0480MediumArrayDynamic ProgrammingMatrix

Working Square Blocks on the Solar Array

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Trains the technique from

LeetCode 1277Count Square Submatrices with All Ones

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 solar array is laid out as a rectangular grid of panels. panels[r][c] is 1 when the panel in row r, column c is generating and 0 when it is dead.

A working block is any square patch of the grid, lined up with the rows and columns, in which every panel is generating. Its side may be any length from one panel upwards, and two working blocks count separately whenever their positions or sides differ, even where they overlap.

Return the total number of working blocks in the array.

Examples

Example 1

Input: panels = [[1, 1, 1], [1, 0, 1], [1, 1, 1]]
Output: 8

Eight panels are generating, each giving a working block of side one. The dead panel in the middle falls inside every possible larger square, so nothing bigger counts.

Example 2

Input: panels = [[1, 1, 1], [1, 1, 1], [1, 1, 1]]
Output: 14

Nine blocks of side one, four of side two and one of side three all count here.

Example 3

Input: panels = [[1, 1, 0, 1], [1, 1, 1, 1], [0, 1, 1, 1]]
Output: 13

Ten panels are generating. Three squares of side two also fit: one in the top-left corner and two more across the lower right of the array.

Constraints

1 <= panels.length <= 300
1 <= panels[0].length <= 300
0 <= panels[r][c] <= 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_working_squares(panels: list[list[int]]) -> int:

Java

public int countWorkingSquares(int[][] panels)

The intended approach

Time O(rows * cols)Space O(cols) for the two rows of figures

Fix a panel and ask how large the biggest working block can be whose bottom-right corner is that panel. If the panel is dead the answer is zero; otherwise it is one more than the smallest of the three figures already worked out for the panel above, the panel to the left and the panel diagonally back. That figure also counts the working blocks whose corner is that panel, one per side length, so adding it up over the grid gives the total. The sweep goes row by row and holds the row above alongside the row being filled, so only two rows of figures are ever alive.

Pythonentry point: count_working_squares

def count_working_squares(panels):
    # For each panel, work out the largest fully-working square whose bottom-right
    # corner it is; that number is also how many working squares end there.
    # Two rows are held at a time: the row above and the row being filled.
    # time O(rows * cols), space O(cols)
    rows = len(panels)
    cols = len(panels[0])
    above = [0] * cols
    total = 0
    for r in range(rows):
        current = [0] * cols
        for c in range(cols):
            if panels[r][c] == 0:
                current[c] = 0
                continue
            if r == 0 or c == 0:
                current[c] = 1
            else:
                small = current[c - 1]
                if above[c] < small:
                    small = above[c]
                if above[c - 1] < small:
                    small = above[c - 1]
                current[c] = small + 1
            total += current[c]
        above = current
    return total

The obvious version

Take every panel as a top-left corner and every side length that still fits inside the grid, then read all the panels of that square to see whether any is dead. Count the squares that come back whole.

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_working_squares(panels):
    rows = len(panels)
    cols = len(panels[0])
    total = 0
    for r in range(rows):
        for c in range(cols):
            reach = rows - r
            if cols - c < reach:
                reach = cols - c
            for side in range(1, reach + 1):
                whole = True
                for i in range(r, r + side):
                    for j in range(c, c + side):
                        if panels[i][j] == 0:
                            whole = False
                            break
                    if not whole:
                        break
                if whole:
                    total += 1
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 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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