What is the time complexity of Largest 1-Bordered Square?

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

What is the space complexity of Largest 1-Bordered Square?

The space complexity is O(m*n).

What data structures are used to solve Largest 1-Bordered 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 Largest 1-Bordered Square?

Insight 1: We need to efficiently determine the length of consecutive 1s horizontally and vertically at each cell to check if a square can be formed. Insight 2: Dynamic programming can be used to precompute the horizontal and vertical lengths of consecutive 1s, enabling faster checks for square existence.

Which companies ask LeetCode 1139. Largest 1-Bordered Square in interviews?

This problem is frequently asked at DE Shaw, Samsung, ZS Associates.

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Largest 1-Bordered Square - LeetCode 1139 Solution

Q: How to solve Largest 1-Bordered Square?

The provided solution uses dynamic programming to precompute the lengths of consecutive 1s to the right (horizontally) and downwards (vertically) for each cell in the grid. Then, it iterates through the grid from the bottom-right corner, checking for possible squares with decreasing side lengths. For each cell, it determines the maximum possible side length of a square that could end at that cell based on the horizontal and vertical lengths. It then checks if a square of that side length exists by looking at the top and left borders. If such a square is found, it updates the maximum side length and continues until a square is found, then breaks to search the remaining grid.

Largest 1-Bordered Square - Complete Solution Guide

Largest 1-Bordered Square is LeetCode problem 1139, 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 a 2D grid of 0 s and 1 s, return the number of elements in the largest square subgrid that has all 1 s on its border , or 0 if such a subgrid doesn't exist in the grid . Example 1: Input: grid = [[1,1,1],[1,0,1],[1,1,1]] Output: 9 Example 2: Input: grid = [[1,1,0,0]] Output: 1 Constraints: 1 <= grid.length <= 100 1 <= grid[0].length <= 100 grid[i][j] is 0 or 1

Detailed Explanation

The problem asks us to find the largest square subgrid within a given 2D grid of 0s and 1s, such that all elements on the border of the square are 1s. The function should return the number of elements in this largest square (which is the side length squared). If no such square exists (i.e., the grid contains no 1s that form a bordered square), the function should return 0.

Solution Approach

The provided solution uses dynamic programming to precompute the lengths of consecutive 1s to the right (horizontally) and downwards (vertically) for each cell in the grid. Then, it iterates through the grid from the bottom-right corner, checking for possible squares with decreasing side lengths. For each cell, it determines the maximum possible side length of a square that could end at that cell based on the horizontal and vertical lengths. It then checks if a square of that side length exists by looking at the top and left borders. If such a square is found, it updates the maximum side length and continues until a square is found, then breaks to search the remaining grid.

Step-by-Step Algorithm

Step 1: Initialize two 2D arrays, `hor` and `ver`, to store the horizontal and vertical lengths of consecutive 1s, respectively. These arrays have the same dimensions as the input `grid`.
Step 2: Iterate through the `grid` from left to right, top to bottom. For each cell `grid[r][c]`, if it's a 1, update `hor[r][c]` with the length of consecutive 1s to its left (including itself) and update `ver[r][c]` with the length of consecutive 1s above it (including itself).
Step 3: Initialize `max_side` to 0, representing the side length of the largest square found so far.
Step 4: Iterate through the `grid` from bottom to top, right to left. For each cell `grid[r][c]`, determine the potential side length `side` based on the minimum of `hor[r][c]` and `ver[r][c]`. The potential side length represents the largest square that could end at that cell.
Step 5: Check if a square of side length `side` exists. This is done by checking if the top border (starting at `grid[r - side + 1][c]`) and the left border (starting at `grid[r][c - side + 1]`) also have at least `side` consecutive 1s. We do this by seeing if hor[r - side + 1][c] >= side and ver[r][c - side + 1] >= side.
Step 6: If a valid square is found, update `max_side` to `side` and break out of the inner `while` loop since you've found the largest square ending at this cell.
Step 7: After iterating through the entire `grid`, return `max_side * max_side`, which is the area of the largest 1-bordered square.

Key Insights

Insight 1: We need to efficiently determine the length of consecutive 1s horizontally and vertically at each cell to check if a square can be formed.
Insight 2: Dynamic programming can be used to precompute the horizontal and vertical lengths of consecutive 1s, enabling faster checks for square existence.
Insight 3: We can iterate through potential square sizes starting from the largest possible and decrementing, which allows us to stop as soon as we find a valid square.

Complexity Analysis

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

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: DE Shaw, Samsung, ZS Associates.

