0198HardArrayDynamic ProgrammingStackMatrixMonotonic Stack

Largest Flawless Panel

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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 scanner inspects a sheet of veneer cell by cell and hands you the reading as panel, a grid of one-character strings. panel[i][j] is "1" when that cell of the sheet is flawless and "0" when the scanner found a blemish there. Every row of the reading has the same length.

A cutting machine can punch out one axis-aligned block of cells: pick a run of consecutive rows and a run of consecutive columns, and the block is every cell where the two runs meet. The block is sellable only if every cell inside it is flawless.

Return the number of cells in the biggest sellable block. If the sheet has no flawless cell at all, return 0.

Examples

Example 1

Input: panel = [["0", "1", "1", "0"], ["1", "1", "1", "1"], ["1", "1", "1", "1"], ["0", "1", "1", "0"]]
Output: 8

Columns 1 and 2 are flawless down all four rows, giving a 4-by-2 block of 8 cells. The two middle rows offer a 2-by-4 block, also 8, and nothing larger exists because the corners are blemished.

Example 2

Input: panel = [["1", "1", "1"], ["1", "1", "1"]]
Output: 6

Nothing is blemished, so the machine punches out the whole sheet.

Example 3

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

Every cell carries a blemish, so no block can be sold.

Constraints

rows == panel.length
cols == panel[i].length
1 <= rows, cols <= 200
panel[i][j] 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 largest_flawless_panel(panel: list[list[str]]) -> int:

Java

public int largestFlawlessPanel(char[][] panel)

The intended approach

Time O(rows * cols)Space O(cols)

Treat each row as the floor of a skyline: above column j stands a bar whose height is the number of flawless cells running upward from that row, which resets to zero the moment a blemish appears. A block resting on this row is exactly a rectangle inside that skyline, so the answer is the best skyline rectangle over all rows. For one skyline, keep a stack of column indices whose bar heights never decrease upward. A bar that is no taller than the stack top closes that top off on the right, and the entry below it on the stack is its left wall, so the top's widest reach is settled and can be scored. A closing bar of height zero drains the stack.

Pythonentry point: largest_flawless_panel

def largest_flawless_panel(panel):
    # Time: O(rows * cols) -- each row builds its skyline in O(cols) and every
    # column index is pushed and popped from the stack at most once per row.
    # Space: O(cols) for the height array plus the stack.
    cols = len(panel[0])
    heights = [0] * cols
    best = 0
    for row in panel:
        for j in range(cols):
            heights[j] = heights[j] + 1 if row[j] == "1" else 0
        stack = []
        for j in range(cols + 1):
            edge = 0 if j == cols else heights[j]
            while stack and heights[stack[-1]] >= edge:
                top = stack.pop()
                left_wall = stack[-1] if stack else -1
                area = heights[top] * (j - left_wall - 1)
                if area > best:
                    best = area
            stack.append(j)
    return best

The obvious version

Fix every pair of top and bottom rows, mark which columns are flawless all the way down that band, and take the longest run of marked columns times the band height. No stack, no skyline.

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_flawless_panel(panel):
    rows = len(panel)
    cols = len(panel[0])
    best = 0
    for top in range(rows):
        for bottom in range(top, rows):
            height = bottom - top + 1
            run = 0
            for j in range(cols):
                clean = True
                for r in range(top, bottom + 1):
                    if panel[r][j] != "1":
                        clean = False
                        break
                run = run + 1 if clean else 0
                if run * height > best:
                    best = run * height
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 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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