0848MediumArrayGreedySortingMatrix

Widest Block After Reordering Columns

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Trains the technique from

LeetCode 1727Largest Submatrix With Rearrangements

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 punch card sheet is a grid of 0s and 1s given as matrix. The columns may be reordered freely, as often as you like, and reordering moves a whole column with all its entries.

A block is a rectangle of cells, taken from consecutive rows and consecutive columns of the reordered sheet, in which every cell is 1. Return the largest area a block can have, where area is the number of cells.

Examples

Example 1

Input: matrix = [[1, 0, 1], [1, 1, 0], [1, 0, 1], [1, 1, 1]]
Output: 4

The first column holds ones all the way down, giving a block one column wide and four rows tall, an area of 4.

Example 2

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

No two cells of ones share a row or a column band, so the largest block is a single cell.

Example 3

Input: matrix = [[1, 1, 1, 1]]
Output: 4

The single row is all ones, so the whole sheet is one block of area 4.

Constraints

1 <= matrix.length <= 10^5
1 <= matrix[0].length <= 10^5
Every row of matrix has the same length, and the sheet holds at most 10^5 cells
0 <= matrix[i][j] <= 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 widest_block(matrix: list[list[int]]) -> int:

Java

public int widestBlock(int[][] matrix)

The intended approach

Time O(R * C log C) for R rows and C columns, dominated by sorting each row's run lengths.Space O(C) for the run lengths.

Because whole columns move together, a block's rows are fixed but its columns can be gathered from anywhere, so what matters for a column is how far the run of ones reaches upward from a given row. Sweep the rows top to bottom keeping, for each column, the length of its unbroken run of ones ending at the current row, which resets to zero on a zero. Sorting those run lengths for the current row descending puts the tallest columns first, and a block of the k tallest columns has height equal to the k-th of them, so its area is that run length times k. Taking the largest such product over every row and every k is the answer.

Pythonentry point: widest_block

def widest_block(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    runs = [0] * cols
    best = 0
    for r in range(rows):
        for c in range(cols):
            runs[c] = runs[c] + 1 if matrix[r][c] == 1 else 0
        tall = sorted(runs, reverse=True)
        for k in range(cols):
            if tall[k] == 0:
                break
            area = tall[k] * (k + 1)
            if area > best:
                best = area
    return best

The obvious version

For every choice of top and bottom row, count how many columns are all ones over that whole band; those columns can be brought together, so the band contributes its height times that count. Keeps the largest. Cubic, but it never reasons about run lengths or sorting.

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_widest_block(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    best = 0
    for top in range(rows):
        for bottom in range(top, rows):
            good = 0
            for c in range(cols):
                solid = True
                for r in range(top, bottom + 1):
                    if matrix[r][c] != 1:
                        solid = False
                        break
                if solid:
                    good += 1
            area = good * (bottom - top + 1)
            if area > best:
                best = area
    return best

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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