0804MediumArrayGreedyBit ManipulationMatrix

Maximum Relay Cabinet Total

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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.

A control cabinet holds a stack of relay racks. Every rack carries the same run of relays side by side, and each relay is either closed, written 1, or open, written 0. racks[i][j] is the state of the relay at position j on rack i.

Each rack reports a reading: read its relays from left to right as the digits of a binary number, so the leftmost relay is the most significant digit. The cabinet total is the sum of all the rack readings.

A maintenance panel offers two kinds of switch, and you may throw as many of them as you like, in any order, including none at all:

a rack switch picks one rack and flips every relay on it, each closed relay becoming open and each open relay becoming closed;
a column switch picks one position and flips the relay at that position on every rack at once.

Return the largest cabinet total you can reach.

Examples

Example 1

Input: racks = [[0, 1, 1, 0], [1, 0, 0, 1], [1, 1, 1, 0]]
Output: 38

Throw the rack switch on rack 0, then the column switches at positions 1 and 2. The racks then read 1111, 1111 and 1000, which are 15, 15 and 8, for a cabinet total of 38.

Example 2

Input: racks = [[1, 0, 1], [1, 1, 0]]
Output: 11

Left alone the two racks read 101 and 110, which are 5 and 6, for a total of 11.

Example 3

Input: racks = [[1, 0, 0, 1, 0]]
Output: 31

With a single rack in the cabinet, throwing the column switches at positions 1, 2 and 4 leaves it reading 11111, which is 31.

Constraints

1 <= racks.length <= 20
1 <= racks[i].length <= 20
0 <= racks[i][j] <= 1
Every rack carries the same number of relays.

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 best_cabinet_total(racks: list[list[int]]) -> int:

Java

public int bestCabinetTotal(int[][] racks)

The intended approach

Time O(m * n) for m racks of n relays: one pass over every relay.Space O(1); only running counts are kept.

The leading relay of a rack outweighs every other relay on it combined, so each rack switch is decided on its own: throw it exactly when that rack's leading relay is open. With those decisions fixed, the leading column is all closed and contributes the number of racks times the leading weight. A column switch after that touches nothing but its own column, whose contribution is the count of closed relays in it times its weight, so for each remaining column count the closed relays under the rack switches already chosen and keep whichever of the two orientations closes more.

Pythonentry point: best_cabinet_total

def best_cabinet_total(racks):
    # The leading relay carries more weight than all the rest put together, so a rack
    # whose leading relay is open is worth flipping outright; that settles every rack
    # switch. Each column switch then only touches its own column, whose contribution
    # is the number of closed relays in it times that column's weight, so take
    # whichever of the two orientations closes more relays.
    # Time O(m * n), space O(1) beyond the input.
    rows = len(racks)
    cols = len(racks[0])
    total = rows * (1 << (cols - 1))
    for c in range(1, cols):
        closed = 0
        for r in range(rows):
            bit = racks[r][c]
            if racks[r][0] == 0:
                bit ^= 1
            closed += bit
        opened = rows - closed
        total += (closed if closed > opened else opened) * (1 << (cols - 1 - c))
    return total

The obvious version

Pack each column into one integer, one bit per rack, so a choice of rack switches is a single bit pattern and xoring it against a column gives that column's states. Try every one of the 2^m rack-switch patterns; for each, a column switch only ever affects its own column, so add the better of that column's two orientations. Keep the best total seen.

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_best_cabinet_total(racks):
    rows = len(racks)
    cols = len(racks[0])
    # one bit per rack for each column, so a rack-switch choice is a single xor
    column = [0] * cols
    for r in range(rows):
        for c in range(cols):
            if racks[r][c]:
                column[c] |= 1 << r
    weight = [1 << (cols - 1 - c) for c in range(cols)]
    best = 0
    for choice in range(1 << rows):
        total = 0
        for c in range(cols):
            closed = bin(column[c] ^ choice).count("1")
            opened = rows - closed
            total += (closed if closed > opened else opened) * weight[c]
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 17 test cases — 17 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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