04 ds and algorithm session_05

Data Structures and Algorithms
Objectives

In this session, you will learn to:
Sort data by using quick sort
Sort data by using merge sort
Search data by using linear search technique
Search data by using binary search technique

Ver. 1.0 Session 5

Sorting Data by Using Quick Sort

Quick sort algorithm:
Is one of the most efficient sorting algorithms
Is based on the divide and conquer approach
Successively divides the problem into smaller parts until the
problems become so small that they can be directly solved

Ver. 1.0 Session 5

Implementing Quick Sort Algorithm

In quick sort algorithm, you:
Select an element from the list called as pivot.
Partition the list into two parts such that:
– All the elements towards the left end of the list are smaller than
the pivot.
– All the elements towards the right end of the list are greater than
the pivot.
Store the pivot at its correct position between the two parts of
the list.
You repeat this process for each of the two sublists created
after partitioning.
This process continues until one element is left in each
sublist.

Ver. 1.0 Session 5

Implementing Quick Sort Algorithm (Contd.)

To understand the implementation of quick sort algorithm,
consider an unsorted list of numbers stored in an array.

0 1 2 3 4 5 6 7
arr 28 55 46 38 16 89 83 30

Ver. 1.0 Session 5


Let us sort this unsorted list.

0 1 2 3 4 5 6 7
arr 28 55 46 38 16 89 83 30

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Select a Pivot.

0 1 2 3 4 5 6 7
arr 28 55 46 38 16 89 83 30

Pivot

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Start from the left end of the list (at index 1).
Move in the left to right direction.
Search for the first element that is greater than the pivot
value.
Greater element

0 1 2 3 4 5 6 7
arr 28 55 46 38 16 89 83 30

Pivot
Greater Value

Ver. 1.0 Session 5


Start from the right end of the list.
Move in the right to left direction.
Search for the first element that is smaller than or equal to
the pivot value.
Smaller element

0 1 2 3 4 5 6 7
arr 28 55 46 38 16 89 83 30

Pivot
Greater Value Smaller Value

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Interchange the greater value with smaller value.

Swap
0 1 2 3 4 5 6 7
arr 28 16
55 46 38 16
55 89 83 30

Pivot

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Continue the search for an element greater than the pivot.
Start from arr[2] and move in the left to right direction.
value.
Greater element

0 1 2 3 4 5 6 7
arr 28 16 46 38 55 89 83 30

Pivot
Greater Value

Ver. 1.0 Session 5


Continue the search for an element smaller than the pivot.
Start from arr[3] and move in the right to left direction.
the pivot value.
Smaller element

0 1 2 3 4 5 6 7
arr 28 16 46 38 55 89 83 30

Pivot
Greater Value
Smaller Value

Ver. 1.0 Session 5


The smaller value is on the left hand side of the greater
value.
Values remain same.

0 1 2 3 4 5 6 7
arr 28 16 46 38 55 89 83 30

Pivot
Greater Value
Smaller Value

Ver. 1.0 Session 5


List is now partitioned into two sublists.
List 1 contains all values less than or equal to the pivot.
List 2 contains all the values greater than the pivot.

0 0 1 1 2 3 4 2 53 6 4 7 5 6 7
arr 28 16 16 46
28 38 55 46 38 55 89
89 83 30 83 30

List 1 List 2
Pivot

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Replace the pivot value with the last element of List 1.
The pivot value, 28 is now placed at its correct position in
the list.

Swap
0 1 2 3 4 5 6 7
arr 28
16 16
28 46 38 55 89 83 30

List 1 List 2

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Truncate the last element, that is, pivot from List 1.

00 1 2 3 4 5 6 7
arr 16 28 46 38 55 89 83 30

List 1 List 2

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List 1 has only one element.
Therefore, no sorting required.

0 2 3 4 5 6 7
arr 16 46 38 55 89 83 30

List 1 List 2

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Sort the second list, List 2.

