#1139

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Largest 1-Bordered Square

Find the largest square of 1s with 1s on its borders in a binary grid using dynamic programming.

Problem Statement

You are given a 2D grid of 0s and 1s. Your task is to find the largest square subgrid within this matrix where all four borders of the square consist of 1s. If no such square exists, return 0.

For example, in a grid where all cells are 1 except for a single 0 in the middle, the largest square would have the side length of 3, and thus, the area would be 9. You are required to determine the maximum possible area for this square.

Examples

Example 1

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]

Output: 9

Example details omitted.

Example 2

Input: grid = [[1,1,0,0]]

Output: 1

Example details omitted.

Constraints

1 <= grid.length <= 100
1 <= grid[0].length <= 100
grid[i][j] is 0 or 1

Solution Approach

Dynamic Programming Table Construction

Construct a dynamic programming table where each entry stores the largest square subgrid with 1s on all borders that ends at that specific cell. You can calculate the size of the square using the minimum of the values from the cells above, to the left, below, and to the right.

State Transitions

For each cell, determine the potential side length of the largest square with a border of 1s by examining the grid above, left, down, and right of the current cell. The dynamic programming table helps track these transitions in O(N^2) time.

Time Complexity

The time complexity of this approach is O(N^2), where N is the maximum dimension of the grid. This is because each cell must be examined in relation to its neighbors to calculate the largest possible square.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(N^2) due to the necessity to examine each cell in the grid and perform constant-time operations for each cell's neighbors. The space complexity is also O(N^2) as it requires storing the results of the dynamic programming table for the grid's size.

What Interviewers Usually Probe

The candidate understands dynamic programming for grid-based problems.
The candidate considers space and time complexity and aims for optimization.
The candidate uses state transitions effectively to track and update valid solutions.

Common Pitfalls or Variants

Common pitfalls

Not considering the boundary conditions properly when constructing the dynamic programming table.
Misunderstanding the problem requirement, where 1s should appear on all four borders of the square.
Ignoring the impact of grid size on the time and space complexity when scaling up the problem.

Follow-up variants

Consider handling grids with varying dimensions or sparse 1s.
Try solving the problem with optimizations to reduce memory usage.
Explore different ways to handle grid cells, such as through greedy approaches or matrix manipulation.

FAQ

How do I approach the Largest 1-Bordered Square problem?

The problem can be solved using dynamic programming. For each cell, track the largest square subgrid with 1s on all four borders, updating the size dynamically based on neighboring cells.

What is the time complexity of the solution?

The time complexity is O(N^2), where N is the largest grid dimension, since each cell is processed once.

Can this problem be solved without dynamic programming?

While it's theoretically possible, dynamic programming offers an efficient solution by avoiding redundant recalculations and enabling faster processing.

What are common pitfalls when solving this problem?

Common pitfalls include neglecting boundary conditions, misunderstanding the border requirement, and failing to optimize space and time complexity.

How can GhostInterview help me prepare for similar dynamic programming questions?

GhostInterview offers structured practice with dynamic programming problems, helping you understand and implement effective state transition strategies.

terminal

Solution

Solution 1: Prefix Sum + Enumeration

We can use the prefix sum method to preprocess the number of consecutive 1s down and to the right of each position, denoted as `down[i][j]` and `right[i][j]`.

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class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        down = [[0] * n for _ in range(m)]
        right = [[0] * n for _ in range(m)]
        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                if grid[i][j]:
                    down[i][j] = down[i + 1][j] + 1 if i + 1 < m else 1
                    right[i][j] = right[i][j + 1] + 1 if j + 1 < n else 1
        for k in range(min(m, n), 0, -1):
            for i in range(m - k + 1):
                for j in range(n - k + 1):
                    if (
                        down[i][j] >= k
                        and right[i][j] >= k
                        and right[i + k - 1][j] >= k
                        and down[i][j + k - 1] >= k
                    ):
                        return k * k
        return 0

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