0355MediumArrayDynamic ProgrammingMatrix

Largest Intact Mosaic Square

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A mosaic panel is surveyed tile by tile into matrix. Each entry is the one-character string "1" when that tile is intact or "0" when it is cracked.

A restorer wants to cut out the largest square block of tiles that contains no cracked tile. The block must be aligned with the grid and every tile inside it must be intact.

Return the number of tiles such a largest square block covers, or 0 when every tile is cracked.

Examples

Example 1

Input: matrix = [["1","0","1","1"],["1","1","1","1"],["0","1","1","1"]]
Output: 4

The tiles in rows 1 and 2 across columns 2 and 3 are all intact, giving a 2 by 2 block that covers 4 tiles. No 3 by 3 block is free of cracks.

Example 2

Input: matrix = [["0","1"],["1","1"]]
Output: 1

Three tiles are intact but they do not form a 2 by 2 block, so the best square is a single tile covering 1 tile.

Example 3

Input: matrix = [["0","0"],["0","0"]]
Output: 0

Every tile is cracked, so no square block can be cut at all.

Constraints

m == matrix.length
n == matrix[r].length
1 <= m, n <= 300
matrix[r][c] is "0" or "1".

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def maximal_square(matrix: list[list[str]]) -> int:

Java

public int maximalSquare(char[][] matrix)

The intended approach

Time O(m * n)Space O(n) for the two rows

For every cell keep the side of the largest all-intact square whose bottom-right corner is that cell. A cracked tile scores zero; an intact tile takes the smallest of the three scores at the cell above, the cell to the left and the cell diagonally up-left, and adds one. Only the previous row and the row being filled are needed, and the answer is the largest side squared.

Pythonentry point: maximal_square

def maximal_square(matrix):
    # side[c] is the side of the largest intact square whose bottom-right corner is
    # the cell in the current row at column c.  An intact cell can extend the squares
    # above it, above-left and to its left, so it takes the smallest of those three
    # sides and adds one.  Two rows are kept: the finished one and the one being built.
    # Time O(m * n).  Space O(n) for the two rows.
    rows = len(matrix)
    cols = len(matrix[0])
    previous = [0] * cols
    best = 0
    for r in range(rows):
        current = [0] * cols
        for c in range(cols):
            if matrix[r][c] == "1":
                if r == 0 or c == 0:
                    current[c] = 1
                else:
                    up = previous[c]
                    left = current[c - 1]
                    corner = previous[c - 1]
                    smallest = up
                    if left < smallest:
                        smallest = left
                    if corner < smallest:
                        smallest = corner
                    current[c] = smallest + 1
                if current[c] > best:
                    best = current[c]
        previous = current
    return best * best

The obvious version

Try every side length and every top-left corner, and check all the tiles inside that block one by one. Keeps the largest side that survives a full check.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_largest_intact_square(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    best = 0
    limit = rows if rows < cols else cols
    for side in range(1, limit + 1):
        for r in range(rows - side + 1):
            for c in range(cols - side + 1):
                whole = True
                for i in range(r, r + side):
                    for j in range(c, c + side):
                        if matrix[i][j] != "1":
                            whole = False
                if whole and side > best:
                    best = side
    return best * best

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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