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Searching and Sorting. Gary Wong. Prerequisite. Time complexity Pseudocode (Recursion). Searching Linear (Sequential) Search Binary Search. Sorting Bubble Sort Merge Sort Quick Sort Counting Sort. Agenda. Linear Search. One by one. Linear Search.

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Prerequisite
Prerequisite

  • Time complexity

  • Pseudocode

  • (Recursion)


Agenda

Searching

Linear (Sequential) Search

Binary Search

Sorting

Bubble Sort

Merge Sort

Quick Sort

Counting Sort

Agenda


Linear search

Linear Search

One by one...


Linear search1
Linear Search

  • Check every element in the list, until the target is found

  • For example, our target is 38:

Not found!

Found!


Linear search2
Linear Search

  • Initilize an index variable i

  • Compare a[i] with target

    • If a[i]==target, found

    • If a[i]!=target,

      • If all have checked already, not found

      • Otherwise, change i into next index and go to step 2


Linear search3
Linear Search

  • Time complexity in worst case?

    • If N is number of elements,

    • Time complexity = O(N)

  • Advantage?

  • Disadvantage?


Binary search

Binary Search

Chop by half...


Binary search1

L

R

Binary Search

  • Given a SORTED list:

  • (Again, our target is 38)

Smaller!

Found!

Larger!


Binary search2
Binary Search

  • Why always in the middle, but not other positions, say one-third of list?

    1) Initialize boundaries L and R

    2) While L is still on the left of R

    • mid = (L+R)/2

    • If a[mid]>Target, set R be m-1 and go to step 2

    • If a[mid]<Target, set L be m+1 and go to step 2

    • If a[mid]==Target, found


Binary search3
Binary Search

  • Time complexity in the worst case?

    • If N is the number of elements,

    • Time complexity = O(lg N)

    • Why?

  • Advantage?

  • Disadvantage?


Example three little pigs
Example: Three Little Pigs

  • HKOI 2006 Final Senior Q1

  • Given three lists, each with M numbers, choose one number from each list such that their sum is maximum, but not greater than N.

  • Constraints:

    • M ≤ 3000

    • Time limit: 1s


Example three little pigs1
Example: Three Little Pigs

  • How many possible triples are there?

  • Why not check them all?

  • Time complexity?

  • Expected score = 50


Example three little pigs2
Example: Three Little Pigs

  • A simpler question: How would you search for the maximum number in ONE SORTED list such that it is not greater than N?

  • Binary search!

    • With slight modification though

    • How?


Example three little pigs3

L

R

Example: Three Little Pigs

  • Say, N=37


Example three little pigs4
Example: Three Little Pigs

  • Let’s go back to original problem

  • If you have chosen two numbers a1[i] and a2[j] already, how would you search for the third number?

  • Recall: How would you search for the maximum number in ONE SORTED list such that it is not greater than N-a1[i]-a2[j]?


Example three little pigs5
Example: Three Little Pigs

  • Overall time complexity?

  • Expected score = 90

  • 


Example three little pigs6
Example: Three Little Pigs

  • Slightly harder: Given TWO lists, each with M numbers, choose one number from each list such that their sum is maximum, but not greater than N.

  • Linear search?

  • Sort one list, then binary search!

  • Time complexity = O(M lg M)

    • O(M2) if less efficient sorting algorithm is used

  • But, can it be better?


Example three little pigs7
Example: Three Little Pigs

  • Why is it so slow if we use linear search?

  • If a1[i] and a2[j] are chosen, and their sum is smaller than N:

    • Will you consider any number in a1 that is smaller than or equal a1[i]?

  • If a1[i] and a2[j] are chosen, and their sum is greater than N:

    • Will you consider any number in a2 that is greater than or equal to a2[j]?


Example three little pigs8
Example: Three Little Pigs

  • Recall: Why is it so slow if we use linear search?

    • Because you use it for too many times!

  • At which number in each list should you begin the linear search?

  • Never look back at those we do not consider!

  • Time complexity?

  • Expected score = 100 


What can you learn
What can you learn?

  • Improve one ‘dimension’ using binary search

  • Linear search for a few times can be more efficient than binary search for many times!

    • DO NOT underestimate linear search!!!


Points to note
Points to note

  • To use binary search, the list MUST BE SORTED (either ascending or decending)

    • NEVER make assumptions yourself

    • Problem setters usually do not sort for you

  • Sorting is the bottleneck of efficiency

  • But... how to sort?


How to sort
How to sort?

  • For C++: sort()

  • Time complexity for sort() is O(N lg N)

    • which is considered as efficient

  • HOWEVER...

    • Problem setters SELDOM test contestants on pure usage of efficient sorting algorithms

    • Special properties of sorting algorithms are essential in problem-solving

  • So, pay attention! 


Bubble sort

Bubble Sort

Smaller? Float! Larger? Sink!


