## Algorithm

Problem Name: 924. Minimize Malware Spread

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.length`
• `n == graph[i].length`
• `2 <= n <= 300`
• `graph[i][j]` is `0` or `1`.
• `graph[i][j] == graph[j][i]`
• `graph[i][i] == 1`
• `1 <= initial.length <= n`
• `0 <= initial[i] <= n - 1`
• All the integers in `initial` are unique.

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const minMalwareSpread = function (graph, initial) {
const l = graph.length
const p = []
const children = []
for (let i = 0; i  <  l; i++) {
p[i] = i
children[i] = [i]
}

for (let i = 0; i  <  l; i++) {
for (let j = i + 1; j  <  l; j++) {
if (graph[i][j] === 1) {
const pi = find(i)
const pj = find(j)
if (pi !== pj) {
union(pi, pj)
}
}
}
}

initial.sort((a, b) => (a > b ? 1 : -1))

const count = {}

let index = initial[0]
let max = 0
// find the index that not unioned with other indexes and with the most number of children
initial.forEach((e) => {
const pe = find(e)
if (!count[pe]) count[pe] = 0
count[pe] += 1
})
initial.forEach((e, i) => {
const pe = find(e)
if (count[pe] === 1 && children[pe].length > max) {
max = children[pe].length
index = e
}
})

return index

function find(x) {
while (p[x] !== x) {
p[x] = p[p[x]]
x = p[x]
}
return x
}

function union(pi, pj) {
p[pj] = pi
//also move the children to the new parent
children[pi] = children[pi].concat(children[pj])
children[pj] = []
}
}
``````
Copy The Code &

Input

cmd
graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def dfs(i):
for j in range(len(graph[i])):
if graph[i][j] and j not in nodes:
dfs(j)
rank, initial = collections.defaultdict(list), set(initial)
for node in sorted(initial):
nodes = set()
dfs(node)
if nodes & initial == {node}:
rank[len(nodes)].append(node)
return rank[max(rank)][0] if rank else min(initial)'
``````
Copy The Code &

Input

cmd
graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]