1 C++ Classes and Data Structures Jeffrey S. Childs Chapter 14 Introduction to Sorting Algorithms Jeffrey S. Childs Clarion University of PA © 2008, Prentice.

2 Sorting Sorting is the process of placing elements in order If elements are objects, they are ordered by a particular data member There are many algorithms for sorting, each having its own advantages No sorting algorithm is better than every other sorting algorithm all the time Choosing the right sorting algorithm for a problem requires careful consideration

3 Heapsort PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] );

4 Heapsort (cont.) PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] ); Provide an initial array into the second constructor of the priority queue – this constructor makes a copy of the array and converts the copy into a heap

5 Heapsort (cont.) PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] ); In the loop, we start with the last array index

6 Heapsort (cont.) PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] ); Then, on the first dequeue, the largest element from the heap will be placed into the last position of the array (where it should be)

7 Heapsort (cont.) PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] ); The index i is decremented, so on the next dequeue, the remaining highest value from the heap will be placed in the next to last array position (where it should be)

8 Heapsort (cont.) PriorityQueue pq( arr ); for ( i = arr.length( ) – 1; i >= 0; i-- ) pq.dequeue( arr[ i ] ); The loop stops when the index becomes less than 0

9 Heapsort Time Complexity Heapsort runs in  ( n lg n ) time on average It has a best-case time of  ( n ) if all elements have the same value –This is an unusual case, but we may want to consider it if many of the elements have the same value

10 Function Templates A function template can be made for heapsort, so that the client can put heapsort into the client program A function template is something like a class template – the compiler makes functions from the function template The client can then use heapsort for different types of arrays, without rewriting the heapsort function

11 Heapsort Function Template (in a Client Program) 1 template 2 void heapsort( Array & arr ) 3 { 4PriorityQueue pq( arr ); 5for ( int i = arr.length( ) - 1; i >= 0; i-- ) 6pq.dequeue( arr[ i ] ); 7 }

12 Example Using the Heapsort Function Template 1 #include 2 #include "PriorityQueue.h" 3 4 using namespace std; 5 6 struct MyStruct { 7int id; 8float amount; 9bool operator >( const MyStruct & right ) 10{ return amount > right.amount; } 11 };

13 Example Using the Heapsort Function Template (cont.) 1 #include 2 #include "PriorityQueue.h" 3 4 using namespace std; 5 6 struct MyStruct { 7int id; 8float amount; 9bool operator >( const MyStruct & right ) 10{ return amount > right.amount; } 11 }; Client will sort an array of MyStruct objects

14 Example Using the Heapsort Function Template (cont.) 1 #include 2 #include "PriorityQueue.h" 3 4 using namespace std; 5 6 struct MyStruct { 7int id; 8float amount; 9bool operator >( const MyStruct & right ) 10{ return amount > right.amount; } 11 }; Apparently, the client would like to sort the MyStruct objects by amount, not by id

15 12 template 13 void heapsort( Array & arr ); 14 15 int main( ) 16 { 17int num; 18 19cout << "How many records do you want to enter? "; 20cin >> num; 21Array objs( num ); Note that the function prototype for heapsort must also be templated Example Using the Heapsort Function Template (cont.)

16 22for ( int i = 0; i < objs.length( ); i++ ) { 23 cout << "Enter id for record " << i + 1 << ": "; 24 cin >> objs[ i ].id; 25 cout << "Enter amount for record " << i + 1 << ": "; 26 cin >> objs[ i ].amount; 27 } Client asks user to enter information for the array of MyStruct objects Example Using the Heapsort Function Template (cont.)

17 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Then, the client calls heapsort to sort the array of MyStruct objects called objs Example Using the Heapsort Function Template (cont.)

18 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) When the compiler sees this line of code, it makes a function out of the function template

19 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) The compiler determines what type objs is…

20 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) then substitutes that type for DataType to make a heapsort function

21 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) The client could also call heapsort on an array of integers in the same program…

22 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) …the compiler would make another heapsort function using int for DataType

23 28heapsort( objs );. 39 template 40 void heapsort( Array & arr ) 41 { 42PriorityQueue pq( arr ); 43for ( int i = arr.length( ) - 1; i >= 0; i-- ) 44pq.dequeue( arr[ i ] ); 45 } Example Using the Heapsort Function Template (cont.) Note that the heapsort function template is placed into the client’s program just like any other function would be

