2023-08-20 11:34:41 +0000 UTC

class Solution:
    def sortItems(self, n, m, group, beforeItems):
        # If an item belongs to zero group, assign it a unique group id.
        group_id = m
        for i in range(n):
            if group[i] == -1:
                group[i] = group_id
                group_id += 1
        
        # Sort all item regardless of group dependencies.
        item_graph = [[] for _ in range(n)]
        item_indegree = [0] * n
        
        # Sort all groups regardless of item dependencies.
        group_graph = [[] for _ in range(group_id)]
        group_indegree = [0] * group_id      
        
        for curr in range(n):
            for prev in beforeItems[curr]:
                # Each (prev -> curr) represents an edge in the item graph.
                item_graph[prev].append(curr)
                item_indegree[curr] += 1
                
                # If they belong to different groups, add an edge in the group graph.
                if group[curr] != group[prev]:
                    group_graph[group[prev]].append(group[curr])
                    group_indegree[group[curr]] += 1      
        
        # Tologlogical sort nodes in graph, return [] if a cycle exists.
        def topologicalSort(graph, indegree):
            visited = []
            stack = [node for node in range(len(graph)) if indegree[node] == 0]
            while stack:
                cur = stack.pop()
                visited.append(cur)
                for neib in graph[cur]:
                    indegree[neib] -= 1
                    if indegree[neib] == 0:
                        stack.append(neib)
            return visited if len(visited) == len(graph) else []

        item_order = topologicalSort(item_graph, item_indegree)
        group_order = topologicalSort(group_graph, group_indegree)
        
        if not item_order or not group_order: 
            return []
        
        # Items are sorted regardless of groups, we need to 
        # differentiate them by the groups they belong to.
        ordered_groups = collections.defaultdict(list)
        for item in item_order:
            ordered_groups[group[item]].append(item)
        
        # Concatenate sorted items in all sorted groups.
        # [group 1, group 2, ... ] -> [(item 1, item 2, ...), (item 1, item 2, ...), ...]
        answer = []
        for group_index in group_order:
            answer += ordered_groups[group_index]
        return answer