What is the time complexity of Largest Submatrix With Rearrangements?

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

What data structures are used to solve Largest Submatrix With Rearrangements?

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

What are the key insights for solving Largest Submatrix With Rearrangements?

Insight 1: Rearranging columns doesn't change the number of consecutive 1s in any row. Thus, for each row, we need to find the best column arrangement to maximize the rectangular area. Insight 2: The height of a submatrix can be thought of as the number of consecutive 1s ending at that row for each column. We can precompute this value for each cell.

Which companies ask LeetCode 1727. Largest Submatrix With Rearrangements in interviews?

This problem is frequently asked at Directi, Samsung.

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Largest Submatrix With Rearrangements - LeetCode 1727 Solution

Largest Submatrix With Rearrangements - Complete Solution Guide

Largest Submatrix With Rearrangements is LeetCode problem 1727, 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

You are given a binary matrix matrix of size m x n , and you are allowed to rearrange the columns of the matrix in any order. Return the area of the largest submatrix within matrix where every element of the submatrix is 1 after reordering the columns optimally. Example 1: Input: matrix = [[0,0,1],[1,1,1],[1,0,1]] Output: 4 Explanation: You can rearrange the columns as shown above. The largest submatrix of 1s, in bold, has an area of 4. Example 2: Input: matrix = [[1,0,1,0,1]] Output: 3 Explanat

Detailed Explanation

The problem asks us to find the largest rectangular submatrix consisting of only 1s in a given binary matrix after rearranging the columns. We are allowed to reorder the columns in any way to maximize the area of such a submatrix. The input is a 2D array (matrix) of 0s and 1s, and the output is the maximum area of a submatrix of 1s achievable after optimal column rearrangement.

Solution Approach

The solution approach involves two main steps. First, we preprocess the input matrix to compute the number of consecutive 1s ending at each cell. Then, for each row, we sort the consecutive 1s counts in descending order. Finally, we iterate through the sorted array and calculate the area of the potential submatrix using the sorted heights and the width determined by the index.

Step-by-Step Algorithm

Step 1: Iterate through the matrix starting from the second row (index 1).
Step 2: For each cell in the current row, check if it's 1. If it is, add the value of the cell directly above it (matrix[i-1][j]) to the current cell (matrix[i][j]). This updates each cell to represent the height of consecutive 1s ending at that cell.
Step 3: Iterate through each row of the modified matrix.
Step 4: Sort the elements of the current row in descending order.
Step 5: Iterate through the sorted row, maintaining an index 'j' representing the width. If we encounter a zero in sorted row, this implies we cannot extend the submatrix width further as only '1's are allowed and we can stop searching for the area.
Step 6: Calculate the area as height * width (matrix[i][j] * (j + 1)).
Step 7: Update the max_area if the current area is larger.
Step 8: Return the max_area after processing all rows.

Key Insights

Insight 1: Rearranging columns doesn't change the number of consecutive 1s in any row. Thus, for each row, we need to find the best column arrangement to maximize the rectangular area.
Insight 2: The height of a submatrix can be thought of as the number of consecutive 1s ending at that row for each column. We can precompute this value for each cell.
Insight 3: Sorting the counts of consecutive 1s in descending order within each row allows us to efficiently calculate the maximum possible area using each column as the right boundary of a potential submatrix.

Complexity Analysis

Time Complexity: O(m*n*log(n))

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting, Matrix.

Companies

Asked at: Directi, Samsung.

