Maximal Square

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

• The input is a 2D binary matrix consisting of only 0s and 1s.
• We need to find the size of the largest square containing only 1s and return its area.
• The square should be formed by consecutive 1s.

2. Analyze the problem: To solve this problem, we can use dynamic programming. We can create a 2D auxiliary matrix to store the lengths of the squares ending at each position. By iterating through the original matrix, we can update the auxiliary matrix to calculate the lengths of squares for each position. Finally, we can find the maximum length and return its square as the area of the largest square.

3. Design an algorithm: Here is the algorithm to solve the problem:

1. Initialize a 2D auxiliary matrix of the same size as the original matrix, filled with zeros.
2. Iterate through each element in the original matrix:
   • If the current element is in the first row or the first column, set the corresponding element in the auxiliary matrix to the current element of the original matrix.
   • Otherwise, if the current element is 1, calculate the length of the square ending at the current position as the minimum of the element above, to the left, and diagonally above-left in the auxiliary matrix, plus 1.
   • Update a variable max_length to keep track of the maximum length seen so far.
3. Return the square of max_length as the area of the largest square.

4. Explain your approach: The approach involves using dynamic programming to find the lengths of squares ending at each position in the matrix. By iterating through the matrix and updating the auxiliary matrix, we can calculate the lengths of squares and find the maximum length. Finally, we return the square of the maximum length as the area of the largest square.

5. Write clean and readable code:

python

def maximalSquare(matrix):
    rows = len(matrix)
    if rows == 0:
        return 0
    cols = len(matrix[0])
    aux = [[0] * cols for _ in range(rows)]
    max_length = 0

    for i in range(rows):
        for j in range(cols):
            if i == 0 or j == 0:
                aux[i][j] = int(matrix[i][j])
            elif matrix[i][j] == '1':
                aux[i][j] = min(aux[i-1][j-1], aux[i-1][j], aux[i][j-1]) + 1
            max_length = max(max_length, aux[i][j])

    return max_length * max_length

6. Test your code: Let's test the code with the following test case:

python

# Example matrix:
# [['1', '0', '1', '0', '0'],
#  ['1', '0', '1', '1', '1'],
#  ['1', '1', '1', '1', '1'],
#  ['1', '0', '0', '1', '0']]

matrix = [
    ['1', '0', '1', '0', '0'],
    ['1', '0', '1', '1', '1'],
    ['1', '1', '1', '1', '1'],
    ['1', '0', '0', '1', '0']
]

result = maximalSquare(matrix)
print(result)

The expected output is:

4

7. Optimize if necessary: The provided solution already has a time complexity of O(mn), where m is the number of rows and n is the number of columns in the matrix. This is because we iterate through each element once. The space complexity is O(mn) as well since we use an auxiliary matrix of the same size as the original matrix.

8. Handle error cases: The code assumes that the input matrix is a valid 2D binary matrix, so there are no specific error handling cases to consider.

9. Discuss complexity analysis: The time complexity of the solution is O(mn), where m is the number of rows and n is the number of columns in the matrix. This is because we iterate through each element once. The space complexity is also O(mn) since we use an auxiliary matrix of the same size as the original matrix to store the lengths of squares ending at each position. The trade-off made in this solution is the use of additional space to store the auxiliary matrix, which allows us to avoid recomputation and achieve a linear time complexity.