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Twin Splitter Cabinets

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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 broadcast site keeps its signal splitters in wall cabinets. One splitter sits at the feed point, and every splitter passes its signal on to at most two further splitters: one hanging off its port A and one hanging off its port B. Each splitter is engraved with a trim in decibels, which runs negative when the splitter cuts the level.

Because the harness passes plain JSON, a cabinet reaches you as an array of engraved trims. The first entry is the feed-point splitter. After that the entries arrive in pairs, giving the port A splitter and then the port B splitter of each splitter already listed, taken in the order they were listed. A port with nothing hanging off it is written null, and a null brings no pair of its own. An empty array means the cabinet is bare.

One consequence of that array form is worth stating: empty ports at the very end may be written out or left off, so a single cabinet can be described by more than one array.

You are given built, the cabinet already on the wall, and spare, a cabinet from the store room. Return true when an engineer could hang the spare in place of the built one, meaning the two hold splitters in exactly the same arrangement with the same trim engraved at every matching position, and false otherwise. Two bare cabinets are interchangeable.

Examples

Example 1

Input: built = [9, -2, 5, null, 7], spare = [9, -2, 5, 7]
Output: false

In the built cabinet the splitter trimmed -2 has nothing on port A and the splitter trimmed 7 on port B. In the spare, that same splitter carries 7 on port A instead, so the arrangements do not line up.

Example 2

Input: built = [4, -6], spare = [4, -6, null]
Output: true

Both cabinets hold a feed splitter trimmed 4 with a splitter trimmed -6 on port A and nothing on port B; the spare array simply writes that empty port out.

Example 3

Input: built = [12, 3, 8], spare = [12, 3, 9]
Output: false

Every position matches up, but the port B splitter is engraved 8 on the wall and 9 in the store room.

Constraints

The number of splitters in each cabinet is in the range [0, 100].
-10^4 <= trim <= 10^4
Every entry is either an integer trim or `null`; the first entry of a non-empty array is an integer.

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 twin_splitter_cabinets(built: list, spare: list) -> bool:

Java

public boolean twinSplitterCabinets(Integer[] built, Integer[] spare)

The intended approach

Time O(n)Space O(n)

Decode each array into splitters wired to their port A and port B followers, then walk the two cabinets side by side from the feed point, pairing each splitter with the splitter sitting at the same position in the other cabinet. A pair fails when one side has a splitter and the other has an empty port, or when both have a splitter but the engraved trims differ; otherwise the two port pairs are pushed on for later comparison. An explicit stack drives the walk, so a lopsided cabinet at the constraint ceiling costs nothing extra.

Pythonentry point: twin_splitter_cabinets

def twin_splitter_cabinets(built, spare):
    # Time: O(n) -- each cabinet is decoded once and walked once.
    # Space: O(n) -- the decoded splitters plus the pairing stack.
    def decode(arr):
        if not arr or arr[0] is None:
            return None
        root = {"trim": arr[0], "a": None, "b": None}
        listed = [root]
        at = 0
        i = 1
        while at < len(listed) and i < len(arr):
            node = listed[at]
            at += 1
            for port in ("a", "b"):
                if i >= len(arr):
                    break
                value = arr[i]
                i += 1
                if value is not None:
                    node[port] = {"trim": value, "a": None, "b": None}
                    listed.append(node[port])
        return root

    stack = [(decode(built), decode(spare))]
    while stack:
        x, y = stack.pop()
        if x is None and y is None:
            continue
        if x is None or y is None:
            return False
        if x["trim"] != y["trim"]:
            return False
        stack.append((x["a"], y["a"]))
        stack.append((x["b"], y["b"]))
    return True

The obvious version

Decode each cabinet into an index-addressed store, spell out a bracketed text form of the whole cabinet with a marker for every empty port, and compare the two strings.

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_twin_splitter_cabinets(built, spare):
    def decode(arr):
        if not arr or arr[0] is None:
            return []
        store = [[arr[0], -1, -1]]
        queue = [0]
        head = 0
        i = 1
        while head < len(queue) and i < len(arr):
            owner = queue[head]
            head += 1
            for slot in (1, 2):
                if i >= len(arr):
                    break
                value = arr[i]
                i += 1
                if value is not None:
                    store.append([value, -1, -1])
                    store[owner][slot] = len(store) - 1
                    queue.append(len(store) - 1)
        return store

    def spell(store, at):
        if at < 0:
            return "."
        node = store[at]
        return "(" + str(node[0]) + spell(store, node[1]) + spell(store, node[2]) + ")"

    a = decode(built)
    b = decode(spare)
    left = "." if not a else spell(a, 0)
    right = "." if not b else spell(b, 0)
    return left == right

How this problem was checked

Reference solution executed against 14 test cases — 14 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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