Largest Submatrix With Rearrangements | LeetCode 1727

Medium

Array Greedy Sorting Matrix

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

Solution Code

Available in:

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

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

        for (int i = 0; i < m; i++) {
            java.util.Arrays.sort(matrix[i]);
            reverse(matrix[i]);
            for (int j = 0; j < n; j++) {
                int height = matrix[i][j];
                if (height == 0) {
                    break;
                }
                int width = j + 1;
                max_area = Math.max(max_area, height * width);
            }
        }
        
        return max_area;
    }

    private void reverse(int[] arr) {
        int start = 0;
        int end = arr.length - 1;
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
}

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Problem Info

Description

Problem 1727: Largest Submatrix With Rearrangements - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Matrix

Companies

DirectiSamsung

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1727

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: Iterate through the matrix starting from the second row (index 1).
2
Step 2: For each cell in the current row, check if it's 1. If it is, add the value of the cell directly above it (matrix[i-1][j]) to the current cell (matrix[i][j]). This updates each cell to represent the height of consecutive 1s ending at that cell.
3
Step 3: Iterate through each row of the modified matrix.
4
Step 4: Sort the elements of the current row in descending order.
5
Step 5: Iterate through the sorted row, maintaining an index 'j' representing the width. If we encounter a zero in sorted row, this implies we cannot extend the submatrix width further as only '1's are allowed and we can stop searching for the area.
6
Step 6: Calculate the area as height * width (matrix[i][j] * (j + 1)).
7
Step 7: Update the max_area if the current area is larger.
8
Step 8: Return the max_area after processing all rows.

💡

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*log(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting step within each row, which takes O(n*log(n)) time. We perform this sorting for each of the 'm' rows, resulting in O(m*n*log(n)) time complexity. The space complexity is O(1) because we modify the input matrix in-place and only use a constant amount of extra space.

📖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 a Priority Queue

Instead of sorting each row, we could use a priority queue (min-heap) to store the consecutive counts of 1s ending at each row. For each column, insert the consecutive 1s into the heap. After processing all rows, we can extract elements from the heap to calculate area. This allows to avoid repetitive sort operations

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

Trade-offs: This approach has the same time complexity as the provided solution, but uses additional space O(n) for the priority queue. It could be beneficial in specific scenarios, but in most cases, the in-place sorting is preferable due to its lower space complexity.

✨

Key Insights

1Insight 1: Rearranging columns doesn't change the number of consecutive 1s in any row. Thus, for each row, we need to find the best column arrangement to maximize the rectangular area.
2Insight 2: The height of a submatrix can be thought of as the number of consecutive 1s ending at that row for each column. We can precompute this value for each cell.
3Insight 3: Sorting the counts of consecutive 1s in descending order within each row allows us to efficiently calculate the maximum possible area using each column as the right boundary of a potential submatrix.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty matrix (m=0 or n=0). The code handles this implicitly as the loops won't execute, and the initial max_area of 0 will be returned.
Edge case 2: Matrix with all 0s. The consecutive 1s will all be 0. The matrix will be sorted, the inner loop breaks immediately for each row, and 0 is returned.
Edge case 3: Matrix with all 1s. The consecutive 1s preprocessing will correctly calculate the height, and after sorting each row will contain all 1s with the height. The algorithm will compute the correct max area. For a m x n matrix with all 1's, the result will be m * n
Edge case 4: Single row matrix. The first loop won't execute and sorting the first row will calculate correct maximum possible area.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not realizing that you can rearrange the columns. Many might try to find the largest submatrix *without* rearrangement, which is a different and more complex problem.

✗Common mistake 2: Incorrectly calculating the area. For instance, forgetting that the 'width' of the submatrix extends from the beginning of the current row to the current column, not from some other index.

✗Common mistake 3: Not handling edge cases such as empty matrices or matrices filled entirely with 0s or 1s.

✅

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: Optimizing resource allocation. Imagine assigning tasks to workers, where matrix[i][j] represents whether worker 'j' can handle task 'i'. Rearranging columns can represent reordering workers to maximize the number of tasks handled by a contiguous group of skilled workers.

🏦Real-world scenario 2: Image processing. Finding the largest contiguous region of a particular color (e.g., '1' representing a specific color) in an image after potentially rearranging the columns of pixels could be used for object detection or image segmentation.

💡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: September 16, 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 14

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

@CodeNinjaJul 7

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

@TechLeadJun 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 are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Largest Submatrix With Rearrangements"?

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

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