column and another to perform the replacement.
2Using an auxiliary array to store column maximum values optimizes the second iteration, avoiding repeated maximum calculations.
3Handling edge cases like empty columns or matrices where no -1 exists needs consideration. The problem statement guarantees at least one non-negative integer per column, simplifying this aspect.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty matrix: The solutions handle this case gracefully because the length checks (`matrix.length` and `matrix[0].length`) prevent errors. If the matrix is empty, the loops won't execute.
Column with all -1's: This edge case is handled by the problem constraint that guarantees at least one non-negative integer per column. So the `max_val` will always be at least 0, not -1.

⚠️

Common Mistakes to Avoid

✗Forgetting to create a deep copy: Simply assigning `answer = matrix` in Python creates a shallow copy. Changes to `answer` will affect `matrix`. This results in incorrect output.

✗Incorrectly calculating column maximums: Failing to properly track the maximum value within each column during the first iteration will lead to wrong replacements.

✅

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?

💼Data preprocessing: This type of matrix manipulation is common in data science, where you might want to fill missing values (represented by -1) with the mean, median, or maximum of a feature (column).

🏦Image processing: In image processing, matrices represent images, and this could be used to fill in missing or corrupted pixel values based on the values of neighboring pixels within the same column.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Modify the Matrix"?

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

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