24 28heapsort( objs ); 29cout << "Records in sorted order:" << endl; 30for ( int i = 0; i < objs.length( ); i++ ) { 31 cout << "record " << i << ":" << endl; 32 cout << "id: " << objs[ i ].id << endl; 33 cout << "amount: " << objs[ i ].amount << endl; 34} 35 36return 0; 37 } Example Using the Heapsort Function Template (cont.) The final part of the main function provides the records in sorted order

25 Insertion Sort 3 18 19 22 17 16 6 1 15 2 sorted sectionunsorted section During the process of insertion sort, the array is divided into a sorted section and an unsorted section – the sorted section grows and the unsorted section shrinks.

26 Insertion Sort (cont.) 3 18 19 22 17 16 6 1 15 2 sorted sectionunsorted section next value to be inserted into sorted section

27 Insertion Sort (cont.) 3 18 19 22 17 16 6 1 15 2 sorted sectionunsorted section

28 Insertion Sort (cont.) 3 18 19 22 17 16 6 1 15 2 sorted sectionunsorted section

29 Insertion Sort (cont.) 3 17 16 6 1 15 2 sorted sectionunsorted section 22 19 18

30 Insertion Sort (cont.) 3 17 16 6 1 15 2 sorted sectionunsorted section 22 19 18

31 Insertion Sort (cont.) sorted sectionunsorted section 3 17 16 6 1 15 2 22 19 18

32 Insertion Sort (cont.) sorted sectionunsorted section On each iteration of insertion sort, the sorted section grows by 1, and the unsorted section shrinks by 1, until the array is sorted. 3 17 16 6 1 15 2 22 19 18

33 Insertion Sort (cont.) sorted sectionunsorted section 16 would be the next value to be “inserted”, giving insertion sort its name 3 17 16 6 1 15 2 22 19 18

34 Insertion Sort (cont.) sorted sectionunsorted section In the algorithm, j is used to mark the index of the next element to be inserted. 3 17 16 6 1 15 2 22 19 18 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i--

35 Insertion Sort (cont.) sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 j 3 17 16 6 1 15 2 22 19 18 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i--

36 Insertion Sort (cont.) sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 j 3 17 16 6 1 15 2 22 19 18 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i--

37 Insertion Sort (cont.) sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 j 3 17 16 6 1 15 2 22 19 18 i 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i--

38 Insertion Sort (cont.) sorted sectionunsorted section 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 j 3 17 16 6 1 15 2 22 19 18 i

72 Insertion Sort (cont.) sorted sectionunsorted section 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 j 3 17 16 6 1 15 2 22 19 18 i etc.

73 Starting it Off 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- Why is it that we start j off at 1 instead of 0?

74 Starting it Off (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- A section of 1 element (at index 0) is already sorted! The index j starts off as the first element in the unsorted section.

75 Inversions An inversion between any two elements when the element that comes first in the array is greater than the element that comes next The array below has three inversions… 4 2 16 5 15 0 1 2 3 4

76 Inversions (cont.) An inversion between any two elements when the element that comes first in the array is greater than the element that comes next The array below has three inversion: 4 2 16 5 15 0 1 2 3 4 one inversion

77 Inversions (cont.) An inversion between any two elements when the element that comes first in the array is greater than the element that comes next The array below has three inversion: 4 2 16 5 15 0 1 2 3 4 a second inversion

78 Inversions (cont.) An inversion between any two elements when the element that comes first in the array is greater than the element that comes next The array below has three inversion: 4 2 16 5 15 0 1 2 3 4 a third inversion

79 Inversions and Swaps 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- A swap in insertion sort removes an inversion…

80 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- because when performed, A[ i ] was greater than A[ i + 1 ]

81 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 16 19 But it can’t remove more than one inversion – those elements inverted with 16 in the shaded part of the array…

82 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 16 19 will still be inverted with 16 after the swap…

86 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 16 19 The same is true with 19

87 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 16 19 So a swap removes one and only one inversion each time it is executed; it is also the only way inversions are removed

88 Inversions and Swaps (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 0 1 2 3 4 5 6 7 8 9 16 19 Therefore, the number of swaps performed in all of insertion sort is equal to the number of inversions in the original array.

