searching and sorting gary wong. prerequisite time complexity pseudocode (recursion)

Searching and SortingGary Wong

Prerequisite

• Time complexity• Pseudocode• (Recursion)

Agenda

• Searching– Linear (Sequential)

Search– Binary Search

• Sorting– Bubble Sort– Merge Sort– Quick Sort– Counting Sort

Linear SearchOne by one...

Linear Search

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

• For example, our target is 38:

i 0 1 2 3 4 5

a[i] 25 14 9 38 77 45

Not found!

Found!

Linear Search

1) Initilize an index variable i2) 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?– If N is number of elements,– Time complexity = O(N)

• Advantage?• Disadvantage?

Binary SearchChop by half...

Binary Search

• Given a SORTED list:• (Again, our target is 38)

i 0 1 2 3 4 5

a[i] 9 14 25 38 45 77

L R

Smaller! Larger!Found!

Binary Search

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

1) Initialize boundaries L and R2) 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?– 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


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


• 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?


• Say, N=37

i 0 1 2 3 4 5

a[i] 9 14 25 38 45 77

L R


• 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]?


• Overall time complexity?• Expected score = 90•


• 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?


• 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]?


• 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!

Bubble SortSmaller? Float! Larger? Sink!

• 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

Bubble Sort

9 20 1118 4577

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-1for j -> 1 to n-i

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

• How to swap?

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

Merge Sort

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

148 22

10 13 29 65

List 1:

List 2:

Temporary list: 8 10 13 14 22 29 65

Combined list:

8 10 13 14 22 29 65

Merge Sort

Merge1) 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

2) If list 1 is not empty, add them into temporary list3) If list 2 is not empty, add them into temporary list4) 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)’?

148 2210 13 29 65L R(L+R)/2 (L+R)/2+1

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

14 8 2210 132965

mergesort(0,6)

mergesort(0,3)

mergesort(0,1)

mergesort(0,0) mergesort(1,1)

mergesort(2,3)


mergesort(4,6)



10 65 291310 13 6529 148 148 2210 13 6529148 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

148 2210 13 29 65

6522 2910 13 8 14

Quick Sort• 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?

148 2210 13 29 65

x y

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

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!

6522 2910 13 8 14

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 SortNo 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 lunchYummy!

searching and sorting gary wong. prerequisite time complexity pseudocode (recursion)

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