Minimize Malware Spread by Removing a Node

There are gNodes servers numbered from 1 through gNodes. The parallel arrays gFrom and gTo describe bidirectional connections: edge i joins gFrom[i] and gTo[i].
The binary array malware describes the initial infection state. Server i is initially infected exactly when malware[i - 1] == 1. An infected server repeatedly infects every directly connected non-infected server until no additional infection is possible.
Before propagation begins, remove exactly one server, whether infected or not, together with all of its incident connections. Return the label of the server whose removal minimizes the number of infected servers remaining after propagation. If several removals produce the same minimum, return the smallest server label.

Function
minimizeMalwareSpread(gNodes: int, gFrom: int[], gTo: int[], malware: int[]) → int

Examples
Example 1
gNodes = 9
gFrom = [1,2,4,6,7]
gTo = [2,3,5,7,8]
malware = [0,0,1,0,1,0,0,0,0]
return = 3
Removing infected server 3 protects servers 1 and 2. Only servers 4 and 5 become infected, which is fewer than for any other removal.
Example 2
gNodes = 6
gFrom = [1,1,1,1,1]
gTo = [2,3,4,5,6]
malware = [0,1,1,0,0,0]
return = 1
Removing the uninfected center server 1 separates the two infected leaves from every other leaf, leaving only servers 2 and 3 infected.

Constraints
1 ≤ gNodes ≤ 500
0 ≤ gFrom.length = gTo.length ≤ 5000
Every edge endpoint is between 1 and gNodes.
malware.length = gNodes.
Every value in malware is 0 or 1, and at least one server is initially infected.

The trick is that nothing clever is required. For each candidate server r from 1 to gNodes, build or reuse an adjacency list, skip r during traversal, and start a multi-source BFS from every infected server that isn't r. Count visited nodes. Track the minimum count, and break ties by smallest label, which a left-to-right loop with a strict less-than gives you for free. The pitfalls are real. Removing an infected server also removes it from the infected count, so don't count it. Don't restrict candidates to infected nodes. Remember the graph may be disconnected and may have duplicate edges. Cost is about N times (N plus E), roughly 500 times 5500, which is easy. If you blank on the loop structure mid-assessment, StealthCoder can hand you the working skeleton so you only verify the tie-breaking and edge cases.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.