What is the time complexity of Largest Magic Square?

The optimal solution has a time complexity of O((min(m,n)^3) * (m * n)).

What is the space complexity of Largest Magic Square?

The space complexity is O(m * n).

What data structures are used to solve Largest Magic Square?

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

What are the key insights for solving Largest Magic Square?

Insight 1: Prefix sums can significantly reduce the time complexity of calculating row and column sums for each potential magic square. Insight 2: Starting the search for magic squares from the largest possible size (min(m, n)) and working downwards allows early termination as soon as a magic square is found.

Which companies ask LeetCode 1895. Largest Magic Square in interviews?

This problem is frequently asked at Wayfair.

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Largest Magic Square - LeetCode 1895 Solution

Largest Magic Square - Complete Solution Guide

Largest Magic Square is LeetCode problem 1895, 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

A k x k magic square is a k x k grid filled with integers such that every row sum, every column sum, and both diagonal sums are all equal . The integers in the magic square do not have to be distinct . Every 1 x 1 grid is trivially a magic square . Given an m x n integer grid , return the size (i.e., the side length k ) of the largest magic square that can be found within this grid . Example 1: Input: grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]] Output: 3 Explanation: The largest mag

Detailed Explanation

The problem asks us to find the largest magic square within a given m x n grid of integers. A k x k magic square is a square grid where the sum of each row, column, and both diagonals is equal. We need to return the size (k) of the largest magic square found within the input grid. If no magic square larger than 1x1 exists, we return 1 since every 1x1 grid is considered a magic square. The integers within the magic square don't have to be distinct, and the constraints limit the grid size to 50x50 with integer values up to 10^6.

Solution Approach

The solution utilizes prefix sums to efficiently calculate the sums of rows and columns within potential magic squares. It iterates through possible square sizes (k) from the largest possible size down to 2. For each size, it iterates through all possible top-left corners of a k x k square within the grid. For each potential square, it first checks if the sums of both diagonals are equal. If not, it moves on to the next potential square. If the diagonals sums are equal, it calculates the sums of all rows and columns using the precomputed prefix sums. If all row, column, and diagonal sums are equal, the current square is a magic square, and the size k is returned. If no magic square larger than 1 is found, the function returns 1.

Step-by-Step Algorithm

Step 1: Calculate prefix sums for each row of the grid. `prefix_row[r][c+1]` stores the sum of `grid[r][0]` to `grid[r][c]` inclusive.
Step 2: Calculate prefix sums for each column of the grid. `prefix_col[r+1][c]` stores the sum of `grid[0][c]` to `grid[r][c]` inclusive.
Step 3: Iterate through possible square sizes `k` from `min(m, n)` down to 2. This ensures we find the *largest* magic square first.
Step 4: Iterate through possible top-left corners `(r, c)` for a `k x k` square within the grid, ensuring the square remains within the grid boundaries.
Step 5: Calculate the sum of the main diagonal (`diag1_sum`) and the anti-diagonal (`diag2_sum`).
Step 6: If `diag1_sum` is not equal to `diag2_sum`, the current `k x k` square cannot be a magic square, so continue to the next potential square.
Step 7: If the diagonal sums are equal, calculate row and column sums using prefix sums. `row_sum = prefix_row[r + i][c + k] - prefix_row[r + i][c]` and `col_sum = prefix_col[r + k][c + i] - prefix_col[r][c + i]`.
Step 8: If any row or column sum is not equal to the diagonal sums (`target_sum`), the square is not a magic square, so continue to the next potential square.
Step 9: If all row, column and diagonal sums are equal, the current `k x k` square is a magic square. Return `k`.
Step 10: If no magic square of size greater than 1 is found, return 1 (since every 1x1 square is a magic square).

Key Insights

Insight 1: Prefix sums can significantly reduce the time complexity of calculating row and column sums for each potential magic square.
Insight 2: Starting the search for magic squares from the largest possible size (min(m, n)) and working downwards allows early termination as soon as a magic square is found.
Insight 3: The diagonal sums must be checked first, as if they are not equal the square cannot be a magic square, and the rows and columns need not be checked saving time.

