Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given a network of n nodes represented as an n x n adjacency matrix graph, where the ith node is directly connected to the jth node if graph[i][j] == 1.
Some nodes initial are initially infected by malware. Whenever two nodes are directly connected, and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.
Suppose M(initial) is the final number of nodes infected with malware in the entire network after the spread of malware stops. We will remove exactly one node from initial.
Return the node that, if removed, would minimize M(initial). If multiple nodes could be removed to minimize M(initial), return such a node with the smallest index.
Note that if a node was removed from the initial list of infected nodes, it might still be infected later due to the malware spread.
Example 1:
Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1] Output: 0
Example 2:
Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2] Output: 0
Example 3:
Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2] Output: 1
Constraints:
n == graph.lengthn == graph[i].length2 <= n <= 300graph[i][j] is 0 or 1.graph[i][j] == graph[j][i]graph[i][i] == 11 <= initial.length <= n0 <= initial[i] <= n - 1initial are unique.Problem summary: You are given a network of n nodes represented as an n x n adjacency matrix graph, where the ith node is directly connected to the jth node if graph[i][j] == 1. Some nodes initial are initially infected by malware. Whenever two nodes are directly connected, and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner. Suppose M(initial) is the final number of nodes infected with malware in the entire network after the spread of malware stops. We will remove exactly one node from initial. Return the node that, if removed, would minimize M(initial). If multiple nodes could be removed to minimize M(initial), return such a node with the smallest index. Note that if a node was removed from the initial list of infected nodes, it might still be infected later due to
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Union-Find
[[1,1,0],[1,1,0],[0,0,1]] [0,1]
[[1,0,0],[0,1,0],[0,0,1]] [0,2]
[[1,1,1],[1,1,1],[1,1,1]] [1,2]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #924: Minimize Malware Spread
class UnionFind {
private final int[] p;
private final int[] size;
public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}
public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
public boolean union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
public int size(int root) {
return size[root];
}
}
class Solution {
public int minMalwareSpread(int[][] graph, int[] initial) {
int n = graph.length;
UnionFind uf = new UnionFind(n);
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (graph[i][j] == 1) {
uf.union(i, j);
}
}
}
int ans = n;
int mi = n, mx = 0;
int[] cnt = new int[n];
for (int x : initial) {
++cnt[uf.find(x)];
mi = Math.min(mi, x);
}
for (int x : initial) {
int root = uf.find(x);
if (cnt[root] == 1) {
int sz = uf.size(root);
if (sz > mx || (sz == mx && x < ans)) {
ans = x;
mx = sz;
}
}
}
return ans == n ? mi : ans;
}
}
// Accepted solution for LeetCode #924: Minimize Malware Spread
type unionFind struct {
p, size []int
}
func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}
func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}
func (uf *unionFind) union(a, b int) bool {
pa, pb := uf.find(a), uf.find(b)
if pa == pb {
return false
}
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
return true
}
func (uf *unionFind) getSize(root int) int {
return uf.size[root]
}
func minMalwareSpread(graph [][]int, initial []int) int {
n := len(graph)
uf := newUnionFind(n)
for i := range graph {
for j := i + 1; j < n; j++ {
if graph[i][j] == 1 {
uf.union(i, j)
}
}
}
cnt := make([]int, n)
ans, mx := n, 0
for _, x := range initial {
cnt[uf.find(x)]++
}
for _, x := range initial {
root := uf.find(x)
if cnt[root] == 1 {
