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Original file line number | Diff line number | Diff line change |
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# 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) where n is the size of the strings array | ||
# Space Complexity: O(n) where n is the size of the strings array | ||
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def grouped_anagrams(strings) | ||
raise NotImplementedError, "Method hasn't been implemented yet!" | ||
return strings if strings.length == 1 || strings.length == 0 | ||
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grouped_words_hash = {} | ||
strings.each do |word| | ||
check_word = word.chars.sort.join | ||
if grouped_words_hash[check_word].nil? | ||
grouped_words_hash[check_word] = [word] | ||
else | ||
grouped_words_hash[check_word].push(word) | ||
end | ||
end | ||
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grouped_words_array = [] | ||
grouped_words_hash.each do |sorted_word, array| | ||
grouped_words_array.push(array) | ||
end | ||
return grouped_words_array | ||
end | ||
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# 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(nlogn) because of the sort_by algorithm | ||
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. That depends on if the number of distinct elements compared to the number of elements overall. So maybe O(n + k log k) where n is the number of elements and k the number of distinct elements. |
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# Space Complexity: O(n) where n is the size of list | ||
def top_k_frequent_elements(list, k) | ||
raise NotImplementedError, "Method hasn't been implemented yet!" | ||
return list if list.length == 0 | ||
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track_common_elements = {} | ||
list.each do |element| | ||
if track_common_elements[element].nil? | ||
track_common_elements[element] = 1 | ||
else | ||
track_common_elements[element] += 1 | ||
end | ||
end | ||
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common_elements_array = track_common_elements.sort_by {|num, occurence| -occurence} | ||
result = [] | ||
i = 0 | ||
common_elements_array.each do |one_num_array| | ||
if i < k | ||
result.push(one_num_array[0]) | ||
end | ||
i += 1 | ||
end | ||
return result | ||
end | ||
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👍