Complexity Analysis

Time Complexity: O((min(m,n)^3) * (m * n))

Space Complexity: O(m * n)

Topics

This problem involves: Array, Matrix, Prefix Sum.

Companies

Asked at: Wayfair.

Largest Magic Square | LeetCode 1895

Medium

Array Matrix Prefix Sum

Time: O((min(m,n)^3) * (m * n))Space: O(m * n)

Solution Code

Available in:

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

        int[][] prefix_row = new int[m][n + 1];
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                prefix_row[r][c + 1] = prefix_row[r][c] + grid[r][c];
            }
        }

        int[][] prefix_col = new int[m + 1][n];
        for (int c = 0; c < n; c++) {
            for (int r = 0; r < m; r++) {
                prefix_col[r + 1][c] = prefix_col[r][c] + grid[r][c];
            }
        }

        for (int k = Math.min(m, n); k > 1; k--) {
            for (int r = 0; r <= m - k; r++) {
                for (int c = 0; c <= n - k; c++) {
                    int diag1_sum = 0;
                    for (int i = 0; i < k; i++) {
                        diag1_sum += grid[r + i][c + i];
                    }
                    int diag2_sum = 0;
                    for (int i = 0; i < k; i++) {
                        diag2_sum += grid[r + i][c + k - 1 - i];
                    }

                    if (diag1_sum != diag2_sum) {
                        continue;
                    }

                    int target_sum = diag1_sum;
                    boolean is_magic = true;

                    for (int i = 0; i < k; i++) {
                        int row_sum = prefix_row[r + i][c + k] - prefix_row[r + i][c];
                        if (row_sum != target_sum) {
                            is_magic = false;
                            break;
                        }

                        int col_sum = prefix_col[r + k][c + i] - prefix_col[r][c + i];
                        if (col_sum != target_sum) {
                            is_magic = false;
                            break;
                        }
                    }

                    if (is_magic) {
                        return k;
                    }
                }
            }
        }

        return 1;
    }
}

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Problem Info

Description

Problem 1895: Largest Magic Square - Detailed explanation and solution approaches.

Topics

Array Matrix Prefix Sum

Companies

Wayfair

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1895

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

10 steps

1
Step 1: Calculate prefix sums for each row of the grid. `prefix_row[r][c+1]` stores the sum of `grid[r][0]` to `grid[r][c]` inclusive.
2
Step 2: Calculate prefix sums for each column of the grid. `prefix_col[r+1][c]` stores the sum of `grid[0][c]` to `grid[r][c]` inclusive.
3
Step 3: Iterate through possible square sizes `k` from `min(m, n)` down to 2. This ensures we find the *largest* magic square first.
4
Step 4: Iterate through possible top-left corners `(r, c)` for a `k x k` square within the grid, ensuring the square remains within the grid boundaries.
5
Step 5: Calculate the sum of the main diagonal (`diag1_sum`) and the anti-diagonal (`diag2_sum`).
6
Step 6: If `diag1_sum` is not equal to `diag2_sum`, the current `k x k` square cannot be a magic square, so continue to the next potential square.
7
Step 7: If the diagonal sums are equal, calculate row and column sums using prefix sums. `row_sum = prefix_row[r + i][c + k] - prefix_row[r + i][c]` and `col_sum = prefix_col[r + k][c + i] - prefix_col[r][c + i]`.
8
Step 8: If any row or column sum is not equal to the diagonal sums (`target_sum`), the square is not a magic square, so continue to the next potential square.
9
Step 9: If all row, column and diagonal sums are equal, the current `k x k` square is a magic square. Return `k`.
10
Step 10: If no magic square of size greater than 1 is found, return 1 (since every 1x1 square is a magic 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((min(m,n)^3) * (m * n))

💾

Space Complexity

O(m * n)

📝

Why?

