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CheezItMan
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Jan 21, 2022
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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) |
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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.
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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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👍 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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| for i in range(k): | ||
| highest_freq_value = max(highest_k_hash, key=highest_k_hash.get) |
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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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