### PROBLEM LINK:

**Author:** RAVIT SINGH MALIK

**Editorialist:** RAVIT SINGH MALIK

### DIFFICULTY:

MEDIUM-HARD

### PREREQUISITES:

BFS , GRAPHS

### PROBLEM:

F(x,y) is a function which determines the performance of the super-computer.

Here (x,y) is that pair of processes (y is always a sub-process of process x) where the difference between the factors of process x and process y is maximum.

You are required to find maximum value of G(x)-G(Y) .

### EXPLANATION:

This problem is completely based on Breadth-first search [BFS], which is used to traversing or searching tree or graph data structures. It starts at the tree root

(or some arbitrary node of a graph, sometimes referred to as a ‘search key’)

and explores the neighbor nodes first, before moving to the next level neighbors.

Pseudocode :

Input: A graph Graph and a starting vertex root of Graph.

Output: All vertices reachable from root labeled as explored.

A non-recursive implementation of breadth-first search:

```
1. Breadth-First-Search(Graph, root):
2.
3. for each node n in Graph:
4. n.distance = INFINITY
5. n.parent = NIL
6.
7. create empty queue Q
8.
9. root.distance = 0
10. Q.enqueue(root)
11.
12. while Q is not empty:
13.
14. current = Q.dequeue()
15.
16. for each node n that is adjacent to current:
17. if n.distance == INFINITY:
18. n.distance = current.distance + 1
19. n.parent = current
20. Q.e
```

Time and space complexity.

The time complexity can be expressed as O(|V|+|E|), since every vertex and every edge will be explored in the worst case.

|V| is the number of vertices and |E| is the number of edges in the graph. Note that O(|E|) may vary between O(1) and O(|V|^{2}), depending on how sparse the input graph is.

So,the given question is an improvisation of Breadth-first search [BFS], you have to just store the maximum difference

of parent to its every childern. and store the maximum value between the parent and child at the child node.

Pseudocode :

```
while(!q.empty())
int s=q.front()
q.pop()
int size=adj[s].size()
for( i=0 ;i < size ; i++)
int num = adj[s][i]
if(visit[num]!=true)
visit[num]=true
if(maxi < a[s] - a[num])
maxi = a[s] - a[num]
if(a[s] > a[num])
a[num]=a[s]
qu.push(adj[s][i])
```

=>> maxi is the required answer.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.