python - How can i get union of 2D list items when there occurs any intersection (in efficient way)? -

i have 2d list in python

list = [[9, 2, 7], [9, 7], [2, 7], [1, 0], [0, 5, 4]] 

i union of list items if there occurs intersection. example [9, 2, 7], [9, 7], [2, 7] has intersection of more 1 digit. union of [9,2,7].

how can final list follows in efficient way ?

finallist = [[9,2,7], [0, 1, 5, 4]] 

n.b. order of numbers not important.

you have graph problem. want build connected components in graph vertices elements of sublists, , 2 vertices have edge between them if they're elements of same sublist. build adjacency-list representation of input , run graph search algorithm on it, or iterate on input , build disjoint sets. here's slightly-modified connected components algorithm wrote a similar question:

import collections  # build adjacency list representation of input graph = collections.defaultdict(set) l in input_list:     if l:         first = l[0]         element in l:             graph[first].add(element)             graph[element].add(first)  # breadth-first search graph produce output output = [] marked = set() # set of nodes connected component known node in graph:     if node not in marked:         # node not in seen connected component         # run breadth-first search determine connected component         frontier = set([node])         connected_component = []         while frontier:             marked |= frontier             connected_component.extend(frontier)              # find unmarked nodes directly connected frontier nodes             # form new frontier             new_frontier = set()             node in frontier:                 new_frontier |= graph[node] - marked             frontier = new_frontier         output.append(tuple(connected_component))