The outer loop iterates through possible sizes of the square, which is O(min(m,n)). The nested loops to find the top left corner iterate at most O(m*n) times. Inside these loops, the code calculates the sums of two diagonals in O(min(m,n)), and then iterates through rows and columns, which is again O(min(m,n)). Calculating the sums for rows and columns using prefix sums takes O(1). However, the initial calculation of the prefixes take O(m*n) time. Therefore, the time complexity is dominated by O(min(m,n) * m * n * (min(m,n) + min(m,n))) which is approximately O((min(m,n)^2)*(m*n)) * O(min(m,n)) = O((min(m,n)^3) * (m * n)). The space complexity arises from the two prefix sum arrays, `prefix_row` and `prefix_col`, both of which take O(m * n) 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

Brute Force

Without using prefix sums, you could calculate the row, column, and diagonal sums for each possible square by iterating through the elements directly. This involves recalculating sums for overlapping regions, leading to inefficiency.

Time: O((min(m, n)^4) * (m * n))Space: O(1)

Trade-offs: This approach is simpler to implement but much slower, especially for larger grids. It avoids the extra space for prefix sums but suffers a significant time penalty. This would likely TLE.

Dynamic Programming

Instead of recalculating the row/col/diag sums every time, you can store intermediate sums to reduce redundancy. However, given that the key to speed is prefix sums which provide O(1) calculation of these sums, DP provides little benefit.

Time: O((min(m,n)^3)*(m*n))Space: O(m*n)

Trade-offs: Likely no performance improvement over prefix sum method, with added code complexity. DP is generally useful when the previous state affects the future state, but that is not the case in this problem.

✨

Key Insights

1Insight 1: Prefix sums can significantly reduce the time complexity of calculating row and column sums for each potential magic square.
2Insight 2: Starting the search for magic squares from the largest possible size (min(m, n)) and working downwards allows early termination as soon as a magic square is found.
3Insight 3: The diagonal sums must be checked first, as if they are not equal the square cannot be a magic square, and the rows and columns need not be checked saving time.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Input grid with only one row or one column. The algorithm correctly identifies that the largest possible magic square is of size 1.
Edge case 2: Empty input grid. While constraints specify m and n >= 1, a robust solution should handle this gracefully, and would likely return 0 or 1 depending on the specification in the prompt. The current solution would throw an error due to accessing `grid[0].length`.
Edge case 3: Input grid where no magic square exists (other than 1x1). The algorithm will correctly iterate through all possibilities and return 1.
Edge case 4: Input grid contains negative numbers. The algorithm functions correctly with negative numbers because it uses addition and subtraction to compute sums.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using prefix sums, leading to a much slower brute-force solution that times out for larger inputs.

✗Common mistake 2: Incorrectly calculating prefix sums, leading to wrong row or column sums and incorrect results.

✗Common mistake 3: Not checking all possible starting positions (top-left corners) for each square size.

✗Common mistake 4: Not starting the search from the largest possible square size, leading to unnecessary iterations and potentially missing the largest magic square.

✗Common mistake 5: Prematurely returning without ensuring *all* conditions of a magic square are met (row, column, and diagonal sums being equal).

✗Common mistake 6: Forgetting the edge case where the largest magic square is just of size 1.

✗Common mistake 7: Integer overflow issues if the data type used is not large enough to hold the sums of rows/columns/diagonals, especially given the constraint 1 <= grid[i][j] <= 10^6. Using `int` is sufficient given the constraints. However, if the constraints were loosened, using `long` may be needed.

✅

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 identifying regions with specific sum properties can be useful for feature extraction or pattern recognition.

🏦Real-world scenario 2: Data analysis where detecting submatrices or subgrids with balanced sums can indicate meaningful correlations or patterns in datasets. For example, identifying a square region in a financial dataset where the sum of revenues across rows and columns is consistent can highlight market stability or growth.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 4, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 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 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 Magic 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 matrix techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?