0876MediumArrayHash TableStringTreeDepth-First SearchBreadth-First SearchBinary LiftingLowest Common Ancestor

Innermost Zone Holding Both Sites

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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 survey is organised into zones. Each entry of regions names a zone first and then the zones lying directly inside it. A zone lies inside another when it is named directly inside it, or inside something that does.

No zone lies directly inside more than one zone, and one zone holds all the others.

Given two distinct zone names region1 and region2, return the name of the innermost zone that holds both of them. A zone counts as holding itself.

Examples

Example 1

Input: regions = [["World", "North", "South"], ["North", "Alba", "Cymru"], ["South", "Kernow"]], region1 = "Alba", region2 = "Cymru"
Output: "North"

Both sites lie directly inside "North", so that is the innermost zone holding them both.

Example 2

Input: regions = [["World", "North", "South"], ["North", "Alba", "Cymru"], ["South", "Kernow"]], region1 = "Alba", region2 = "Kernow"
Output: "World"

One site lies inside "North" and the other inside "South", and the innermost zone holding both of those is "World".

Example 3

Input: regions = [["World", "North", "South"], ["North", "Alba", "Cymru"], ["South", "Kernow"]], region1 = "Alba", region2 = "North"
Output: "North"

A zone counts as holding itself, and "Alba" lies inside "North", so "North" holds them both.

Constraints

2 <= regions.length <= 10^4
2 <= regions[i].length <= 20
1 <= regions[i][j].length <= 20
1 <= region1.length <= 20
1 <= region2.length <= 20
region1 and region2 are different names
Every name consists of English letters only
No zone lies directly inside more than one zone, and one zone holds all the others

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 innermost_holder(regions: list[list[str]], region1: str, region2: str) -> str:

Java

public String innermostHolder(List<List<String>> regions, String region1, String region2)

The intended approach

Time O(total names) to record the nesting, then O(depth) for each walk.Space O(total names) for the nesting map and the first chain.

Record, for each zone named as lying inside another, which zone that is, which the entries give directly. Then walk outward from the first site collecting every zone that holds it, itself included, into a set. Walking outward from the second site and stopping at the first zone already in that set gives the innermost zone holding both, because the walk from the second site meets the first site's chain at the deepest point they share. Both walks are plain loops, so a deeply nested survey is no problem.

Pythonentry point: innermost_holder

def innermost_holder(regions, region1, region2):
    outside = {}
    for entry in regions:
        holder = entry[0]
        for inner in entry[1:]:
            outside[inner] = holder
    chain = set()
    here = region1
    while True:
        chain.add(here)
        if here not in outside:
            break
        here = outside[here]
    here = region2
    while here not in chain:
        here = outside[here]
    return here

The obvious version

Build the full list of zones holding each site by walking outward, then compare the two lists from their outermost ends inwards, keeping the last zone they agree on. Materialises both chains instead of stopping at the first shared zone.

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_innermost_holder(regions, region1, region2):
    outside = {}
    for entry in regions:
        for inner in entry[1:]:
            outside[inner] = entry[0]

    def chain(name):
        out = [name]
        while out[-1] in outside:
            out.append(outside[out[-1]])
        out.reverse()
        return out

    a = chain(region1)
    b = chain(region2)
    best = a[0]
    i = 0
    while i < len(a) and i < len(b) and a[i] == b[i]:
        best = a[i]
        i += 1
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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