89 Worst Number of Inversions The worst number of inversions occurs when an array is in descending order Each element is inverted with every other element (can’t get any worse) 23 19 16 12 10 0 1 2 3 4

90 Worst Number of Inversions (cont.) How many inversions are there? There are n elements, and each is inverted with (n – 1) elements, so n( n – 1 )? 23 19 16 12 10 0 1 2 3 4

91 Worst Number of Inversions (cont.) Not n( n – 1 ), because we are counting each inversion twice –19 inverted with 16 and 16 inverted with 19 The number of inversions is n( n – 1 ) / 2 23 19 16 12 10 0 1 2 3 4

92 Worst Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- Therefore, in the worst case, the number of swaps performed is n( n – 1 ) / 2 …

93 Worst Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- Therefore, in the worst case, the TOTAL number of swaps performed is n( n – 1 ) / 2 = ½ n 2 – ½ n which gives us the worst-case time complexity of  ( n 2 ) for insertion sort.

94 Average Number of Inversions 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- In the average case, we might assume that each possible inversion has a 50% chance of occurring. Therefore, on average, we would expect that the number of inversions is 50% of n( n – 1 ) / 2

95 Average Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 50% of n( n – 1 ) / 2 = ½ * n ( n – 1 ) / 2 = n( n – 1 ) / 4 (still  ( n 2 ) on average)

96 Best Number of Inversions 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 There are no inversions in an array that is already sorted, so the condition in the while loop is never true (no swaps)

97 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 The outer loop still executes n – 1 times (the algorithm does not know it is already sorted)

98 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 Thus, in the best case, insertion sort runs in  ( n ) time

99 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 Does an array have to be already sorted in order for insertion sort to run in  ( n ) time?

100 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 Consider the case where we have n inversions, so we have a total of n swaps

101 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 For each iteration of the for loop, a swap is executed an average of 1 time (approximately)

102 Best Number of Inversions (cont.) 1 for each j, from 1 to the length of A – 1 2 i = j – 1 3 while i is greater than -1 and A[ i ] is greater than A[ i + 1 ] 4 swap( A[ i ], A[ i + 1 ] ) 5 i-- 21 25 30 32 40 0 1 2 3 4 Hence, we would still have  ( n ) time if we have n inversions or less

103 Another Advantage of Insertion Sort Insertion sort is often used to sort a small number of elements Consider, in the average case, that we have n( n – 1 ) / 4 inversions However, n = n( n – 1 ) / 4 has a solution of n = 5 Therefore, at around 5 elements, insertion sort behaves like a  ( n ) sorting algorithm

104 Speeding up Insertion Sort We can use the same idea that we used in the heap to perform “one-assignment” swaps instead of full swaps We first save the element at index j to a temporary variable to hold it An iteration of the for loop would then behave like this…

105 One-Assignment Swap 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 With sorting in progress, let’s suppose j is set to 6 here

106 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 j

107 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 j

108 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 j temp: 16

109 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 j temp: 16

110 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 16 1 15 2 22 19 18 j temp: 16 i

113 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 1 2 22 19 18 j temp: 16 i

114 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 1 2 22 19 18 j temp: 16 i

115 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i— 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 1 2 22 19 18 j temp: 16 i 22

116 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i-- 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 1 2 22 19 18 j temp: 16 i 22

120 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i-- 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 17 15 1 2 19 18 j temp: 16 i 22

147 One-Assignment Swap (cont.) 1 for each j, from 1 to the length of A – 1 2 temp = A[ j ] 3 i = j – 1 4 while i is greater than -1 and A[ i ] is greater than temp 5 A[ i + 1 ] = A[ i ] 6 i-- 7 A[ i + 1] = temp sorted sectionunsorted section 0 1 2 3 4 5 6 7 8 9 3 16 15 1 2 19 18 17 j temp: 16 i 22 etc.

148 Quicksort Time complexities of Quicksort –best:  ( n lg n ) –average:  ( n lg n ) –worst:  ( n 2 ) The average case usually dominates over the worst case in sorting algorithms, and in Quicksort, the coefficient of the n lg n term is small –makes Quicksort one of the most popular general- purpose sorting algorithms

149 Functions of Quicksort A recursive function, called quicksort A nonrecursive function, usually called partition quicksort calls partition

150 Partition 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 In the partition function, the last element is chosen as a pivot – a special element used for comparison purposes pivot

151 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 Each element is compared to the pivot. When partition is done, it produces the result shown next… pivot

152 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 All elements less than or equal to the pivot are on its left

153 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 All elements greater than the pivot are on its right side

154 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 The pivot is where it will be in the final sorted array

155 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition is called from quicksort more than once, but when called again, it will work with a smaller section of the array.