0 2 3 4 5 6 7
arr 16 46 38 55 89 83 30

List 1 List 2

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Select a pivot.
The pivot in this case will be arr[2], that is, 46.

0 2 3 4 5 6 7
arr 16 46 38 55 89 83 30

List 1 List 2
Pivot

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Start from the left end of the list (at index 3).
Move in the left to right direction.
value.
Greater element

0 2 3 4 5 6 7
arr 16 46 38 55 89 83 30

List 1 List 2
Pivot
Greater Value

Ver. 1.0 Session 5


Start from the right end of the list (at index 7).
Move in the right to left direction.
the pivot value.
Smaller element

0 2 3 4 5 6 7
arr 16 46 38 55 89 83 30

List 1 List 2
Pivot

Ver. 1.0 Session 5


Interchange the greater value with smaller value.

Swap
0 2 3 4 5 6 7
arr 16 46 38 30
55 89 83 30
55

List 1 List 2
Pivot

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Continue the search for an element greater than the pivot.
Start from arr[5] and move in the left to right direction.
value.
Greater element

0 2 3 4 5 6 7
arr 16 46 38 30 89 83 55

List 1 List 2
Pivot
Greater Value

Ver. 1.0 Session 5


Continue the search for an element smaller than the pivot.
Start from arr[6] and move in the right to left direction.
Search for the first element that is smaller than the pivot
value.
Smaller element

0 2 3 4 5 6 7
arr 16 46 38 30 89 83 55

List 1 List 2
Pivot
Greater Value
Smaller Value

Ver. 1.0 Session 5


The smaller value is on the left hand side of the greater
value.
Values remain same.

0 2 3 4 5 6 7
arr 16 46 38 30 89 83 55

List 1 List 2
Pivot
Greater Value
Smaller Value

Ver. 1.0 Session 5


Divide the list into two sublists.
Sublist 1 contains all values less than or equal to the pivot.
Sublist 2 contains all the values greater than the pivot.

0 1 2 3 4 5 6 7
arr 16 28 46 38 30 89 83 55

Ver. 1.0 Session 5


Replace the pivot value with the last element of Sublist 1.
The pivot value, 46 is now placed at its correct position in
the list.
This process is repeated until all elements reach their
correct position.
Swap
0 1 2 3 4 5 6 7
arr 16 28 46
30 38 30
46 89 83 55

Sublist 1 Sublist 2

Ver. 1.0 Session 5


Write an algorithm to implement quick sort:
QuickSort(low,high)
1. If (low > high):
a. Return
2. Set pivot = arr[low]
3. Set i = low + 1
4. Set j = high
5. Repeat step 6 until i > high or arr[i] > pivot // Search for an

// element greater than
//
pivot
6. Increment i by 1
7. Repeat step 8 until j < low or arr[j] < pivot // Search for an element

// smaller than pivot
8. Decrement j by 1
9. If i < j: // If greater element is on the left of smaller element
Ver. 1.0 Session 5


10. If i <= j:
a. Go to step 5 // Continue the search
11. If low < j:
a. Swap arr[low] with arr[j] // Swap pivot with last element in
// first part of the list
12. QuickSort(low, j – 1) // Apply quicksort on list left to pivot
13. QuickSort(j + 1, high) // Apply quicksort on list right to pivot

Ver. 1.0 Session 5

Determining the Efficiency of Quick Sort Algorithm

• The total time taken by this sorting algorithm depends on
the position of the pivot value.
• The worst case occurs when the list is already sorted.
• If the first element is chosen as the pivot, it leads to a worst
2
case efficiency of O(n ).
• If you select the median of all values as the pivot, the
efficiency would be O(n log n).

Ver. 1.0 Session 5

Just a minute

What is the total number of comparisons for an average
case in a quick sort algorithm?

Answer:
O(n log n)

Ver. 1.0 Session 5

Activity: Sorting Data by Using Quick Sort Algorithm

Problem Statement:
Write a program that stores 10 numbers in an array, and sorts
them by using the quick sort algorithm.

