0912MediumArrayMatrixPrefix Sum

Corner Blocks With Matching Marks

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Trains the technique from

LeetCode 3212Count Submatrices With Equal Frequency of X and Y

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 board is given as grid, each cell holding 'X', 'Y' or '.'.

A corner block is a rectangle of cells whose top-left cell is the board's own top-left cell. A corner block balances when it holds as many 'X' marks as 'Y' marks and holds at least one 'X'.

Return how many corner blocks balance.

Examples

Example 1

Input: grid = [["X", ".", "Y", "X"], [".", "Y", "X", "."], ["Y", "X", ".", "Y"]]
Output: 7

Each corner block is named by its bottom-right cell. Those whose X and Y counts come out level, with at least one X present, are the ones counted.

Example 2

Input: grid = [["X", "Y"]]
Output: 1

The block ending at the first cell holds one X and no Y, so it does not balance. The block covering both cells holds one of each, so it does.

Example 3

Input: grid = [["."]]
Output: 0

The only block holds no X at all, so nothing balances.

Constraints

1 <= grid.length <= 1000
1 <= grid[i].length <= 1000
Every row of grid has the same length
grid[i][j] is 'X', 'Y' or '.'

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 number_of_submatrices(grid: list[list[str]]) -> int:

Java

public int numberOfSubmatrices(char[][] grid)

The intended approach

Time O(m * n) for an m by n board, one sweep.Space O(m * n) for the two running tallies, or O(n) keeping only the previous row.

A corner block is fixed by its bottom-right cell, so there is exactly one block per cell and the answer is a count over cells. What each block needs is its own tally of the two marks, and those tallies grow in a simple way: the block ending at a cell is the block above it plus the block to its left, less the block above-left counted twice over, plus the cell itself. Carrying two such running tallies across the board gives every block's counts in one sweep. A block then balances when its two counts agree and its `'X'` count is not zero, which is one comparison per cell. Note that once the counts agree, demanding an `'X'` and demanding a `'Y'` come to the same thing, so either check rules out the blocks holding nothing but dots.

Pythonentry point: number_of_submatrices

def number_of_submatrices(grid):
    m = len(grid)
    n = len(grid[0])
    xs = [[0] * (n + 1) for _ in range(m + 1)]
    ys = [[0] * (n + 1) for _ in range(m + 1)]
    total = 0
    for r in range(m):
        for c in range(n):
            cell = grid[r][c]
            xs[r + 1][c + 1] = (xs[r][c + 1] + xs[r + 1][c] - xs[r][c]
                                + (1 if cell == "X" else 0))
            ys[r + 1][c + 1] = (ys[r][c + 1] + ys[r + 1][c] - ys[r][c]
                                + (1 if cell == "Y" else 0))
            if xs[r + 1][c + 1] > 0 and xs[r + 1][c + 1] == ys[r + 1][c + 1]:
                total += 1
    return total

The obvious version

Take every corner block in turn and count its marks by walking all of its cells. Quartic in the board's side lengths and it carries nothing from one block to the next.

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_number_of_submatrices(grid):
    m = len(grid)
    n = len(grid[0])
    total = 0
    for bottom in range(m):
        for right in range(n):
            x_count = 0
            y_count = 0
            for r in range(bottom + 1):
                for c in range(right + 1):
                    if grid[r][c] == "X":
                        x_count += 1
                    elif grid[r][c] == "Y":
                        y_count += 1
            if x_count > 0 and x_count == y_count:
                total += 1
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 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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