156 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition is called from quicksort more than once, but when called again, it will work with a smaller section of the array. Partition (cont.)

159 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition is called from quicksort more than once, but when called again, it will work with a smaller section of the array. Partition (cont.)

160 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition (cont.) The next time partition is called, for example, it will only work with this section to the left of the previous pivot.

161 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition (cont.) The next time partition is called, for example, it will only work with this section to the left of the previous pivot.

162 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 38 47 11 3 6 4 35 46 Partition (cont.) pivot

163 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) pivot

164 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) Each section of the array separated by a previous pivot will eventually have partition called for it

165 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) Except that partition is not called for a section of just one element – it is already a pivot and it is where it belongs

166 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) Partition is not called for a section of just one element – it is already a pivot and it is where it belongs

167 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) Partition is called for this section

169 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) pivot All elements are smaller and stay to the left of the pivot

170 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 11 14 6 28 35 46 Partition (cont.) Partition is called for this section

173 0 1 2 3 4 5 6 7 8 9 10 11 3 10 4 28 38 47 14 11 6 28 35 46 Partition (cont.) partition

176 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46 Partition (cont.) partition not called for this

180 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 47 14 11 10 28 35 46 Partition (cont.) partition called for this

183 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 38 35 14 11 10 28 46 47 Partition (cont.) partition called for this

186 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47 Partition (cont.) Partition not called for this

190 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47 Partition (cont.) At this point, the array is sorted

191 0 1 2 3 4 5 6 7 8 9 10 11 3 6 4 28 35 38 14 11 10 28 46 47 Partition (cont.) But to achieve this, what steps does the algorithm for partition go through?

192 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 2 3 38 11 47 46 35 6 4 28 pivot Partition has a loop which iterates through each element, comparing it to the pivot

193 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot When partition is in progress, there is a partitioned section on the left, and an unpartitioned section on the right partitioned section unpartitioned section

194 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot The dividing line in the partitioned section means that all elements to the left are less than or equal to the pivot; all elements to the right are greater than the pivot partitioned section unpartitioned section

195 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot On each iteration, the partitioned section grows by one and the unpartitioned section shrinks by one, until the array is fully partitioned partitioned section unpartitioned section

196 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot i is an index used to mark the last value in the small side of the partition, while j is used to mark the first value in the unpartitioned section. partitioned section unpartitioned section

197 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot i is an index used to mark the last value in the small side of the partition, while j is used to mark the first value in the unpartitioned section. partitioned section unpartitioned section ij

198 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot On each iteration of partition, the value at j is compared to the pivot. partitioned section unpartitioned section ij

199 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot On each iteration of partition, the value at j is compared to the pivot. partitioned section unpartitioned section ij

200 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot If the value at j is greater, j is just incremented (the partitioned section grows by one) partitioned section unpartitioned section ij

201 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot If the value at j is greater, j is just incremented (the partitioned section grows by one) partitioned section unpartitioned section ij

202 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot If, on an iteration, the value at j is less than the pivot… partitioned section unpartitioned section ij

203 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot then i is incremented… partitioned section unpartitioned section ij

204 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot then i is incremented… partitioned section unpartitioned section ij

205 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot and the value at i is swapped with the value at j… partitioned section unpartitioned section ij

211 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot then j is incremented partitioned section unpartitioned section ij

212 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 pivot then j is incremented partitioned section unpartitioned section ij

213 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 Partition starts off by passing in the array arr…

214 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 the beginning index of the array section it is working with…

215 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 and the ending index of the array section it is working with

216 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 i is used to mark the end of the small-side part of the partition

217 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 initially, there isn’t one, so i is set to p – 1 (-1 if p is 0)

218 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 this loop iterates through the elements…

219 Partition (cont.) 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 comparing them to the pivot

220 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 j is currently 7, with partition having already iterated through elements 0 through 6 p r

221 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 35 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 p r

259 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 p r 35

262 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 46 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 p r 35 j isn’t incremented past r - 1

