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Scissors - Aida R. #54
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Original file line number | Diff line number | Diff line change |
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@@ -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) | ||
""" | ||
pass | ||
basket = {} | ||
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for value in strings: | ||
key = ''.join(sorted(value)) | ||
if key in basket.keys(): | ||
basket[key].append(value) | ||
else: | ||
basket[key] = [] | ||
basket[key].append(value) | ||
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result = [] | ||
for key,value in basket.items(): | ||
result.append(value) | ||
return result | ||
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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
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There was a problem hiding this comment. Choose a reason for hiding this commentThe 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. |
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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 | ||
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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 | ||
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def valid_sudoku(table): | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more.
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@@ -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 | ||
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# 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 | ||
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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 |
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👍 , 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.