1099MediumTreeDepth-First SearchBinary TreeDP on Trees

Pods Whose Cluster Reads the Same

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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 cluster hangs from a single top pod. Every pod holds at most two pods below it in a first and a second slot, and either slot may be empty. Each pod shows a reading.

A flat list is read level by level: its first entry is the top pod's reading, 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. An empty list means there is no cluster at all.

A pod is plain when every pod hanging at or below it shows the same reading as the pod itself. A pod with both slots empty is therefore always plain.

Return how many pods are plain.

Examples

Example 1

Input: cluster = [1, 1, 1, 2, 1, 1, 1]
Output: 5

The four pods at the bottom are plain on their own. The pod in the top pod's second slot has both its pods reading 1 like itself, so it is plain too. The pod in the first slot has a 2 hanging under it, which spoils that pod and the top pod with it.

Example 2

Input: cluster = [1, 2]
Output: 1

The pod hanging in the first slot is plain on its own. The top pod is not, because the reading below it differs.

Example 3

Input: cluster = [7, 7, 7, 7, 7, 7, 7]
Output: 7

Every pod shows the same reading, so every one of the seven is plain.

Constraints

0 <= cluster.length <= 2000
The cluster holds between 0 and 1000 pods.
-1000 <= cluster[i] <= 1000
The first entry of the 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 count_unival_subtrees(cluster: list) -> int:

Java

public int countUnivalSubtrees(Integer[] cluster)

The intended approach

Time O(n) over the pods.Space O(n) for the slot links and the plainness flags.

Being plain is a fact about a pod and everything under it, and it settles from the bottom upwards. A pod with both slots empty is plain outright. Otherwise the pod is plain exactly when each pod directly below it is itself plain and shows the same reading as the pod. That second condition is the whole trick: a matching reading directly below is not enough on its own, because something deeper down may differ, and a plain pod below is not enough either, because its reading may differ from the pod above. Both have to hold. So read the flat list into slot links, walk the pods in reverse of the order they were discovered, which puts every pod after everything below it, and settle each pod's plainness from its slots. Count the ones that come out plain.

Pythonentry point: count_unival_subtrees

def count_unival_subtrees(cluster):

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

    if n == 0:
        return 0
    plain = {}
    total = 0
    for pos in reversed(order):
        ok = True
        for child in (left.get(pos), right.get(pos)):
            if child is not None:
                if not plain[child] or cluster[child] != cluster[pos]:
                    ok = False
        plain[pos] = ok
        if ok:
            total += 1
    return total

The obvious version

Take each pod in turn and walk everything hanging at or below it, comparing every reading found against that pod's own and stopping the walk on the first difference. Costs the pod count squared rather than one sweep upwards.

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_count_unival_subtrees(cluster):

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

    if n == 0:
        return 0
    total = 0
    for pos in order:
        stack = [pos]
        same = True
        while stack:
            cur = stack.pop()
            if cluster[cur] != cluster[pos]:
                same = False
                break
            for child in (left.get(cur), right.get(cur)):
                if child is not None:
                    stack.append(child)
        if same:
            total += 1
    return total

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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