fromtypingimportListclassSolution:deffindRedundantDirectedConnection(self,edges:List[List[int]])-List[int]:nlen(edges)# Step 1: Find a node with two parents (in-degree 2)parent[0]*(n1)candidate1None# earlier edge u1 - vcandidate2None# later edge u2 - vforu,vinedges:ifparent[v]0:parent[v]uelse:candidate1[parent[v],v]candidate2[u,v]break# Step 2: Union-Find to detect a cycleuflist(range(n1))deffind(x:int)-int:whileuf[x]!x:uf[x]uf[uf[x]]# path halvingxuf[x]returnxforu,vinedges:# Temporarily ignore candidate2ifcandidate2isnotNoneand[u,v]candidate2:continueru,rvfind(u),find(v)ifrurv:# Cycle detectedifcandidate1isnotNone:# Removing candidate2 still leaves a cycle,# so candidate1 must be removedreturncandidate1# No node with two parents: this edge forms the cyclereturn[u,v]uf[ru]rv# No cycle after removing candidate2 - candidate2 is redundantreturncandidate2ApproachThere are only two possible kinds of invalid edges in this problem:A node has two parents — one of its two incoming edges must be removed.There is a cycle — the edge that closes the cycle must be removed.The algorithm:· First, scan edges to find a node v with two parents. Store the earlier edge as candidate1 and the later edge as candidate2.· Run Union-Find while skipping candidate2:· If a cycle still appears, then candidate1 is the redundant edge.· If no cycle appears, then candidate2 is the redundant edge.· If no node has two parents, simply return the first edge that creates a cycle.Complexity· Time: O(n · α(n)), effectively O(n) with union-find.· Space: O(n) for the union-find and parent arrays.
fromtypingimportListclassSolution:deffindRedundantDirectedConnection(self,edges:List[List[int]])-List[int]:nlen(edges)# Step 1: Find a node with two parents (in-degree 2)parent[0]*(n1)candidate1None# earlier edge u1 - vcandidate2None# later edge u2 - vforu,vinedges:ifparent[v]0:parent[v]uelse:candidate1[parent[v],v]candidate2[u,v]break# Step 2: Union-Find to detect a cycleuflist(range(n1))deffind(x:int)-int:whileuf[x]!x:uf[x]uf[uf[x]]# path halvingxuf[x]returnxforu,vinedges:# Temporarily ignore candidate2ifcandidate2isnotNoneand[u,v]candidate2:continueru,rvfind(u),find(v)ifrurv:# Cycle detectedifcandidate1isnotNone:# Removing candidate2 still leaves a cycle,# so candidate1 must be removedreturncandidate1# No node with two parents: this edge forms the cyclereturn[u,v]uf[ru]rv# No cycle after removing candidate2 - candidate2 is redundantreturncandidate2ApproachThere are only two possible kinds of invalid edges in this problem:A node has two parents — one of its two incoming edges must be removed.There is a cycle — the edge that closes the cycle must be removed.The algorithm:· First, scan edges to find a node v with two parents. Store the earlier edge as candidate1 and the later edge as candidate2.· Run Union-Find while skipping candidate2:· If a cycle still appears, then candidate1 is the redundant edge.· If no cycle appears, then candidate2 is the redundant edge.· If no node has two parents, simply return the first edge that creates a cycle.Complexity· Time: O(n · α(n)), effectively O(n) with union-find.· Space: O(n) for the union-find and parent arrays.
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