0985HardArrayHash TableDepth-First SearchBreadth-First SearchUnion-FindGraph Theory

Which Faulty Unit to Take Offline

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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 rack of units is given as graph, where graph[i][j] is 1 when units i and j are wired together. Every unit is wired to itself, and the wiring runs both ways.

The units listed in initial start out faulty. A fault spreads along the wiring until every unit reachable from a faulty one is faulty too.

Exactly one unit is taken off the fault list before anything spreads; it stays in the rack and still carries the fault onward if something reaches it. Choose the one that leaves the fewest units faulty at the end, and return it. When two choices leave the same number faulty, return the smaller unit number.

Examples

Example 1

Input: graph = [[1, 1, 0, 0, 0], [1, 1, 0, 0, 0], [0, 0, 1, 1, 1], [0, 0, 1, 1, 1], [0, 0, 1, 1, 1]], initial = [1, 2]
Output: 2

Unit 1 sits alone with unit 0 in a group of two, and unit 2 sits in a group of three. Each group holds one faulty unit, so taking off unit 2 saves three while taking off unit 1 saves only two.

Example 2

Input: graph = [[1, 1], [1, 1]], initial = [0, 1]
Output: 0

Both units are wired together and both are faulty, so whichever is taken off the list the other still fills the group. The tie falls to the smaller number.

Example 3

Input: graph = [[1, 0, 0], [0, 1, 0], [0, 0, 1]], initial = [0, 1, 2]
Output: 0

No unit is wired to any other, so taking one off the list saves exactly that one, and all three choices are equal.

Constraints

2 <= graph.length <= 300
graph.length == graph[i].length
graph[i][j] is 0 or 1
graph[i][j] == graph[j][i]
graph[i][i] == 1
1 <= initial.length <= graph.length
0 <= initial[i] <= graph.length - 1
The unit numbers in initial are all different

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 min_malware_spread(graph: list[list[int]], initial: list[int]) -> int:

Java

public int minMalwareSpread(int[][] graph, int[] initial)

The intended approach

Time O(n^2) for n units: reading the wiring dominates.Space O(n) for the groups.

A fault fills whatever group of wired-together units it lands in, so the final count is the total size of every group holding at least one faulty unit. Which unit inside a group started it makes no difference. So find the groups first by joining wired units together. Then a group holding exactly one faulty unit is saved entirely by taking that unit off the list, and a group holding two or more is doomed whatever is taken off, since another faulty unit remains inside it. The best choice is therefore the single faulty unit whose group is largest, breaking a tie by the smaller unit number. When no group holds exactly one faulty unit, nothing can be saved and the rule falls back to the smallest number on the list.

Pythonentry point: min_malware_spread

def min_malware_spread(graph, initial):
    n = len(graph)
    owner = list(range(n))

    def find(x):
        while owner[x] != x:
            owner[x] = owner[owner[x]]
            x = owner[x]
        return x

    for i in range(n):
        for j in range(i + 1, n):
            if graph[i][j] == 1:
                a = find(i)
                b = find(j)
                if a != b:
                    owner[a] = b
    size = {}
    for i in range(n):
        root = find(i)
        size[root] = size.get(root, 0) + 1
    faulty = {}
    for unit in initial:
        root = find(unit)
        faulty[root] = faulty.get(root, 0) + 1
    best = None
    best_saved = -1
    for unit in sorted(initial):
        root = find(unit)
        saved = size[root] if faulty[root] == 1 else 0
        if saved > best_saved:
            best_saved = saved
            best = unit
    return best

The obvious version

Try taking each faulty unit off the list in turn, spread the fault from the rest by searching outwards through the wiring, count what ends up faulty, and keep the choice that leaves the fewest. Costs a full spread per candidate and groups nothing in advance.

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_min_malware_spread(graph, initial):
    from collections import deque
    n = len(graph)
    best = None
    best_count = None
    for removed in sorted(initial):
        start = [u for u in initial if u != removed]
        seen = set(start)
        queue = deque(start)
        while queue:
            node = queue.popleft()
            for nxt in range(n):
                if nxt == node:
                    continue
                if graph[node][nxt] == 1 and nxt not in seen:
                    seen.add(nxt)
                    queue.append(nxt)
        if best_count is None or len(seen) < best_count:
            best_count = len(seen)
            best = removed
    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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