sz := uf.getSize(root)
if sz > mx || sz == mx && x < ans {
ans, mx = x, sz
}
}
}
if ans == n {
return slices.Min(initial)
}
return ans
}
# Accepted solution for LeetCode #924: Minimize Malware Spread
class UnionFind:
__slots__ = "p", "size"
def __init__(self, n: int):
self.p = list(range(n))
self.size = [1] * n
def find(self, x: int) -> int:
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a: int, b: int) -> bool:
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
def get_size(self, root: int) -> int:
return self.size[root]
class Solution:
def minMalwareSpread(self, graph: List[List[int]], initial: List[int]) -> int:
n = len(graph)
uf = UnionFind(n)
for i in range(n):
for j in range(i + 1, n):
graph[i][j] and uf.union(i, j)
cnt = Counter(uf.find(x) for x in initial)
ans, mx = n, 0
for x in initial:
root = uf.find(x)
if cnt[root] > 1:
continue
sz = uf.get_size(root)
if sz > mx or (sz == mx and x < ans):
ans = x
mx = sz
return min(initial) if ans == n else ans
// Accepted solution for LeetCode #924: Minimize Malware Spread
struct Solution;
struct UnionFind {
parent: Vec<usize>,
n: usize,
}
impl UnionFind {
fn new(n: usize) -> Self {
let parent = (0..n).collect();
UnionFind { parent, n }
}
fn find(&mut self, i: usize) -> usize {
let j = self.parent[i];
if i == j {
i
} else {
let k = self.find(j);
self.parent[i] = k;
k
}
}
fn union(&mut self, mut i: usize, mut j: usize) {
i = self.find(i);
j = self.find(j);
if i != j {
self.parent[i] = j;
self.n -= 1;
}
}
}
impl Solution {
fn min_malware_spread(graph: Vec<Vec<i32>>, initial: Vec<i32>) -> i32 {
let n = graph.len();
let mut uf = UnionFind::new(n);
let mut initial: Vec<usize> = initial.into_iter().map(|i| i as usize).collect();
for i in 0..n {
for j in i + 1..n {
if graph[i][j] == 1 {
uf.union(i, j);
}
}
}
let mut group_size: Vec<usize> = vec![0; n];
for i in 0..n {
group_size[uf.find(i)] += 1;
}
let mut initial_groups: Vec<usize> = vec![0; n];
for &i in initial.iter() {
let gid = uf.find(i as usize);
initial_groups[gid] += 1;
}
initial.sort_unstable();
let mut res = (0, initial[0]);
for i in initial {
let gid = uf.find(i as usize);
if group_size[gid] > res.0 && initial_groups[gid] == 1 {
res = (group_size[gid], i);
}
}
res.1 as i32
}
}
#[test]
fn test() {
let graph = vec_vec_i32![[1, 1, 0], [1, 1, 0], [0, 0, 1]];
let initial = vec![0, 1];
let res = 0;
assert_eq!(Solution::min_malware_spread(graph, initial), res);
let graph = vec_vec_i32![[1, 0, 0], [0, 1, 0], [0, 0, 1]];
let initial = vec![0, 2];
let res = 0;
assert_eq!(Solution::min_malware_spread(graph, initial), res);
let graph = vec_vec_i32![[1, 1, 1], [1, 1, 1], [1, 1, 1]];
let initial = vec![1, 2];
let res = 1;
assert_eq!(Solution::min_malware_spread(graph, initial), res);
}
// Accepted solution for LeetCode #924: Minimize Malware Spread
class UnionFind {
p: number[];
size: number[];
constructor(n: number) {
this.p = Array(n)
.fill(0)
.map((_, i) => i);
this.size = Array(n).fill(1);
}
find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}
union(a: number, b: number): boolean {
const [pa, pb] = [this.find(a), this.find(b)];
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}
getSize(root: number): number {
return this.size[root];
}
}
function minMalwareSpread(graph: number[][], initial: number[]): number {
const n = graph.length;
const uf = new UnionFind(n);
for (let i = 0; i < n; ++i) {
for (let j = i + 1; j < n; ++j) {
graph[i][j] && uf.union(i, j);
}
}
let [ans, mx] = [n, 0];
const cnt: number[] = Array(n).fill(0);
for (const x of initial) {
++cnt[uf.find(x)];
}
for (const x of initial) {
const root = uf.find(x);
if (cnt[root] === 1) {
const sz = uf.getSize(root);
if (sz > mx || (sz === mx && x < ans)) {
[ans, mx] = [x, sz];
}
}
}
return ans === n ? Math.min(...initial) : ans;
}
Use this to step through a reusable interview workflow for this problem.
Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.
With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.