0106HardHash TableTreeDepth-First SearchBreadth-First SearchSortingBinary Tree

Notch Readout of a Hanging Mobile

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Trains the technique from

LeetCode 987Vertical Order Traversal of a Binary Tree

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 studio suspends a mobile from one hook. Every piece of the mobile carries a weight and dangles at most two further pieces beneath it, one on a left arm and one on a right arm. An arm carries its piece one notch sideways and one notch further down: a left arm one notch to the left, a right arm one notch to the right. The piece on the hook occupies notch 0 at the top level.

The sculpture reaches you as pieces, a listing that walks the mobile one level at a time from the hook downward and, within a level, from left to right. Slot 0 holds the piece on the hook. Each piece that appears in the listing claims the next two free slots, its left arm first and its right arm second. A slot holding null carries no piece and claims no slots of its own; null slots at the very end of the listing may be dropped.

A photographer wants the mobile written out notch by notch. Produce one group per occupied notch, and place the groups from the leftmost notch to the rightmost. Inside a group, write the weights from the highest piece down to the lowest. Should two pieces of one group hang at the same level, write the lighter weight ahead of the heavier one.

Examples

Example 1

Input: pieces = [12, 5, 40, null, null, 33, 9]
Output: [[5], [12, 33], [40], [9]]

Notch -1 holds only the piece weighing 5. Notch 0 holds the hook piece 12 at the top level and 33 two levels down. Notches 1 and 2 hold 40 and 9.

Example 2

Input: pieces = [4, 7, 7, 2, 6, 1, 3]
Output: [[2], [7], [4, 1, 6], [7], [3]]

Notch 0 collects the hook piece 4 and, two levels down, both 6 and 1. Those two hang at the same level in the same notch, so the lighter 1 is written ahead of 6.

Example 3

Input: pieces = [806]
Output: [[806]]

A single piece hangs on the hook, so one notch is occupied and it holds one weight.

Constraints

The number of pieces in the mobile is in the range [1, 1000].
0 <= weight of a piece <= 1000
`pieces` is the level-by-level listing described above, so slot 0 always holds a piece.

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 mobile_column_readout(pieces: list[int | None]) -> list[list[int]]:

Java

public List<List<Integer>> mobileColumnReadout(Integer[] pieces)

The intended approach

Time O(n log n)Space O(n)

Rebuild the arm links from the level-by-level listing, then walk the mobile once while carrying each piece's notch and level, recording a triple of notch, level and weight. Sorting those triples lexicographically puts notches in left-to-right order, orders each notch from top to bottom, and settles same-level ties by the lighter weight, so a single pass over the sorted triples splits them into groups whenever the notch changes.

Pythonentry point: mobile_column_readout

def mobile_column_readout(pieces):
    n = len(pieces)
    left = [-1] * n
    right = [-1] * n
    listed = [0]
    slot = 1
    head = 0
    while head < len(listed):
        cur = listed[head]
        head += 1
        if slot < n:
            if pieces[slot] is not None:
                left[cur] = slot
                listed.append(slot)
            slot += 1
        if slot < n:
            if pieces[slot] is not None:
                right[cur] = slot
                listed.append(slot)
            slot += 1
    entries = []
    stack = [(0, 0, 0)]
    while stack:
        idx, notch, depth = stack.pop()
        entries.append((notch, depth, pieces[idx]))
        if left[idx] >= 0:
            stack.append((left[idx], notch - 1, depth + 1))
        if right[idx] >= 0:
            stack.append((right[idx], notch + 1, depth + 1))
    entries.sort()
    groups = []
    current = None
    for notch, depth, weight in entries:
        if notch != current:
            groups.append([])
            current = notch
        groups[-1].append(weight)
    return groups

The obvious version

Recursive walk to collect notch, level and weight for every piece, then build the answer by repeated linear scans: find the smallest remaining notch, split off its records, and pull them out one at a time by selecting the smallest level-then-weight record each round. O(n^2).

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_mobile_columns(pieces):
    n = len(pieces)
    kids = {}
    listed = [0]
    slot = 1
    head = 0
    while head < len(listed):
        cur = listed[head]
        head += 1
        kids[cur] = [-1, -1]
        for side in (0, 1):
            if slot < n:
                if pieces[slot] is not None:
                    kids[cur][side] = slot
                    listed.append(slot)
                slot += 1
    records = []

    def walk(idx, notch, depth):
        records.append([notch, depth, pieces[idx]])
        for side in (0, 1):
            nxt = kids[idx][side]
            if nxt >= 0:
                walk(nxt, notch - 1 if side == 0 else notch + 1, depth + 1)

    walk(0, 0, 0)
    remaining = list(records)
    groups = []
    while remaining:
        low = remaining[0][0]
        for rec in remaining:
            if rec[0] < low:
                low = rec[0]
        column = []
        rest = []
        for rec in remaining:
            if rec[0] == low:
                column.append(rec)
            else:
                rest.append(rec)
        remaining = rest
        picked = []
        while column:
            best = 0
            for i in range(1, len(column)):
                if (column[i][1], column[i][2]) < (column[best][1], column[best][2]):
                    best = i
            picked.append(column.pop(best)[2])
        groups.append(picked)
    return groups

How this problem was checked

Reference solution executed against 8 test cases — 8 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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