Merge pull request royalpranjal#19 from me-ydv-5/patch-2

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Author: Pranjal

Graphs/LargestDistanceBetweenNodesOfATree.cpp

@@ -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;
+}