Skip to content

Facebook, Google #19

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
royalpranjal merged 1 commit into from
Oct 14, 2018
Merged

Facebook, Google #19

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
58 changes: 58 additions & 0 deletions Graphs/LargestDistanceBetweenNodesOfATree.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,58 @@
pair<int, int> process(queue<pair<int, int> >& q, vector<int>& visited, unordered_map<int, vector<int> >& x){
// Typical BFS with keeping track of the longest distance and farthest element.
int maxi = INT_MIN, farthest = 0;
while(!q.empty()){
int node = q.front().first;
int distance = q.front().second;
q.pop();
if(distance > maxi){
maxi = distance;
farthest = node;
}
visited[node] = 1;
for(int i = 0; i < x[node].size(); i++){
if(!visited[x[node][i]]){
visited[x[node][i]] = 1;
q.push({x[node][i], distance+1});
}
}
}
return {maxi, farthest};
}

int Solution::solve(vector<int> &A) {
// Base Case
if(A.size() <= 1)return 0;

// Visited array to keep track of visited elements
vector<int> visited(A.size(), 0);

// Map to make the graph from given array
unordered_map<int, vector<int> > x;
x.clear();

for(int i = 0; i < A.size(); i++){
// If the value is -1, it doesn't have any parent.
// Make the pairs in the adj. list type map
if( A[i] != -1){
x[A[i]].push_back(i);
x[i].push_back(A[i]);
}
}

// Start from the first element in the array and mark it visited.
queue<pair<int, int> > q;
q.push({0, 0});
visited[0] = 1;

// Gets the farthest element from the first element
int farthest = process(q, visited, x).second;

vector<int> vis(A.size(), 0);

// Now apply BFS on the farthest element found and return the distance
// of the farthest element found so far.
q.push({farthest, 0});
vis[farthest] = 1;
return process(q, vis, x).first;
}