Problem Description

The problem presents us with a 2-dimensional grid made up of 0s and 1s. The task is to determine the size of the largest square subgrid where all the cells on its border are filled with 1s. The size of a subgrid is defined by the number of elements it contains. If no such subgrid exists, we are to return a size of 0. This is a classic problem for exploring the concepts of dynamic programming in a 2D grid structure.

Intuition

The solution is built upon a dynamic programming approach, where we prepare auxiliary arrays down and right to store the maximum number of consecutive 1s below each cell (including the cell itself) and to the right of each cell (including the cell itself), respectively.

The process is as follows:

• We iterate over the grid from bottom to top and from right to left. This reverse order helps us easily calculate the consecutive 1s for the down and right arrays as we can simply add 1 to the count found in the next row or column.
• After populating the down and right arrays, we look for the largest square subgrid with all 1s on its border. We do this by trying to find a square of size k (starting from the largest possible size and decreasing the size) where in the top left corner of this square, both the number of consecutive 1s downward and rightward from this cell is at least k.
• We also need to ensure that the bottom-left and top-right corners of the potential subgrid have at least k consecutive 1s upwards and leftwards to ensure the border is made entirely of 1s.
• We do this by iterating over all possible positions of the grid where a square of size k can fit. If we find such a square, we immediately return its size (which is k*k).
• If no such square is found after checking all potential positions for all potential sizes, we return 0.

Using this approach, the algorithm efficiently identifies the largest square subgrid with all 1s on its border without checking every possible subgrid exhaustively, thus reducing the time complexity significantly compared to a naive approach.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation of the solution employs dynamic programming. We create two matrices called down and right. Both matrices have the same dimensions as the original grid. Each entry in down[i][j] represents the number of consecutive 1s below (and including) grid cell (i, j), extending all the way down to the bottom of the grid or until a 0 is encountered. Similarly, right[i][j] represents the number of consecutive 1s to the right (and including) grid cell (i, j), extending all the way to the right side of the grid or until a 0 is encountered.

Here is how the implementation approach breaks down:

1. Iterate over the grid from the bottom-right corner to the top-left corner. This iteration order is critical because it allows us to build the down and right matrices by looking at the cells that have already been processed.

2. For each cell (i, j), we check if it contains a 1. If it does, we set down[i][j] to down[i + 1][j] + 1 and right[i][j] to right[i][j + 1] + 1. If the cell is at the last row or column, we simply set down[i][j] or right[i][j] to 1. This is because there are no cells below or to the right to continue the consecutive sequence.

3. Once the down and right matrices are filled, we iterate over all potential squares starting from the largest possible size k and decrease the size until we either find a valid square or finish checking all sizes.

4. We look for a square starting at cell (i, j) with size k. To verify if the square's borders consist entirely of 1s, we check four conditions:

   • down[i][j] should be greater than or equal to k.
   • right[i][j] should be greater than or equal to k.
   • right[i + k - 1][j] (bottom-left corner of the square) should be greater than or equal to k.
   • down[i][j + k - 1] (top-right corner of the square) should be greater than or equal to k.

5. If a square satisfying these conditions is found, we immediately return the size of the square, k * k, since we are iterating from the largest to the smallest possible size, ensuring that the first valid square found is also the largest.

6. If no valid square is found after checking all possibilities, which means no square with borders made entirely of 1s exists, we return 0.

This approach uses the dynamic programming paradigm to build solutions to subproblems (the number of consecutive 1s extending downward and rightward from each cell) and uses those solutions to solve the bigger problem (finding the largest square with 1s on its border).

Example Walkthrough

Let's illustrate the solution approach with a small example grid:

1 1 1 1 0
1 1 1 1 1
1 1 0 1 1
1 1 1 1 1
0 1 1 1 1

Following the steps of the implementation approach:

1. Iterate over the grid from the bottom-right corner to the top-left corner.

   We start from the bottom-right cell (4, 4) of the grid and move towards the top-left cell (0, 0), calculating the down and right values for each cell.

2. For each cell (i, j), we check if it contains a 1.

   For instance, at cell (4, 4), it contains a 1. Since it's in the last row and last column, down[4][4] and right[4][4] will both be set to 1.

3. Fill the down and right matrices.

   Here's what the matrices look like after being filled based on step 2:

   down matrix:

   1 3 3 1 0
   1 2 2 1 1
   1 1 0 1 1
   1 1 1 1 1
   0 1 1 1 1

   right matrix:

   4 3 2 1 0
   4 3 2 2 1
   3 2 0 2 1
   2 1 1 1 1
   1 1 1 1 1

4. Iterate over all potential squares

   Starting with k = 4 (the largest possible square based on the smallest grid dimension), check if a square of that size can exist beginning from each cell (i, j).

   From the right and down matrices, we can see that there's no cell that has both down and right values of 4. The maximum value is 3, so we can't form a square of size 4.

   Decrease k to 3 and repeat the process. Now we have multiple cells with both down and right values of at least 3. We choose one, say cell (1, 1), and check the conditions for the square's bottom-left and top-right corners as well. Since the bottom-left (3, 1) has a down value of 1, which is less than 3, this cell does not form a valid square of size 3.

   We continue to check other potential top-left corners of squares of size 3, eventually finding that cell (0, 1) satisfies all conditions:

   • down[0][1] >= 3
   • right[0][1] >= 3
   • right[0 + 3 - 1][1] (bottom-left corner) = right[2][1] >= 3
   • down[0][1 + 3 - 1] (top-right corner) = down[0][3] >= 3

5. Return the size of the found square

   Since we've found a valid square of size 3 whose all border cells are 1, we return the size of this square: 3 * 3 which is 9.

6. If no valid square is found

   However, if we were unable to find a valid square of size 3, we would decrease k to 2, then to 1, and if still no valid square is found, we would return 0. In this example, we found a square, so we return 9.

Through this process, we efficiently determined the largest square subgrid with all 1s on its border without needing to inspect every possible subgrid. This dynamic programming approach reduces the time complexity and provides a methodical way to find the solution.

Solution Implementation

Time and Space Complexity

The given Python code is for finding the largest 1-bounded square in a 2D binary grid.

Time Complexity:

The time complexity of the code can be analyzed as follows:

• First, we have two nested loops that iterate over the entire grid of size m * n to populate the down and right matrices. This step has a complexity of O(m * n).
• Next, we have a triple nested loop where the outer loop decreases k from min(m, n) to 1, and the two inner loops iterate over the m and n dimensions. In the worst case where k equals min(m, n), the complexity is O(m * n * min(m, n)).

Therefore, the overall time complexity is O(m * n) + O(m * n * min(m, n)) => O(m * n * min(m, n)), which is dominated by the triple nested loop.

Space Complexity:

The space complexity can be analyzed by considering the additional space used:

• Two auxiliary matrices down and right, each of size m * n, are created. Hence, the space required for these matrices is 2 * m * n.
• Apart from these two matrices, only a constant amount of additional space is used for indices and temporary variables.

Thus, the overall space complexity is O(m * n) because the space used for the down and right matrices dominates the space requirement.

Learn more about how to find time and space complexity quickly using problem constraints.