Runtime: 22 ms (Top 29.21%) | Memory: 9.9 MB (Top 41.59%)

Author: AnasImloul

scripts/algorithms/R/Redundant Connection II/Redundant Connection II.cpp

@@ -1,20 +1,21 @@
+// Runtime: 22 ms (Top 29.21%) | Memory: 9.9 MB (Top 41.59%)
 class Solution {
 public:
     /*
     There are two cases for the tree structure to be invalid.
     1) A node having two parents;
     2) A circle exists
-    
-    If there are both invalid conditions, which means there is a node which has 2 parents and there is also a circle even after we invalid edgeB. 
+
+    If there are both invalid conditions, which means there is a node which has 2 parents and there is also a circle even after we invalid edgeB.
     In this case, we have to return edgeA. only in this way can we avoid both double parents and circle for the tree.
-    */        
-	
+    */
+
     int find(vector<int>& par, int n){
         if(par[n]==n || par[n]<0) return n;
         par[n] = find(par, par[n]);
         return par[n];
-    }    
-    
+    }
+
     void findSpecialEdge(vector<vector<int>>& edges, vector<int>& edgeA, vector<int>& edgeB){
         int len = edges.size();
         vector<int>par(len+1, 0);
@@ -23,26 +24,26 @@ class Solution {
             if(par[c]){
                 edgeA = {par[c], c};
                 edgeB = e;
-                return;    
+                return;
             }else{
                 par[c] = p;
             }
         }
     }
-    
+
     vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
         int len = edges.size();
         vector<int>parent(len+1, 0);
         for(int i=0; i<=len; i++) parent[i]=i;
-        
+
         vector<int>edgeA={}; // 1st candidate
         vector<int>edgeB={}; // 2nd candidate
         findSpecialEdge(edges, edgeA, edgeB);
-        
+
         for(vector<int> e : edges){
             int n1 = e[0];
             int n2 = e[1];
-            if(edgeB.size()>0 && (edgeB[0]==n1 && edgeB[1]==n2)){  // invalidate edgeB
+            if(edgeB.size()>0 && (edgeB[0]==n1 && edgeB[1]==n2)){ // invalidate edgeB
                 continue;
             }
             int p1 = find(parent, n1);
@@ -56,7 +57,7 @@ class Solution {
             } else{
                 parent[p2] = p1;
             }
-        }         
+        }
         return edgeB;
     }
-};
+};