269 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 p r 35 46

272 Partition (cont.) 0 1 2 3 4 5 6 7 8 9 10 11 14 10 28 3 38 11 47 6 4 28 ij 1 partition( arr, p, r ) 2i = p – 1 3for all j from p to r – 1 4if arr[ j ] <= arr[ r ] 5i++ 6swap( arr[ i ], arr[ j ] ) 7swap ( arr[ i + 1 ] and arr[ r ] ) 8return i + 1 p r 35 46 The final step of partition returns the INDEX of the pivot back to the quicksort function that called it

273 The Quicksort Function 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) Since quicksort is called recursively, it also passes in the array arr, the beginning index of the section it is working with, and the ending section it is working with

274 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) If p < r, those same indexes are used in the call to partition (in such a case, a call to quicksort always produces a matching call to partition)

275 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) The index of the pivot is returned from partition and assigned to q

276 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) The index of the pivot is returned from partition and assigned to q pr q

277 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) quicksort is called recursively here, working with the section on the left of the pivot pr q

278 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) quicksort is called recursively here, working with the section on the left of the pivot pr q q-1

279 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) and quicksort is called recursively here, working with the section on the right of the pivot pr q q-1

280 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) and quicksort is called recursively here, working with the section on the right of the pivot pr q q-1q+1

281 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) if the array sections are at least 2 elements in size, these quicksort functions will call partition for the same array section it is working with

282 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) and partition will break the sections into even smaller sections

283 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) The recursive case occurs if p < r (this means the array section has at least two elements, the one at p and the one at r)

284 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) If the recursive case occurs, a call to quicksort will match a call to partition

285 The Quicksort Function (cont.) 1quicksort( arr, p, r ) 2if p < r 3q = partition( arr, p, r ) 4quicksort( arr, p, q – 1 ) 5quicksort( arr, q + 1, r ) If p == r, the base case occurs – there is only one element in the array section (recall that partition is not called for a section that just has one element)

286 Counting Sort The time complexity of counting sort depends on both n and also the range of the elements –for example, if the elements range in value from 25 to 30, the range is 30 – 25 + 1 = 6 If the range is smaller than n, counting sort is very fast, running in  ( n ) time

287 Counting Sort (cont.) The main counting sort algorithm works with elements that range in value from 0 to some positive integer k However, with a little modification, counting sort can work with nearly any data, still maintaining a  ( n ) time complexity Let’s look at the main counting sort algorithm…

288 Counting Sort Step 1 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 This is the array A to be sorted. It’s values range from 0 to 7, giving a range of 8. 8 < n (since n = 10), so counting sort should be fast on this array. A

289 Counting Sort Step 1 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0

290 Counting Sort Step 1 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 0 1 2 3 4 5 6 7 C

291 Counting Sort Step 1 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 0 1 2 3 4 5 6 7 C

292 Counting Sort Step 1 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 C

293 Counting Sort Step 2 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 C

294 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 C j

296 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 C j A[ j ] is 4, so this is really C[4]++

306 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 0 0 0 0 1 2 0 0 0 1 2 3 4 5 6 7 C j Notice that the C array is being used to “count” the number of values in the A array…

307 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 0 0 0 0 1 2 0 0 0 1 2 3 4 5 6 7 C j Since C[ 5 ] is 2, it means we’ve found 2 5’s in the A array so far

308 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 0 0 0 0 1 2 0 0 0 1 2 3 4 5 6 7 C j Thus, at the end of this loop, the C array will look like this…

309 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j Thus, at the end of this loop, the C array will look like this

310 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j The total of the values in C is equal to n (since it a total count of the number of elements).

311 Counting Sort Step 2 (cont.) 4 5 5 2 6 5 5 7 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j Notice that we don’t need the values in the A array any more – all information is contained in C.

312 Counting Sort Step 2 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C Notice that we don’t need the values in the A array any more – all information is contained in C.

313 Counting Sort Step 2 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C Now, it is just a matter of writing the appropriate indexes of C into the A array (one 0 goes in the first spot, two 2’s go in the next two places, etc.

