{"componentChunkName":"component---src-templates-post-template-tsx","path":"/posts/leetcode-154/","result":{"data":{"markdownRemark":{"id":"a88b8711-09ce-5f78-9f59-de6992cde364","html":"

Sort Items by Groups Respecting Dependencies

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There are n items each belonging to zero or one of m groups where group[i] is the group that the i-th item belongs to and it’s equal to -1 if the i-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.

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Return a sorted list of the items such that:

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Return any solution if there is more than one solution and return an empty list if there is no solution.

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Keyword: Graph, Topological sort, DAG

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Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological Sorting for a graph is not possible if the graph is not a DAG. A directed graph doesn’t have topological sorting, if and only if it contains a cycle.

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In this problem, we need to do a cycle check and topological sorting between groups, and inside each group need another check and topological sorting.

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Steps:

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  1. Get the in_degree map for groups and for items
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  2. Sort groups. By sorting here, we get all nodes whose in degree is 0 (which means no dependency), add them in a stack and update in_degree map, and do the process again until there are no nodes. Then pop nodes from stack and get the topological sort result.
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  3. Sort inside each group
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Reference:

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Critical Connections in a Network

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There are n servers numbered from 0 to n-1 connected by undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. Any server can reach any other server directly or indirectly through the network.

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A critical connection is a connection that, if removed, will make some server unable to reach some other server.

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Return all critical connections in the network in any order.

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Keyword: Graph, DFS

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Basically it’s a DFS over all the nodes, the difficult part is to check whether an edge is a ‘critical connection’ or not. To define this edge [a,b], it’s critical if and only if there is no other way from a to reach b except this edge (not in a cycle).

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Solution:

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from __future__ import print_function\n\nfrom collections import defaultdict\n\n\nclass Solution(object):\n    def criticalConnections(self, n, connections):\n        \"\"\"\n        :type n: int\n        :type connections: List[List[int]]\n        :rtype: List[List[int]]\n        \"\"\"\n        map = defaultdict(list)\n        for [n1, n2] in connections:\n            map[n1].append(n2)\n            map[n2].append(n1)\n\n        visited = defaultdict(lambda: False)\n        # node with lowest index that this node can reach\n        lowestNode = defaultdict(lambda: float('-inf'))\n        # time that dfs went through this node, or depth in DFS\n        discoverTime = defaultdict(lambda: float('-inf'))\n\n        parentMap = defaultdict(lambda: -1)\n        results = []\n\n        def dfs(node, time):\n            if visited[node]: return\n            visited[node] = True\n            lowestNode[node] = time\n            discoverTime[node] = time\n\n            for t in map[node]:\n                # dfs over all adjacent nodes\n                if not visited[t]:\n                    parentMap[t] = node\n                    dfs(t, time + 1)\n                    lowestNode[node] = min(lowestNode[t], lowestNode[node])\n\n                    if lowestNode[t] > discoverTime[node]:\n                        results.append([node, t])\n                else:\n                    if t != parentMap[node]:\n                        lowestNode[node] = min(lowestNode[node], discoverTime[t])\n\n        dfs(0, 0)\n        return results\n\n\nn = 4\nconnections = [[0, 1], [1, 2], [2, 0], [1, 3]]\nprint(Solution().criticalConnections(n, connections))
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