0640MediumArrayHash Table

Cable Drop Through The Panel Partition

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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.

An office partition is described by wall, listed from the top row downwards. Row i is the list of panel widths in that row, read from the left edge to the right edge, and every row spans the same total width.

A cable is dropped straight down the partition, from the top of the top row to the bottom of the bottom row, at one position strictly inside the partition: the two outer edges are not allowed. In each row the cable either slips down the seam where two neighbouring panels meet, touching neither of them, or it passes through the body of one panel. Return the smallest number of panels a single cable can pass through.

Examples

Example 1

Input: wall = [[4, 4], [2, 6], [2, 6], [3, 5]]
Output: 2

Every row is 8 wide. A cable dropped 2 from the left edge slips down the seam of row 1 and the seam of row 2, and passes through a panel in row 0 and in row 3, giving 2 panels.

Example 2

Input: wall = [[1, 1, 1], [3], [1, 2]]
Output: 1

Every row is 3 wide. A cable dropped 1 from the left edge slips down a seam in row 0 and a seam in row 2, and passes through the single panel of row 1, giving 1 panel.

Example 3

Input: wall = [[6], [6], [6]]
Output: 3

Each row holds one panel 6 wide and has no seam, and the cable is not allowed on either outer edge, so it passes through a panel in all 3 rows.

Constraints

n == wall.length
1 <= n <= 10^4
1 <= wall[i].length <= 10^4
1 <= sum of all wall[i].length <= 2 * 10^4
Every row of wall has the same total width.
1 <= wall[i][j] <= 2^31 - 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 least_bricks(wall: list[list[int]]) -> int:

Java

public int leastBricks(List<List<Integer>> wall)

The intended approach

Time O(P) where P is the total number of panels in the partition.Space O(S) for the tally, with S the number of distinct seam distances.

Walk each row accumulating panel widths, which gives the distance from the left edge to every seam in that row, and tally each such distance across the whole partition. The last accumulation of a row lands on the right edge of the partition, which is not an allowed cable position, so that one is skipped. A cable dropped at a tallied distance avoids one panel in every row that has a seam there and passes through one panel in every other row, so the answer is the number of rows minus the largest tally. When no seam exists the largest tally stays at zero and every row is crossed.

Pythonentry point: least_bricks

def least_bricks(wall):
    # O(total panels) time, O(distinct seams) space: tally, over all rows, how
    # many rows have a seam at each distance from the left edge. The seam at the
    # far right edge is skipped because the cable may not be dropped there. A
    # cable dropped on a seam misses that row's panels, so a cable on the most
    # popular seam passes through the row count minus that tally, and a wall with
    # no seam at all leaves a tally of zero.
    seams = {}
    for row in wall:
        pos = 0
        for j in range(len(row) - 1):
            pos += row[j]
            seams[pos] = seams.get(pos, 0) + 1
    best = 0
    for pos in seams:
        if seams[pos] > best:
            best = seams[pos]
    return len(wall) - best

The obvious version

Collect every seam distance appearing anywhere in the partition as a candidate cable position. For each candidate, walk every row from its left edge, accumulating widths, and decide whether the candidate lands on a seam of that row or inside a panel. Keep the candidate with the fewest crossings, and fall back to crossing every row when there is no candidate at all.

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_least_bricks(wall):
    width = 0
    for w in wall[0]:
        width += w
    cands = set()
    for row in wall:
        pos = 0
        for w in row:
            pos += w
            if pos < width:
                cands.add(pos)
    if not cands:
        return len(wall)
    best = None
    for cut in cands:
        crossed = 0
        for row in wall:
            pos = 0
            through = True
            for w in row:
                pos += w
                if pos == cut:
                    through = False
                    break
                if pos > cut:
                    break
            if through:
                crossed += 1
        if best is None or crossed < best:
            best = crossed
    return best

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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