Ver. 1.0 Session 5

Sorting Data by Using Merge Sort

Merge sort algorithm:
Is based on the divide and conquer approach
Divides the list into two sublists of sizes as nearly equal as
possible
Sorts the two sublists separately by using merge sort
Merges the sorted sublists into one single list

Ver. 1.0 Session 5

Implementing Merge Sort Algorithm

To understand the implementation of merge sort algorithm,
consider an unsorted list of numbers stored in an array.

0 1 2 3 4 5 6
arr 53 10 30 76 3 57 24

Ver. 1.0 Session 5

Implementing Merge Sort Algorithm (Contd.)

Let us sort this unsorted list.

0 1 2 3 4 5 6
arr 53 10 30 76 3 57 24

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The first step to sort data by using merge sort is to split the
list into two parts.

0 1 2 3 4 5 6
arr 53 10 30 76 3 57 24

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The list has odd number of elements, therefore, the left
sublist is longer than the right sublist by one entry.

0 1 2 3 4 5 6
arr 53 10 30 76 3 57 24

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Further divide the two sublists into nearly equal parts.

0 1 2 2 3 3 4 4 5 5 6 6
arr 53 10
10 30 76
30 76 3 3 5757 24 24

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Further divide the sublists.

0 11 22 3 3 4 45 56 6
arr 53 1010 30 30 76 76 3
10 30 76 357 24
57 24

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There is a single element left in each sublist.
Sublists with one element require no sorting.

0 1 2 3 4 5 6
arr 53 10 30 76 3 57 24

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Start merging the sublists to obtain a sorted list.

0 11 22 3 3 4 5
4 6
5 6
arr 10
53 5310 3030 76 76 3 57
3 57
24
57 24

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Further merge the sublists.

0 1 22 3 3 4 4 55 6 6
arr 10 30
53 53 7676
30 3 3 2457 57 57
24

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Again, merge the sublists.

0
0 1
1 2
2 3 44 55 6 6
arr 10
3 30
10 53 76
24 30 53 5724 7657
3

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The list is now sorted.

0 1 2 3 4 5 6
arr 3 10 24 30 53 57 76

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Write an algorithm to implement merge sort:
MergeSort(low,high)
1. If (low >= high):
a. Return
2. Set mid = (low + high)/2
3. Divide the list into two sublists of nearly equal lengths, and sort
each sublist by using merge sort. The steps to do this are as
follows:
a. MergeSort(low, mid
b. MergeSort(mid + 1, high)
4. Merge the two sorted sublists:
a. Set i = low
b. Set j = mid + 1
c. Set k = low
d. Repeat until i > mid or j > high: // This loop will terminate when
// you reach the
end of one of the
// two sublists.

Ver. 1.0 Session 5


i. If (arr[i] <= arr[j])
Store arr[i] at index k in array B
Increment i by 1
Else
Store arr[j] at index k in array B
Increment j by 1
ii. Increment k by 1
e. Repeat until j > high: // If there are still some elements in the
// second sublist append them to the new list

i. Store arr[j] at index k in array B
ii. Increment j by 1
iii. Increment k by 1
f. Repeat until i > mid: // If there are still some elements in the
// first sublist append them to the new list
i. Store arr[i] at index k in array B
ii. Increment I by 1
iii. Increment k by 1
5. Copy all elements from the sorted array B into the original array arr

Ver. 1.0 Session 5

Determining the Efficiency of Merge Sort Algorithm

To sort the list by using merge sort algorithm, you need to
recursively divide the list into two nearly equal sublists until
each sublist contains only one element.
To divide the list into sublists of size one requires log n
passes.
In each pass, a maximum of n comparisons are performed.
Therefore, the total number of comparisons will be a
maximum of n × log n.
The efficiency of merge sort is equal to O(n log n)
There is no distinction between best, average, and worst
case efficiencies of merge sort because all of them require
the same amount of time.

