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Seal The Trapped Pockets

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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 gasket sheet is inspected cell by cell and handed to you as sheet, a grid of rows rows and cols columns. Every cell carries one character: "X" for solid material, "O" for an open channel.

Two open cells belong to the same channel network when they share a side, so up, down, left or right. Meeting only at a corner joins nothing.

A network holding at least one cell on the outer rim of the sheet can vent to the air outside, and the plant leaves it alone. A network with no cell on the rim is a trapped pocket, and every cell of such a network is filled with solid material.

Do the filling where the sheet already sits. Overwrite the cells of sheet itself instead of assembling a second grid of the same size, then hand sheet back.

Examples

Example 1

Input: sheet = [["O", "X", "X"], ["X", "O", "X"], ["X", "X", "O"]]
Output: [["O", "X", "X"], ["X", "X", "X"], ["X", "X", "O"]]

The two corner cells sit on the rim and stay open. The middle cell only meets them at corners, which joins nothing, so it is a pocket of its own and gets filled.

Example 2

Input: sheet = [["X", "X", "X", "X"], ["X", "O", "O", "X"], ["X", "O", "O", "X"], ["X", "X", "X", "X"]]
Output: [["X", "X", "X", "X"], ["X", "X", "X", "X"], ["X", "X", "X", "X"], ["X", "X", "X", "X"]]

The four open cells form one network ringed by solid material on all sides, so the whole pocket is filled and the sheet ends up solid.

Example 3

Input: sheet = [["X", "O", "X"], ["O", "O", "X"], ["X", "X", "X"]]
Output: [["X", "O", "X"], ["O", "O", "X"], ["X", "X", "X"]]

All three open cells are one network, and two of them lie on the rim, so the network vents and the sheet is returned untouched.

Constraints

rows == sheet.length
cols == sheet[0].length
1 <= rows, cols <= 200
sheet[r][c] is either "X" or "O".
The sheet must be filled in place: the grid returned is the grid that was passed in, not a copy.

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 seal_pockets(sheet: list[list[str]]) -> list[list[str]]:

Java

public char[][] sealPockets(char[][] sheet)

The intended approach

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

Hunting for pockets directly means proving a negative, so invert the question and hunt for the vented networks instead. Push every open cell on the rim onto a stack and spread from them through side-sharing open cells, stamping each one with a scratch character so it is only visited once. Anything still reading as open after that spread never reached the rim, so a final sweep fills those cells and turns the stamped cells back into open channels.

Pythonentry point: seal_pockets

def seal_pockets(sheet):
    # Time O(rows * cols): every cell is stamped at most once and swept twice.
    # Space O(rows * cols) worst case for the explicit stack of one network;
    # no second grid of the sheet's size is ever built.
    rows = len(sheet)
    cols = len(sheet[0])
    stack = []
    for r in range(rows):
        for c in range(cols):
            on_rim = r == 0 or c == 0 or r == rows - 1 or c == cols - 1
            if on_rim and sheet[r][c] == "O":
                sheet[r][c] = "#"
                stack.append((r, c))
    while stack:
        r, c = stack.pop()
        for nr, nc in ((r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)):
            if 0 <= nr < rows and 0 <= nc < cols and sheet[nr][nc] == "O":
                sheet[nr][nc] = "#"
                stack.append((nr, nc))
    for r in range(rows):
        for c in range(cols):
            if sheet[r][c] == "#":
                sheet[r][c] = "O"
            elif sheet[r][c] == "O":
                sheet[r][c] = "X"
    return sheet

The obvious version

Union every side-sharing pair of open cells with a disjoint set structure, plus one extra node standing for the air outside that every rim open cell joins. Then build a fresh grid where an open cell survives only if it shares a set with the outside node.

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_seal_pockets(sheet):
    rows = len(sheet)
    cols = len(sheet[0])
    parent = list(range(rows * cols + 1))
    outside = rows * cols

    def find(x):
        while parent[x] != x:
            parent[x] = parent[parent[x]]
            x = parent[x]
        return x

    def join(a, b):
        ra = find(a)
        rb = find(b)
        if ra != rb:
            parent[ra] = rb

    for r in range(rows):
        for c in range(cols):
            if sheet[r][c] != "O":
                continue
            here = r * cols + c
            if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:
                join(here, outside)
            if r + 1 < rows and sheet[r + 1][c] == "O":
                join(here, (r + 1) * cols + c)
            if c + 1 < cols and sheet[r][c + 1] == "O":
                join(here, r * cols + c + 1)

    filled = []
    for r in range(rows):
        row = []
        for c in range(cols):
            if sheet[r][c] == "O" and find(r * cols + c) != find(outside):
                row.append("X")
            else:
                row.append(sheet[r][c])
        filled.append(row)
    return filled

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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