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CheezItMan
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Well done Whit, you hit the learning goals here. Well done.
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| 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: ? | ||
| Each subarray will have strings which are anagrams of each other. | ||
| Time Complexity: O(n^2) -- or is it O(n) bc of the "continue"s? | ||
| Space Complexity: O(n) | ||
| """ |
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👍 this works, but can you figure out a way to do this in O(n) time complexity using a dictionary?
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| freqs = __create_indexed_freq_map(nums) | ||
| reverse_freqs = __create_reverse_indexed_freq_map(freqs) |
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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: ? | ||
| Time Complexity: O(n x m), where n is length of original list and m is number of repeats for each repeated value | ||
| Space Complexity: O(n) |
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👍 Just note that you are sorting, what does that do to the time/space complexity?
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| if y < 3 and x < 3: | ||
| quadrants[0].append(board[i][j]) | ||
| elif y < 3 and 3 <= x and x < 6: | ||
| quadrants[1].append(board[i][j]) | ||
| elif y < 3 and 6 <= x: | ||
| quadrants[2].append(board[i][j]) | ||
| elif 3 <= y and y < 6 and x < 3: | ||
| quadrants[3].append(board[i][j]) | ||
| elif 3 <= y and y < 6 and 3 <= x and x < 6: | ||
| quadrants[4].append(board[i][j]) | ||
| elif 3 <= y and y < 6 and x >= 6: | ||
| quadrants[5].append(board[i][j]) | ||
| elif 6 <= y and x < 3: | ||
| quadrants[6].append(board[i][j]) | ||
| elif 6 <= y and 3 <= x and x < 6: | ||
| quadrants[7].append(board[i][j]) | ||
| elif 6 <= y and x >= 6: | ||
| quadrants[8].append(board[i][j]) | ||
|
|
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Could you do this with some kind of loop?
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| def valid_sudoku(board): | ||
| """ This method will return the true if the table is still | ||
| a 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(n^2) | ||
| Space Complexity: O(n) ? its a little complicated and im tired |
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Just a note that the time complexity is really O(1) since the size of the input never changes. The grid is always 9x9.
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