Ver. 1.0 Session 5

Just a minute

Which algorithm uses the following procedure to sort a
given list of elements?
1. Select an element from the list called a pivot.
2. Partition the list into two parts such that one part contains
elements lesser than the pivot, and the other part contains
elements greater than the pivot.
3. Place the pivot at its correct position between the two lists.
4. Sort the two parts of the list using the same algorithm.

Answer:
Quick sort

Ver. 1.0 Session 5

Just a minute

On which algorithm design technique are quick sort and
merge sort based?

Answer:
Quick sort and merge sort are based on the divide and conquer
technique.

Ver. 1.0 Session 5

Performing Linear Search

Linear Search:
Is the simplest searching method
Is also referred to as sequential search
Involves comparing the items sequentially with the elements in
the list

Ver. 1.0 Session 5

Implementing Linear Search

The linear search would begin by comparing the required
element with the first element in the list.
If the values do not match:
The required element is compared with the second element in
the list.
If the values still do not match:
The required element is compared with the third element in the
list.
This process continues, until:
The required element is found or the end of the list is reached.

Ver. 1.0 Session 5

Implementing Linear Search (Contd.)

Write an algorithm to search for a given employee ID in a
list of employee records by using linear search algorithm:
1. Read the employee ID to be searched
2. Set i = 0
3. Repeat step 4 until i > n or arr[i] = employee ID
4. Increment i by 1
5. If i > n:
Display “Not Found”
Else
Display “Found”

Ver. 1.0 Session 5

Determining the Efficiency of Linear Search

The efficiency of a searching algorithm is determined by the
running time of the algorithm.
In the best case scenario:
The element is found at the first position in the list.
The number of comparisons in this case is 1.
The best case efficiency of linear search is therefore, O(1).
In the worst case scenario:
The element is found at the last position of the list or does not
exists in the list.
The number of comparisons in this case is equal to the number
of elements.
The worst case efficiency of linear search is therefore, O(n).

Ver. 1.0 Session 5

Determining the Efficiency of Linear Search (Contd.)

In the average case scenario:
The number of comparisons for linear search can be
determined by finding the average of the number of
comparisons in the best and worst case.
The average case efficiency of linear search is 1/2(n + 1).

Ver. 1.0 Session 5

Just a minute

You have to apply linear search to search for an element in
an array containing 5,000 elements. If, at the end of the
search, you find that the element is not present in the array,
how many comparisons you would have made to search the
required element in the given list?

Answer:
5,000

Ver. 1.0 Session 5

Activity: Performing Linear Search

Problem Statement:
Write a program to search a given number in an array that
contains a maximum of 20 numbers by using the linear search
algorithm. If there are more than one occurrences of the
element to be searched, then the program should display the
position of the first occurrence. The program should also
display the total number of comparisons made.

Ver. 1.0 Session 5

Performing Binary Search

Binary search algorithm:
Is used for searching large lists
Searches the element in very few comparisons
Can be used only if the list to be searched is sorted

Ver. 1.0 Session 5

Implementing Binary Search

Consider an example.
You have to search for the name Steve in a telephone
directory that is sorted alphabetically.
To search the name Steve by using binary search algorithm:
– You open the telephone directory at the middle to determine which
half contains the name.
– Open that half at the middle to determine which quarter of the
directory contains the name.
– Repeat this process until the name Steve is not found.
– Binary search reduces the number of pages to be searched by
half each time.

Ver. 1.0 Session 5

Implementing Binary Search (Contd.)

Consider a list of 9 elements in a sorted array.

0 1 2 3 4 5 6 7 8
arr 9 13 17 19 25 29 39 40 47

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You have to search an element 13 in the given list.

