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All finished #43
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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: ? | ||
Time Complexity: O(n) | ||
Space Complexity: O(n log n) | ||
""" | ||
pass | ||
letters_hash = {} | ||
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for word in strings: | ||
anagram_sorted = "".join(sorted(word)) | ||
if anagram_sorted in letters_hash: | ||
letters_hash[anagram_sorted].append(word) | ||
else: | ||
letters_hash[anagram_sorted] = [word] | ||
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return list(letters_hash.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) | ||
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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. 👍 However your time complexity is O(nk) because you're looping through the words and k times finding the maximum word (most common). |
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""" | ||
pass | ||
highest_k_hash = {} | ||
if nums == []: | ||
return nums | ||
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for num in nums: | ||
if num in highest_k_hash: | ||
highest_k_hash[num] += 1 | ||
else: | ||
highest_k_hash[num] = 1 | ||
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sorted_k_values = [] | ||
for i in range(k): | ||
highest_freq_value = max(highest_k_hash, key=highest_k_hash.get) | ||
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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. Note that you have a loop here which runs k times and inside it you're going through the dictionary (of max n elements) and finding the maximum entry. So O(nk) |
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sorted_k_values.append(highest_freq_value) | ||
highest_k_hash.pop(highest_freq_value) | ||
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return sorted_k_values | ||
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def valid_sudoku(table): | ||
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👍 The time complexity is right if the words are all of limited length (like english words).
The space complexity is O(n) because you're building a dictionary of n words.