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Mariela Cruz - Scissors #41

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46 changes: 39 additions & 7 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
@@ -1,20 +1,52 @@

def grouped_anagrams(strings):
def grouped_anagrams(d1):
""" This method will return an array of arrays.
Each subarray will have strings which are anagrams of each other
Time Complexity: ?
Space Complexity: ?
"""
Comment on lines +2 to 7

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👍 Space/time complexity?

pass
# empty dictionary for anagrams together
dict = {}

# traversal
for val in d1:
# sorts list
key = ''.join(sorted(val))

if key in dict.keys():
dict[key].append(val)
else:
dict[key] = []
dict[key].append(val)

# traverse dictionary and join keys together
result = []
for key,value in dict.items():
# result = result + ' '.join(value) + ' '
result.append(value)
return result

def top_k_frequent_elements(nums, k):
def top_k_frequent_elements(numbers, k):
""" This method will return the k most common elements
In the case of a tie it will select the first occuring element.
Time Complexity: ?
Space Complexity: ?
Time Complexity: o(n)
Space Complexity: o(n)
"""
Comment on lines +29 to 34

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👍 However because you're finding max of a list of n elements k times, this is O(nk) for time complexity.

pass

freq_map = {}
if numbers == []:
return numbers
for i in numbers:
if i in freq_map:
freq_map[i] += 1
elif i not in freq_map:
freq_map[i] = 1

new_list = []
for i in range(k):
highest_value = max(freq_map, key=freq_map.get)
new_list.append(highest_value)
freq_map.pop(highest_value)
return new_list

def valid_sudoku(table):
""" This method will return the true if the table is still
Expand Down
2 changes: 1 addition & 1 deletion tests/test_top_k_frequent_elements.py
Original file line number Diff line number Diff line change
Expand Up @@ -7,7 +7,7 @@ def test_it_works_with_example_1():
k = 2

# Act
answer = top_k_frequent_elements(nums=numbers, k=k)
answer = top_k_frequent_elements(numbers=numbers, k=k)

# Assert
answer.sort()
Expand Down