Heather #15
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CheezItMan
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Sep 18, 2019
| # Each subarray will have strings which are anagrams of each other | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: Would this by O(nlog n) for sorting of the string? |
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I would say O( n * m log m) where n is the number of words, and m the length of each word. If you can assume that the words are small, then you could say this is O(n)
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: Would this by O(nlog n) for sorting of the string? | ||
| # Space Complexity: O(n + m), n is the size of the array and m is the size of the hash |
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Since the size of the hash depends on the size of strings I would say O(n)
| # in the case of a tie it will select the first occuring element. | ||
| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(nlog n) because of the sorted array is at least |
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True! Could this be faster without sorting?
| sorted_frequencies = freq_hash.values.sort | ||
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| answer_arr = [] | ||
| k.times do |
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This looks like a loop that runs k times and a nested loop which runs maybe n times.
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