0519EasyArrayMatrix

Isolated Pallets on the Rack

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Trains the technique from

LeetCode 1582Special Positions in a Binary Matrix

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 storage rack is laid out as a grid with m aisles and n columns. rack[i][j] is 1 when a pallet sits in the slot at aisle i, column j, and 0 when that slot is empty.

A pallet is isolated when no other pallet shares its aisle and no other pallet shares its column. Return how many isolated pallets the rack holds.

Examples

Example 1

Input: rack = [[1, 0, 0], [0, 0, 0], [0, 0, 1]]
Output: 2

The pallet at aisle 0, column 0 has its aisle and its column to itself, and so does the pallet at aisle 2, column 2, so both are isolated.

Example 2

Input: rack = [[1, 0, 0, 0], [0, 0, 1, 0]]
Output: 2

One pallet sits at aisle 0, column 0 and the other at aisle 1, column 2. Neither shares an aisle or a column with the other, so both are isolated.

Example 3

Input: rack = [[1, 0], [1, 0]]
Output: 0

Both pallets sit in column 0, so each of them shares its column with the other and neither is isolated.

Constraints

m == rack.length
n == rack[i].length
1 <= m, n <= 100
rack[i][j] is either 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 count_isolated_pallets(rack: list[list[int]]) -> int:

Java

public int countIsolatedPallets(int[][] rack)

The intended approach

Time O(m * n): two sweeps over the grid.Space O(m + n): one total per aisle and per column.

Sweep the rack once to total the pallets in each aisle and the pallets in each column. A pallet at aisle i, column j is isolated exactly when the aisle total for i and the column total for j are both one, so a second sweep over the slots that hold a pallet counts the isolated ones with no further scanning.

Pythonentry point: count_isolated_pallets

def count_isolated_pallets(rack):
    # Total each aisle and each column once, then a pallet is isolated exactly when
    # the totals for its own aisle and column are both one.
    # O(m * n) time, O(m + n) space.
    rows = len(rack)
    cols = len(rack[0])
    per_row = [0] * rows
    per_col = [0] * cols
    for i in range(rows):
        line = rack[i]
        for j in range(cols):
            if line[j] == 1:
                per_row[i] += 1
                per_col[j] += 1
    found = 0
    for i in range(rows):
        if per_row[i] != 1:
            continue
        for j in range(cols):
            if rack[i][j] == 1 and per_col[j] == 1:
                found += 1
    return found

The obvious version

For each slot holding a pallet, walk its entire aisle and its entire column looking for another pallet, and count the slot when none is found. O(m * n * (m + n)).

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_count_isolated_pallets(rack):
    rows = len(rack)
    cols = len(rack[0])
    found = 0
    for i in range(rows):
        for j in range(cols):
            if rack[i][j] != 1:
                continue
            alone = True
            for jj in range(cols):
                if jj != j and rack[i][jj] == 1:
                    alone = False
            for ii in range(rows):
                if ii != i and rack[ii][j] == 1:
                    alone = False
            if alone:
                found += 1
    return found

How this problem was checked

Reference solution executed against 13 test cases — 13 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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