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Hunter Smith - Sprint-Challenge--Data-Structures-Python #1

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652df0a
initial commit
Hunter315 Jan 25, 2019
2580204
depth_first_for_each
Hunter315 Jan 25, 2019
73a2aeb
breadth_first_ffor_each
Hunter315 Jan 25, 2019
c8c6ad6
names.py using sets complete in 0.031 seconds
Hunter315 Jan 25, 2019
eece83e
finished complexities answers (MVP MET)
Hunter315 Jan 25, 2019
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names.py using sets complete in 0.031 seconds
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@Hunter315
Hunter315 committed Jan 25, 2019
commit c8c6ad6989e4da934b6af8c9916028b14d0a51a6
12 changes: 8 additions & 4 deletions names/names.py
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Original file line number Diff line number Diff line change
Expand Up @@ -11,10 +11,14 @@
f.close()

duplicates = []
for name_1 in names_1:
for name_2 in names_2:
if name_1 == name_2:
duplicates.append(name_1)
# for name_1 in names_1:
# for name_2 in names_2:
# if name_1 == name_2:
# duplicates.append(name_1)
a_set = set(names_1)
b_set = set(names_2)
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@Tulf Tulf Jan 25, 2019

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nice you can also use intersection or a for loop with an if all of these would be O(n)

if (a_set & b_set):
duplicates.append(repr(a_set & b_set))

end_time = time.time()
print (f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
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1 change: 1 addition & 0 deletions search/binary_search_tree.py
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Expand Up @@ -15,6 +15,7 @@ def depth_first_for_each(self, cb):

def breadth_first_for_each(self, cb):
nodes = [self]
#while loop going through every node
while len(nodes) > 0:
current = nodes.pop(0)
cb(current.value)
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