What is the time complexity of Maximal Square?

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

What is the space complexity of Maximal Square?

The space complexity is O(m*n).

What data structures are used to solve Maximal Square?

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

What are the key insights for solving Maximal Square?

Insight 1: Dynamic programming is well-suited because the size of the largest square ending at a given cell depends on the sizes of the largest squares ending at its top, left, and top-left neighbors. Insight 2: The side length of the maximal square ending at matrix[i][j] is 1 + min(side length of square ending at matrix[i-1][j], side length of square ending at matrix[i][j-1], side length of square ending at matrix[i-1][j-1]).

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Maximal Square - LeetCode 221 Solution

Q: How to solve Maximal Square?

The solution employs dynamic programming to build a DP table (dp) where dp[i][j] represents the side length of the largest square ending at matrix[i-1][j-1]. We iterate through the matrix, and if matrix[i-1][j-1] is '1', we calculate dp[i][j] using the formula mentioned above (Insight 2). We keep track of the maximum side length encountered, and finally return its square as the area.

Q: Which companies ask LeetCode 221. Maximal Square in interviews?

This problem is frequently asked at Airbnb, BharatPe, Booking.com, ByteDance, Citadel and 15 more companies.

Maximal Square - Complete Solution Guide

Maximal Square is LeetCode problem 221, 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 filled with 0 's and 1 's, find the largest square containing only 1 's and return its area . Example 1: Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] Output: 4 Example 2: Input: matrix = [["0","1"],["1","0"]] Output: 1 Example 3: Input: matrix = [["0"]] Output: 0 Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 300 matrix[i][j] is '0' or '1' .

Detailed Explanation

The problem asks us to find the largest square submatrix within a given binary matrix (containing only '0's and '1's) that contains only '1's. We need to return the area of this largest square.

Solution Approach

The solution employs dynamic programming to build a DP table (dp) where dp[i][j] represents the side length of the largest square ending at matrix[i-1][j-1]. We iterate through the matrix, and if matrix[i-1][j-1] is '1', we calculate dp[i][j] using the formula mentioned above (Insight 2). We keep track of the maximum side length encountered, and finally return its square as the area.

Step-by-Step Algorithm

Step 1: Initialize a DP table `dp` of size (m+1) x (n+1) with all values set to 0. This table will store the side length of the largest square ending at each cell.
Step 2: Initialize `max_side` to 0. This variable will store the side length of the largest square found so far.
Step 3: Iterate through the input `matrix` from row 1 to m and column 1 to n (using `i` and `j` as indices).
Step 4: Inside the inner loop, check if `matrix[i-1][j-1]` is equal to '1'.
Step 5: If `matrix[i-1][j-1]` is '1', calculate `dp[i][j]` as 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]). This uses the dynamic programming relation to determine the size of the square ending at the current cell.
Step 6: Update `max_side` with the maximum value between the current `max_side` and `dp[i][j]`.
Step 7: After the loops finish, return `max_side * max_side`. This returns the area of the largest square.

Key Insights

Insight 1: Dynamic programming is well-suited because the size of the largest square ending at a given cell depends on the sizes of the largest squares ending at its top, left, and top-left neighbors.
Insight 2: The side length of the maximal square ending at matrix[i][j] is 1 + min(side length of square ending at matrix[i-1][j], side length of square ending at matrix[i][j-1], side length of square ending at matrix[i-1][j-1]).
Insight 3: The area is simply the square of the side length of the largest square.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Airbnb, BharatPe, Booking.com, ByteDance, Citadel, DE Shaw, GSA Capital, Goldman Sachs, Karat, Myntra, Nvidia, Oracle, PayPal, PhonePe, SAP, Salesforce, ServiceNow, Walmart Labs, Wise, eBay.

Maximal Square | LeetCode 221

Medium

Array Dynamic Programming Matrix

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

Solution Code

Available in:

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

        int[][] dp = new int[m + 1][n + 1];
        int max_side = 0;

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (matrix[i-1][j-1] == '1') {
                    dp[i][j] = 1 + Math.min(dp[i-1][j], Math.min(dp[i][j-1], dp[i-1][j-1]));
                    max_side = Math.max(max_side, dp[i][j]);
                }
            }
        }

        return max_side * max_side;
    }
}

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Problem Info

Description

Problem 221: Maximal Square - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

AirbnbBharatPeBooking.comByteDanceCitadelDE ShawGSA CapitalGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:221

Problem Explanation

The problem asks us to find the largest square submatrix within a given binary matrix (containing only '0's and '1's) that contains only '1's. We need to return the area of this largest square.

💡

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
Step 1: Initialize a DP table `dp` of size (m+1) x (n+1) with all values set to 0. This table will store the side length of the largest square ending at each cell.
2
Step 2: Initialize `max_side` to 0. This variable will store the side length of the largest square found so far.
3
Step 3: Iterate through the input `matrix` from row 1 to m and column 1 to n (using `i` and `j` as indices).
4
Step 4: Inside the inner loop, check if `matrix[i-1][j-1]` is equal to '1'.
5
Step 5: If `matrix[i-1][j-1]` is '1', calculate `dp[i][j]` as 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]). This uses the dynamic programming relation to determine the size of the square ending at the current cell.
6
Step 6: Update `max_side` with the maximum value between the current `max_side` and `dp[i][j]`.
7
Step 7: After the loops finish, return `max_side * max_side`. This returns the area of the largest square.

💡

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 nested loops iterate over each element of the input matrix once, resulting in O(m*n) time complexity. The DP table's size is directly proportional to the size of the input matrix, leading to O(m*n) space complexity.

📖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

Space Optimized DP

Instead of using an O(m*n) DP table, we can optimize space to O(n) by only storing the previous row's DP values. This is because the current row's DP values only depend on the previous row and the current column.

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

Trade-offs: This approach reduces space complexity but might be slightly harder to understand and implement. It is preferable when memory usage is a primary concern.

✨

Key Insights

1Insight 1: Dynamic programming is well-suited because the size of the largest square ending at a given cell depends on the sizes of the largest squares ending at its top, left, and top-left neighbors.
2Insight 2: The side length of the maximal square ending at matrix[i][j] is 1 + min(side length of square ending at matrix[i-1][j], side length of square ending at matrix[i][j-1], side length of square ending at matrix[i-1][j-1]).
3Insight 3: The area is simply the square of the side length of the largest square.

📚

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: Empty matrix (m=0 or n=0). The C code handles this explicitly by checking for `matrixSize == 0 || matrixColSize == NULL || *matrixColSize == 0` and returns 0. Other language solutions implicitly handle this due to loop conditions.
Edge case 2: Matrix containing only '0's. The algorithm correctly identifies that there are no squares of '1's, so `max_side` remains 0, and the function returns 0.
Edge case 3: Matrix with a single '1'. The algorithm correctly identifies the single '1' as a square of size 1, sets `max_side` to 1 and returns 1 * 1 = 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to add 1 when calculating dp[i][j]. The formula should be `1 + min(...)`, not just `min(...)`.

✗Common mistake 2: Returning `max_side` instead of `max_side * max_side`. The problem asks for the area, not the side length.

✗Common mistake 3: Not handling the base cases correctly, especially the first row and column in the DP table. Initialization is crucial for dynamic programming solutions.

✅

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 - Finding the largest connected region of a specific color or feature in an image.

🏦Real-world scenario 2: Data mining - Identifying the largest contiguous block of relevant data points in a dataset, such as a sequence of successful transactions in financial analysis.

💡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: July 12, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

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

@CodeNinjaJun 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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Maximal Square"?

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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