314 Counting Sort Step 3 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 7j = 0

315 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 7j = 0 j

316 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++

317 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i

318 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i

319 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i m counts the number of times i has to be written into A[ j ]

320 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one time

321 Counting Sort Step 3 (cont.) 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one time

322 Counting Sort Step 3 (cont.) 0 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one time

325 Counting Sort Step 3 (cont.) 0 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i

328 Counting Sort Step 3 (cont.) 0 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i 0 times

332 Counting Sort Step 3 (cont.) 0 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i two times

333 Counting Sort Step 3 (cont.) 0 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i two times

334 Counting Sort Step 3 (cont.) 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i two times

337 Counting Sort Step 3 (cont.) 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i

338 Counting Sort Step 3 (cont.) 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one more time

339 Counting Sort Step 3 (cont.) 0 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one more time

340 Counting Sort Step 3 (cont.) 0 2 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C j 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one more time

341 Counting Sort Step 3 (cont.) 0 2 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one more time j

342 Counting Sort Step 3 (cont.) 0 2 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i one more time j

343 Counting Sort Step 3 (cont.) 0 2 2 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i etc. j

344 Counting Sort Step 3 (cont.) 0 2 2 4 5 5 5 5 6 7 0 1 2 3 4 5 6 7 8 9 A 1 0 2 0 1 4 1 1 0 1 2 3 4 5 6 7 C 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ i etc. j

345 Analysis of Counting Sort 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++

346 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( 1 )

347 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( k )  ( 1 )

348 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( k )  ( 1 )  ( n )

349 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( k )  ( 1 )  ( n )  ( 1 )

350 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( k )  ( 1 )  ( n )  ( 1 ) So far, we have  ( 1 ) +  ( k ) +  ( n ) +  ( 1 ), which is  ( n ) +  ( k )

351 Analysis of Counting Sort (cont.) 1COUNTING-SORT( A ) 2make an Array C of length k + 1 3for each i, from 0 to k 4C[ i ] = 0 5for each j, from 0 to the length of A - 1 6C[ A[ j ] ]++ 7j = 0 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++  ( k )  ( 1 )  ( n )  ( 1 ) What about this nested loop?

352 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++

353 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ This line is executed n times – if any less, array A wouldn’t get filled in – if any more, we would be writing values past the end of the array!

354 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ Thus, lines 8-11 would appear to have a  ( n ) time complexity…but there is more to it

355 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ What about all of the times that C[ i ] is 0? In these cases, the body of the inner loop doesn’t execute, but this line still executes a number of times

356 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ C[ i ] is checked to see if it is empty k + 1 times

357 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ Hence, these lines execute in  ( k ) +  ( n ) time and this is the time complexity of counting sort

358 Analysis of Counting Sort (cont.) 8for each i from 0 to k 9for each m from 1 to C[ i ] 10A[ j ] = i 11j++ If k < n, then counting sort is considered to run in  ( n ) time If k is much greater than n, we may not want to use it

359 Modifications to Counting Sort What if the least value in A is some minimum value min, where min is not equal to 0? We could adjust the A array by first subtracting min from each value (this would make min 0) Then, run counting sort Then, add min to each value to restore all the values The adjustments before and after counting sort take only  ( n ) time

360 Modifications to Counting Sort (cont.) What if we do not know the maximum or minimum values? These can also be found in  ( n ) time: min = A[0] max = A[0] for each j, from 1 to the length of A - 1 if A[ j ] > max then max = A[ j ] else if A[ j ] < min then min = A[ j ]

361 Modifications to Counting Sort (cont.) We may even want to run a test to see which sorting algorithm we should use: if max – min < n CountingSort( A ) else Quicksort( A, 0, n – 1)

362 Modifications to Counting Sort (cont.) What if we have floating-point numbers? Is counting sort out of the question? Not really. Consider monetary values. We could multiply each value by 100 to make an array of integers After running counting sort, we multiply each value by 0.01 to adjust it back

363 When Counting Sort Isn’t Practical The main thing that can kill the practicality of counting sort is if the range (measured in terms of the unit of precision) is much greater than n –unfortunately, very often the case

364 Sorting a Linked List Linked lists can be sorted too Consider going through a linked list, element by element, assigning each value to an array (  ( n ) time ) Then, sort the array Then, go through the linked list, assigning each value of the array to the linked list

365 Sorting a Linked List (cont.) A sorting algorithm must take at least  ( n ) time if nothing about the elements is known ahead of time –it takes this long just to look at the elements Therefore, assigning elements from linked list to array and back to linked list again will not affect the time complexity of a sorting algorithm

366 Copying Elements Recall that elements in linked lists will tend to be large –a reason for using linked lists Therefore, it would be very time- consuming to do this element copying We can sort a linked list without copying elements

367 Array of Pointers We use a dynamic array of pointers into the linked list Declared like this: Node ** arr = new Node * [numElements]; Why?

