1073EasyTreeDepth-First SearchString MatchingBinary TreeHash Function

Does the Rig Hold This Section

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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.

Two cargo rigs are given as the flat lists rig and section. Each hangs from a single top joint, every joint carries at most two joints below it in a first and a second slot, and either slot may be empty.

A flat list is read level by level: its first entry is the top joint's load, and reading left to right, every entry that is not null claims the next two unused positions as its first and second slot in that order, while null marks an empty slot and claims no positions of its own.

Return whether some joint of rig, taken together with everything hanging below it, matches section exactly: the same shape, slot for slot, and the same loads throughout.

Examples

Example 1

Input: rig = [1, 2, 3, 4, 5], section = [2, 4, 5]
Output: true

The joint loaded 2 carries 4 in its first slot and 5 in its second, which is exactly the section asked about.

Example 2

Input: rig = [1, 2], section = [1]
Output: false

The section is a single bare joint loaded 1, but the rig's only joint loaded 1 carries another joint below it, so the shapes differ.

Example 3

Input: rig = [7], section = [7]
Output: true

The whole rig is itself a single bare joint matching the section.

Constraints

1 <= rig.length <= 6000
1 <= section.length <= 3000
-10000 <= rig[i] <= 10000
-10000 <= section[i] <= 10000
The first entry of each list is not null.

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 is_subtree(rig: list, section: list) -> bool:

Java

public boolean isSubtree(Integer[] rig, Integer[] section)

The intended approach

Time O(n * m) for rigs of n and m joints.Space O(n + m) for the slots and the walk's depth.

Read both lists into slots, level by level, so each joint knows where its two slots sit. That is one pass apiece. Then try every joint of the larger rig as the place the section might hang, and for each one walk the two rigs together from that joint and the section's top. A pair matches when both slots are empty, or both are filled with the same load and the two pairs of slots below them match in turn. Any other case fails at once, including one slot being empty while the other is filled, which is what makes shape count as much as the loads. Trying each joint costs the section's size at worst, so the whole thing is the two sizes multiplied together, which is comfortable at these lengths.

Pythonentry point: is_subtree

def is_subtree(rig, section):

    def read(flat):
        n = len(flat)
        left = {}
        right = {}
        if n == 0:
            return left, right, None, []
        order = [0]
        level = [0]
        at = 1
        while level and at < n:
            nxt = []
            for owner in level:
                if at < n:
                    if flat[at] is not None:
                        left[owner] = at
                        nxt.append(at)
                    at += 1
                if at < n:
                    if flat[at] is not None:
                        right[owner] = at
                        nxt.append(at)
                    at += 1
            order.extend(nxt)
            level = nxt
        return left, right, 0, order

    a_left, a_right, a_top, a_order = read(rig)
    b_left, b_right, b_top, _b_order = read(section)
    if b_top is None:
        return True
    if a_top is None:
        return False
    def alike(x, y):
        if x is None and y is None:
            return True
        if x is None or y is None:
            return False
        if rig[x] != section[y]:
            return False
        if not alike(a_left.get(x), b_left.get(y)):
            return False
        return alike(a_right.get(x), b_right.get(y))
    for pos in a_order:
        if alike(pos, b_top):
            return True
    return False

The obvious version

Spell each section out as a string instead of walking two rigs together. A joint's spelling is its load followed by the spellings of its two slots, with a marker for an empty slot, all in brackets so the pieces cannot run together. Two sections match exactly when their spellings match, so spell the wanted section once and compare it against the spelling of every joint of the larger rig. Costs the length of the spellings rather than an early exit on the first mismatch.

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_is_subtree(rig, section):

    def read(flat):
        n = len(flat)
        left = {}
        right = {}
        if n == 0:
            return left, right, None, []
        order = [0]
        level = [0]
        at = 1
        while level and at < n:
            nxt = []
            for owner in level:
                if at < n:
                    if flat[at] is not None:
                        left[owner] = at
                        nxt.append(at)
                    at += 1
                if at < n:
                    if flat[at] is not None:
                        right[owner] = at
                        nxt.append(at)
                    at += 1
            order.extend(nxt)
            level = nxt
        return left, right, 0, order

    a_left, a_right, a_top, a_order = read(rig)
    b_left, b_right, b_top, _b_order = read(section)
    if b_top is None:
        return True
    if a_top is None:
        return False
    def spell(flat, left, right, pos):
        if pos is None:
            return "#"
        return "(" + str(flat[pos]) + spell(flat, left, right, left.get(pos)) + \
            spell(flat, left, right, right.get(pos)) + ")"
    want = spell(section, b_left, b_right, b_top)
    for pos in a_order:
        if spell(rig, a_left, a_right, pos) == want:
            return True
    return False

How this problem was checked

Reference solution executed against 15 test cases — 15 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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