Bubble sort1
Bubble Sort

  • Suppose we need to sort in ascending order...

  • Repeatedly check adjacent pairs sequentially, swap if not in correct order

  • Example:

  • The last number is always the largest

18

9

20

11

77

45

Incorrect order, swap!

Correct order, pass!


Bubble sort2
Bubble Sort

  • Fix the last number, do the same procedures for first N-1 numbers again...

  • Fix the last 2 numbers, do the same procedures for first N-2 numbers again...

  • ...

  • Fix the last N-2 numbers, do the same procedures for first 2 numbers again...


Bubble sort3
Bubble Sort

for i -> 1 to n-1

for j -> 1 to n-i

if a[j]>a[j+1], swap them

  • How to swap?


Merge sort quick sort

Merge Sort & Quick Sort

Many a little makes a mickle...


Merge sort
Merge Sort

  • Now given two SORTED list, how would you ‘merge’ them to form ONE SORTED list?

List 1:

8

14

22

List 2:

10

13

29

65

Temporary list:

8

8

10

10

13

13

14

14

22

22

29

29

65

65

Combined list:


Merge sort1
Merge Sort

Merge

  • While both lists have numbers still not yet considered

    • Compare the current first number in two lists

    • Put the smaller number into temporary list, then discard it

  • If list 1 is not empty, add them into temporary list

  • If list 2 is not empty, add them into temporary list

  • Put the numbers in temporary list back to the desired list


Merge sort2
Merge Sort

  • Suppose you are given a ‘function’ called ‘mergesort(L,R)’, which can sort the left half and right half of list from L to R:

  • How to sort the whole list?

  • Merge them!

  • How can we sort the left and right half?

  • Why not making use of ‘mergesort(L,R)’?

10

13

29

65

8

14

22

L

(L+R)/2

(L+R)/2+1

R


Merge sort3
Merge Sort

mergesort(L,R){

If L is equal to R, done;

Otherwise,

m=(L+R)/2;

mergesort(L,M);

mergesort(M+1,R);

Merge the lists [L,M] and [M+1,R];

}


Merge sort4
Merge Sort

mergesort(0,6)

mergesort(0,1)

mergesort(2,3)

mergesort(4,5)

mergesort(6,6)

65

10

29

13

14

8

22

mergesort(0,0)

mergesort(1,1)

mergesort(2,2)

mergesort(3,3)

mergesort(4,4)

mergesort(5,5)

mergesort(0,3)

mergesort(4,6)

8

10

10

65

13

10

13

29

13

14

29

65

22

8

8

14

29

14

65

22


Merge sort5
Merge Sort

  • Time complexity?

  • O(N lg N)

  • Why?


Quick sort
Quick Sort

  • Choose a number as a ‘pivot’

  • Put all numbers smaller than ‘pivot’ on its left side

  • Put all numbers greater than (or equal to) ‘pivot’ on its right side

10

13

29

65

8

14

22

10

13

8

14

22

65

29


Quick sort1

x

y

Quick Sort

a[y] < pivot! shift x, swap!

  • How?

  • y shifts to right by 1 unit in every round

  • Check if a[y] < pivot

    • If yes, shift x to right by 1 unit and swap a[x] and a[y]

  • If y is at 2nd last position, swap a[x+1] and pivot

  • Time complexity?

10

13

29

65

8

14

22


Quick sort2
Quick Sort

  • Use similar idea as in merge sort

  • If we have a function called ‘quicksort(L,R)’...

    • Make use of ‘quicksort(L,R)’ to sort the two halves!

10

13

8

14

22

65

29


Quick sort3
Quick Sort

quicksort(L,R){

If L is equal to R, done;

Otherwise,

Choose a number as a pivot, say a[p];

Perform pivoting action;

quicksort(L,p-1);

quicksort(p+1,R);

}


Quick sort4
Quick Sort

  • Time complexity in worst case?

  • O(N2) 

  • What is the worst case?

  • Preventions:

    • Choose a random number as pivot

    • Pick 3 numbers, choose their median

  • Under randomization, expected time complexity is O(N lg N)


Counting sort

Counting Sort

No more comparison...


Counting sort1
Counting Sort

  • Create a list, then tally!

  • Count the occurences of elements

  • Example:

    • Sort {2,6,1,4,2,1,9,6,4,1} in ascending order

    • A list from 1 to 9 (or upper bound of input no.)

    • Three ‘1’s, two ‘2’s, two ‘4’s, two ‘6’s and one ‘9’

    • List them all!


Counting sort2
Counting Sort

  • Time complexity?

  • O(range * N)

  • Good for numbers with small range


More...

If we have time...


More...

  • Lower bound of sorting efficiency?!

  • Guess the answer by binary search

  • Count the number of ‘inversions’

  • Find the kth largest number

  • Other sorting algorithms



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