0 1 2 3 4 5 6 7 8
arr 9 13 17 19 25 29 39 40 47

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Determine the index of the middlemost element in the list:
Mid = (Lower bound + Upper bound)/2
= (0 + 8)/2
=4

0 1 2 3 4 5 6 7 8
arr 9 13 17 19 25 29 39 40 47

Lower bound Middle element Upper bound

Ver. 1.0 Session 5


13 is not equal to the middle element, therefore, again
divide the list into two halves:
Mid = (Lower bound + Upper bound)/2
= (0 + 3)/2
=1

0 1 2 3 4 5 6 7 8
arr 9 13 17 19 25 29 39 40 47

Lower bound Upper bound Upper bound

Middle element Middle element

Ver. 1.0 Session 5


13 is equal to middle element.
Element found at index 1.

0 1 2 3 4 5 6 7 8
arr 9 13 17 19 25 29 39 40 47

Lower bound Upper bound

Element found

Ver. 1.0 Session 5


Write an algorithm to implement binary search algorithm.
1. Accept the element to be searched
2. Set lowerbound = 0
3. Set upperbound = n – 1
4. Set mid = (lowerbound + upperbound)/2
5. If arr[mid] = desired element:
a. Display “Found”
b. Go to step 10
6. If desired element < arr[mid]:
a. Set upperbound = mid – 1

Ver. 1.0 Session 5


1. If desired element > arr[mid]:
a. Set lowerbound = mid + 1
2. If lowerbound <= upperbound:
a. Go to step 4
3. Display “Not Found”
4. Exit

Ver. 1.0 Session 5

Determining the Efficiency of Binary Search

In binary search, with every step, the search area is
reduced to half.
In the best case scenario, the element to be search is found
at the middlemost position of the list:
The number of comparisons in this case is 1.
In the worst case scenario, the element is not found in the
list:
– After the first bisection, the search space is reduced to n/2
elements, where n is the number of elements in the original list.
– After the second bisection, the search space is reduced to n/4
2
elements, that is, n/2 elements.
– After ith bisections, the number of comparisons would be n/2i
elements.

Ver. 1.0 Session 5

Just a minute

In ___________ search algorithm, you begin at one end of
the list and scan the list until the desired item is found or the
end of the list is reached.

Answer:
linear

Ver. 1.0 Session 5

Just a minute

To implement __________ search algorithm, the list should
be sorted.

Answer:
binary

Ver. 1.0 Session 5

Activity: Performing Binary Search

Problem Statement:
Write a program to search a number in an array that contains a
maximum of 20 elements by using binary search. Assume that
the array elements are entered in ascending order. If the
number to be searched is present at more than one location in
the array, the search should stop when one match is found.
The program should also display the total number of
comparisons made.

Ver. 1.0 Session 5

Summary

In this session, you learned that:
Quick sort and merge sort algorithms are based on the divide
and conquer technique.
To sort a list of items by using the quick sort algorithm, you
need to:
Select a pivot value.
Partition the list into two sublists such that one sublist contains
all items less than the pivot, and the second sublist contains all
items greater than the pivot.
Place the pivot at its correct position between the two sublists.
Sort the two sublists by using quick sort.

Ver. 1.0 Session 5

Summary (Contd.)

– The total time taken by the quick sort algorithm depends on
the position of the pivot value and the initial ordering of
elements.
2
– The worst case efficiency of the quick sort algorithm is O(n ).
– The best case efficiency of the quick sort algorithm is O(n log
n).
– To sort a list of items by using merge sort, you need to:
– Divide the list into two sublists.
– Sort each sublist by using merge sort.
– Merge the two sorted sublists.
– The merge sort algorithm has an efficiency of O(n log n).

Ver. 1.0 Session 5

Summary (Contd.)

The best case efficiency of linear search is O(1) and the worst
case efficiency of linear search is O(n).
To apply binary search algorithm, you should ensure that the
list to be searched is sorted.
The best case efficiency of binary search is O(1) and the
worst case efficiency of binary search is O(log n).

Ver. 1.0 Session 5

04 ds and algorithm session_05

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