Largest 1-Bordered Square | LeetCode 1139

Medium

Array Dynamic Programming Matrix

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

Solution Code

Available in:

class Solution {
    public int largest1BorderedSquare(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        int[][] hor = new int[m][n];
        int[][] ver = new int[m][n];

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] == 1) {
                    hor[r][c] = (c > 0 ? hor[r][c - 1] : 0) + 1;
                    ver[r][c] = (r > 0 ? ver[r - 1][c] : 0) + 1;
                }
            }
        }

        int max_side = 0;
        for (int r = m - 1; r >= 0; r--) {
            for (int c = n - 1; c >= 0; c--) {
                int side = Math.min(hor[r][c], ver[r][c]);
                while (side > max_side) {
                    // Check if the top and left borders can form a square of this side length
                    if (hor[r - side + 1][c] >= side && ver[r][c - side + 1] >= side) {
                        max_side = side;
                        break;
                    }
                    side--;
                }
            }
        }

        return max_side * max_side;
    }
}

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Problem Info

Description

Problem 1139: Largest 1-Bordered Square - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

DE ShawSamsungZS Associates

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1139

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
Step 1: Initialize two 2D arrays, `hor` and `ver`, to store the horizontal and vertical lengths of consecutive 1s, respectively. These arrays have the same dimensions as the input `grid`.
2
Step 2: Iterate through the `grid` from left to right, top to bottom. For each cell `grid[r][c]`, if it's a 1, update `hor[r][c]` with the length of consecutive 1s to its left (including itself) and update `ver[r][c]` with the length of consecutive 1s above it (including itself).
3
Step 3: Initialize `max_side` to 0, representing the side length of the largest square found so far.
4
Step 4: Iterate through the `grid` from bottom to top, right to left. For each cell `grid[r][c]`, determine the potential side length `side` based on the minimum of `hor[r][c]` and `ver[r][c]`. The potential side length represents the largest square that could end at that cell.
5
Step 5: Check if a square of side length `side` exists. This is done by checking if the top border (starting at `grid[r - side + 1][c]`) and the left border (starting at `grid[r][c - side + 1]`) also have at least `side` consecutive 1s. We do this by seeing if hor[r - side + 1][c] >= side and ver[r][c - side + 1] >= side.
6
Step 6: If a valid square is found, update `max_side` to `side` and break out of the inner `while` loop since you've found the largest square ending at this cell.
7
Step 7: After iterating through the entire `grid`, return `max_side * max_side`, which is the area of the largest 1-bordered 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*min(m,n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is determined by the nested loops that iterate through the grid. Inside the inner loop, there is a `while` loop which, in the worst-case scenario, may iterate up to `min(m, n)` times. Therefore, the overall time complexity is O(m * n * min(m, n)). The space complexity comes from the `hor` and `ver` arrays, each of size `m * n`, 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

Brute Force

Iterate through all possible square sizes and starting positions, then check if the square's border consists of all 1s.

Time: O(m^3 * n^3) or O((m*n)^2) in the worst case.Space: O(1)

Trade-offs: The brute force approach has significantly worse time complexity. The provided DP approach trades off space for time. The brute force approach avoids using extra space.

✨

Key Insights

1Insight 1: We need to efficiently determine the length of consecutive 1s horizontally and vertically at each cell to check if a square can be formed.
2Insight 2: Dynamic programming can be used to precompute the horizontal and vertical lengths of consecutive 1s, enabling faster checks for square existence.
3Insight 3: We can iterate through potential square sizes starting from the largest possible and decrementing, which allows us to stop as soon as we find a valid 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: An empty grid or a grid with all 0s. In this case, the solution will correctly return 0 because the hor and ver matrices will be all 0's, and max_side will remain 0.
Edge case 2: A grid containing only a single '1'. The algorithm correctly identifies this as a square of side 1 and returns 1. The hor and ver values will be 1 at the coordinate of the 1.
Edge case 3: A grid consisting of a single row or column of 1s. The solution correctly finds the largest square (side=1) or determines there is no square possible.
Edge case 4: A grid where the largest square is at the top-left corner. The algorithm correctly identifies this since it iterates from the bottom right, so if finds a result the side length is updated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the horizontal and vertical lengths. Ensuring that the `hor` and `ver` arrays accurately reflect the consecutive lengths of 1s is crucial.

✗Common mistake 2: Off-by-one errors when checking the top and left borders of potential squares. Carefully managing the indices is essential to avoid incorrect checks.

✗Common mistake 3: Not considering the case where no square exists and failing to return 0.

✅

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 - Identifying objects with certain border characteristics in a binary image.

🏦Real-world scenario 2: Resource allocation - Finding the largest contiguous block of available resources (represented by 1s) in a grid, where the block must be square and have a border of availability.

💡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 30, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 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 "Largest 1-Bordered 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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