Searching and Sorting Gary Wong

Prerequisite Time complexity Pseudocode (Recursion)

Agenda Searching Sorting Linear (Sequential) Search Binary Search Bubble Sort Merge Sort Quick Sort Counting Sort

Linear Search One by one...

Linear Search Check every element in the list, until the target is found For example, our target is 38: i 1 2 3 4 5 a[i] 25 14 9 38 77 45 Not found! Found!

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 Search Time complexity in worst case? Advantage? Disadvantage? If N is number of elements, Time complexity = O(N) Advantage? Disadvantage?

Binary Search Chop by half...

Binary Search Given a SORTED list: (Again, our target is 38) i 1 2 3 4 Smaller! Found! Larger! i 1 2 3 4 5 a[i] 9 14 25 38 45 77 L R

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 Search Time complexity in the worst case? Advantage? If N is the number of elements, Time complexity = O(lg N) Why? Advantage? Disadvantage?

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 Pigs How many possible triples are there? Why not check them all? Time complexity? Expected score = 50

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 Pigs Say, N=37 i 1 2 3 4 5 a[i] 9 14 25 38 45 77 L R

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 Pigs Overall time complexity? Expected score = 90 

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 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 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? 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 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? 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! 

Smaller? Float! Larger? Sink! Bubble Sort Smaller? Float! Larger? Sink!

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 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 Sort for i -> 1 to n-1 for j -> 1 to n-i if a[j]>a[j+1], swap them How to swap?

Many a little makes a mickle... Merge Sort & Quick Sort Many a little makes a mickle...

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 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 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 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 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 13 10 65 29 13 13 14 65 29 22 8 8 29 14 14 65 22

Merge Sort Time complexity? O(N lg N) Why?

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

a[y] < pivot! shift x, swap! 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 x y In fact, initial positions of x and y should be shifted to the left by 1 unit.

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 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 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 No more comparison...

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 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

Time for lunch Yummy!