368 Array of Pointers (cont.) We use a dynamic array of pointers into the linked list Declared like this: Node ** arr = new Node * [numElements]; arr will point to the first element and the first element is a pointer

369 Array of Pointers (cont.) We use a dynamic array of pointers into the linked list Declared like this: Node ** arr = new Node * [numElements]; Therefore, arr is a pointer to a pointer

370 numElements We need numElements to make the declaration Set to 0 inside the constructor Increment when adding an element Decrement when removing Set to 0 when makeEmpty is called

371 Step 1 i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 6 3 8 7 start Note: Only the data members to be sorted are shown in the linked list – the objects may be quite large

372 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start 6 3 8 7

373 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start i 6 3 8 7

374 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start i 6 3 8 7

375 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start i ptr 6 3 8 7

391 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start i ptr etc. 6 3 8 7

392 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start ptr : NULL etc. 6 3 8 7

393 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start ptr : NULL This loop is a  ( n ) loop and did not copy any elements 6 3 8 7

394 Step 1 (cont.) i = 0 ptr = start while ptr is not equal to NULL arr[ i ] = ptr ptr = ptr->next i++ 0 1 2 3 start ptr : NULL Step 2 uses a modified sorting algorithm – instead of moving elements, we move array pointers We’ll use insertion sort… 6 3 8 7

395 Step 2 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr

396 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j

397 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j In the original insertion sort, temp held an element; now it holds an address

398 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j In the original insertion sort, temp held an element; now it holds an address tempPtr

399 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr

400 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i

401 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i This overloaded operator should pass its parameter by const reference…

402 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i to completely avoid copying these large elements

406 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i : -1

410 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i : -1 Note that we’ve swapped addresses (much faster than swapping elements)

419 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i no effect

434 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i The array is now sorted according to the pointers

435 Step 2 (cont.) 0 1 2 3 start 6 3 8 7 for each j, from 1 to the length of arr – 1 tempPtr = arr[ j ] i = j – 1 while i > -1 and arr[ i ]->info > tempPtr->info arr[ i + 1 ] = arr[ i ] i— arr[ i + 1] = tempPtr j tempPtr i Now, in step 3, we just relink the linked list

436 Step 3 0 1 2 3 start 6 3 8 7 for all i from 0 to n – 2 A[ i ]->next = A[ i + 1 ] start = A[ 0 ] A[ n – 1 ]->next = NULL

437 Step 3 (cont.) 0 1 2 3 start 6 3 8 7 for all i from 0 to n – 2 A[ i ]->next = A[ i + 1 ] start = A[ 0 ] A[ n – 1 ]->next = NULL i

452 Step 3 (cont.) 0 1 2 3 start 6 3 8 7 for all i from 0 to n – 2 A[ i ]->next = A[ i + 1 ] start = A[ 0 ] A[ n – 1 ]->next = NULL i The list has been relinked in order, without copying elements

453 Step 3 (cont.) 0 1 2 3 start 6 3 8 7 for all i from 0 to n – 2 A[ i ]->next = A[ i + 1 ] start = A[ 0 ] A[ n – 1 ]->next = NULL i This step also takes  ( n ) time

454 Modifying a Sorting Algorithm for a Linked List Those variables that were assigned elements in the sorting algorithm should now be assigned addresses (they become pointers) Use ->info when element values need to be compared

455 Modifying Counting Sort for a Linked List Each element of the C array does not need to be an integer An alternative is to use struct objects as elements of the C array One data member can be a count for the C array index Another data member can be an array of pointers to linked list nodes, relevant to that index of the C array (the data member for which the list is being sorted)

1 C++ Classes and Data Structures Jeffrey S. Childs Chapter 14 Introduction to Sorting Algorithms Jeffrey S. Childs Clarion University of PA © 2008, Prentice.

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1 C++ Classes and Data Structures Jeffrey S. Childs Chapter 14 Introduction to Sorting Algorithms Jeffrey S. Childs Clarion University of PA © 2008, Prentice.

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Presentation on theme: "1 C++ Classes and Data Structures Jeffrey S. Childs Chapter 14 Introduction to Sorting Algorithms Jeffrey S. Childs Clarion University of PA © 2008, Prentice."— Presentation transcript:

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