Mergesort

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Mergesort
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Mergesort Algorithm

• void mergesort( Elem a, Elem temp, int Ieft,
  int right )
• int I, j, k, mid (leftright)/2
• if( left right ) return
• mergesort( a, temp, left, mid)
• mergesort( a, temp, mid1, right)
• // do the merge operation
• for( i left i lt mid i )
  tempi ai
• for( j 1 j lt right-mid j ) temp
  right-j1 a jmid
• // merge sublists back to array
• for( ileft, jright, kleft kltright k
  )
• if( temp i lt temp j ) ak
  temp i
• else aktemp j--

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Mergesort Illustration
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Mergesort Illustration

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Mergesort Illustration

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Mergesort Time complexity

• Best, worst, average-case
• Each merge operation takes 0(k) time for 2 lists
  each k/2 elements long (merged into one list k
  elements long)
• There will be log2n levels
• 1st level 2 n/2 long lists to be merged into 1
  n long list
• 2nd level 4 n/4 long lists to be merged into 2
  n/2 long lists
• 3rd level 8 n/8 long lists to be merged into 4
  n/4 long lists
• Time Complexity O(nlog2n)

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Heapsort Algorithm

• void heapify(Elem a, 1, n )
• for( i n/2 i gt 0 i--)
• buildheap(a, i, n )
• void buildheap(Elem a, i, n )
• int j 2i
• int b ai
• while( j lt n)
• if( j1 ltn a j lt a j1 ) j
  j1
• if( b lt a j ) ai a j
• i j
• j 2i
• a j/2 b
• void Heapsort( Elem a, 1, n)
• heapify( a, 1, n )

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Heapsort Illustration
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Heapsort Time complexity

• heapify() takes O(n) time
• n buildheap()s each take O( log2n ) time
• Best, worst, average-case
• O(nlog2n)

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Binsort Algorithm

• void binsort( Elem a, int n )
• List B new List MAX_KEY_VALUE
• for( i0 Iltn I )
• B ai.key .append( ai )
• int index 0
• for( i0 ilt MAX_KEY_VALUE i)
• for( Bi.first() Bi.isInList()
  Bi.next() )
• ai Bi.currentValue()

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Bucketsort Algorithm

• Put a range of values in bins and then sort the
  contents of each bin and then output the values
  of each bin in order.

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Radixsort Algorithm

• void radixsort( Elem a, Elem B, int n, int k,
  int r, Elem count)
• // counti stores the number of records in
  bini
• for( int i0, rtok1 iltk i, rtokr) //for k
  digits
• for( int j0 jltr j) countj0
  //initialize count
• //Count the no. of records for each bin on this
  pass
• for( j0 jltn j ) count (aj.key /
  rtok) r
• // Index B count j will be index for last
  slot of bin j
• for( j1 jltr j ) countj countj-1
  countj
• // put records into bins working from bottom of
  each bin
• // since bins fill from the bottom, j counts
  downwards
• for( jn-1 jgt0 j--) B--count (aj.key /
  rtok) r aj
• for( j0 jltn j ) aj Bj //copy B
  back into a