35 lines
1.1 KiB
Ruby
35 lines
1.1 KiB
Ruby
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#Given an array ,sort the array by descending order of count of element,
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#if count is same then the element which comes in array first comes first in sorted array.
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#Time-complexity: O(nlogn) , Auxiliary-space: O(n) {for hash}
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def sort_by_frequency(a)
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len=a.length
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freq=Hash.new() #Hash to store element,count and starting index
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for i in 0...len
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# Storing count and index in hash with element as the key
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unless freq[a[i]]
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freq[a[i]]={}
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freq[a[i]][:count]=1
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freq[a[i]][:index]=i
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else
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freq[a[i]][:count]+=1
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end
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end
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freq=freq.sort_by{|k,v| [-v[:count],v[:index]]} # Sort the hash by decreasing count value and if count is same then by index(starting index)
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p=0 # Index to iterate over array
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# Iterate on hash
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freq.each do |k,v|
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for i in 0...v[:count]
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a[p]=k
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p+=1
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end
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end
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return a
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end
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sort_by_frequency([1,3,3,3,4,4,4,2,2,2,5]) # => [3, 3, 3, 4, 4, 4, 2, 2, 2, 1, 5]
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sort_by_frequency([1,3,3,3,4,4,4,2,2,2,5,5,6,7,7,7,8,8]) # => [3, 3, 3, 4, 4, 4, 2, 2, 2, 7, 7, 7, 5, 5, 8, 8, 1, 6]
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