Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Scissors - Aida R. #54

Open
wants to merge 1 commit into
base: master
Choose a base branch
from
Open
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
68 changes: 60 additions & 8 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,46 @@
def grouped_anagrams(strings):
""" This method will return an array of arrays.
Each subarray will have strings which are anagrams of each other
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(n)
Space Complexity: O(n)
"""
Comment on lines 2 to 7

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

👍 , Time/space are correct if the words are limited in length (like English words) Otherwise it would be O(n * m log m) where m is the length of each word.

pass
basket = {}

for value in strings:
key = ''.join(sorted(value))
if key in basket.keys():
basket[key].append(value)
else:
basket[key] = []
basket[key].append(value)

result = []
for key,value in basket.items():
result.append(value)
return result


def top_k_frequent_elements(nums, 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: ?
Space Complexity: O(n)
"""
Comment on lines 24 to 29

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

👍 Because of the loop in lines 40-43, the time complexity is O(nk) because you have a loop going k times and each time finding the max element from the frequency map.

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

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


def valid_sudoku(table):

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

⚠️ Not quite working as the board dictionary is not reset when checking the columns or subgrids.

Expand All @@ -22,8 +50,32 @@ def valid_sudoku(table):
Each element can either be a ".", or a digit 1-9
The same digit cannot appear twice or more in the same
row, column or 3x3 subgrid
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(n2)
Space Complexity: O(1)
"""
pass
for i in range(9):
board = {}
for j in range(9):
if table[i][j] in board:
return False
elif table[i][j] is not '.':
board[table[i][j]]
return True

# Check columns
for j in range(9):
for i in range(9):
if table[i][j] in board:
return False
elif table[i][j] is not '.':
board[table[i][j]]
return True

for subI in range(0, 9, 3):
for subJ in range(0, 9, 3):
if table[i][j] in board:
return False
elif table[i][j] is not '.':
board[table[i][j]]
return True
return True