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CheezItMan
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Jun 26, 2021
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Nice work Olga, this is a solid BFS solution, although you can write one which is O(n * E) instead.
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| while !q.empty? | ||
| current_pup, current_group = q.shift |
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Just a note that q.shift is an O(n) operation making this algorithm O(n^2 * E) Where n is the number of nodes and E the number of edges.
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