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Kayla Pine #59
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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^2)? The fucntion only loops through the length of strings for | ||
| the number of letters in each string? | ||
| Space Complexity: O(n)? Theres only 2 varaibles being stored...sortedWord and anagramsHash? | ||
|
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. 🪐 Yes! |
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| """ | ||
| pass | ||
| anagramsHash = {} | ||
| for string in strings: | ||
| sortedWord = "".join(sorted(string)) | ||
| if sortedWord in anagramsHash: | ||
| anagramsHash[sortedWord].append(string) | ||
| else: | ||
| anagramsHash[sortedWord] = [string] | ||
| return list(anagramsHash.values()) | ||
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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) | ||
| Space Complexity: O(n^2)? | ||
|
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. Space complexity would be O(n). |
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| """ | ||
| pass | ||
| freqHash = {} | ||
| finalList = [] | ||
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| if len(nums) == 0: | ||
| return finalList | ||
| for number in nums: | ||
| if number not in freqHash: | ||
| freqHash[number] = 0 | ||
| else: | ||
| freqHash[number] += 1 | ||
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| count = 1 | ||
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| while count <= k: | ||
| finalList.append(sorted(freqHash, key=freqHash.get)[-count]) | ||
| count += 1 | ||
| return finalList | ||
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| def valid_sudoku(table): | ||
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There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
⏱ Sorting is an O(n) operation, however if we assume all strings are valid english words, we can say this is an O(1) operation because n will never be very large.
Therefore you just have the for loop through the elements of
stringswhich would make this O(n)