Lecture 3 data structures & algorithms - sorting techniques - http://techieme.in
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This document provides a 90-minute discussion on algorithms including quicksort, order statistics, searching, and substring searching. It begins with an overview of the topics and then provides details on quicksort, including the divide and conquer approach and partitioning elements around a pivot. It also describes algorithms for order statistics to find the kth smallest element, binary search, and a basic substring searching approach. Special cases and better solutions like Boyer-Moore are also mentioned.
![Partition(A, p, q) //A[p , q]
X ← A[p]
i ← p
for j ← p+1 to q
do if A[j] <= x
then i ← i+1
exchange A[i] ↔ A[j]
exchange A[p] ↔ A[i];
return i
4
A
p qi j](https://image.slidesharecdn.com/lecture3-datastructuresalgorithms-sortingtechniques-130521015559-phpapp01/85/Lecture-3-data-structures-algorithms-sorting-techniques-http-techieme-in-4-320.jpg)
![Algorithm:
OrderStat(A,p,q,k) // means kth smallest number in A between
index p and q
if p==q
return A[p]
r <- Partition(A,p,q)
i <- r – p + 1;
if k == i
return A[r]
if k < i
return OrderStat(A,p,r-1,k)
else
return OrderStat(A,r+1,q,k-i)
8](https://image.slidesharecdn.com/lecture3-datastructuresalgorithms-sortingtechniques-130521015559-phpapp01/85/Lecture-3-data-structures-algorithms-sorting-techniques-http-techieme-in-8-320.jpg)
![Searching:
Basic Idea: In an array A[a1,a2,a3,a4,…,an] find the index k
such that p = A[k]
Naïve Solution: Traverse through the array in a loop and
compare each element with the given number. If the number
matches, return the index of the number else return null.
Algorithm:
Search (A[1.. n], p)
for i<- 1 to n
do if A[i] == p
return i
return null
9](https://image.slidesharecdn.com/lecture3-datastructuresalgorithms-sortingtechniques-130521015559-phpapp01/85/Lecture-3-data-structures-algorithms-sorting-techniques-http-techieme-in-9-320.jpg)
![Searching:
Basic Idea: In an array A[a1,a2,a3,a4,…,an] find the index k such
that p = A[k]
Binary Search Solution: Only if the array is sorted. Divide it into two
halves, check the element at the center, if it is less than what we
are searching, look into the upper half else look into the lower
half. Repeat till you find the number or the array exhausts.
Algorithm:
BinarySearch (A, p, low,high)
middle = (low+high)/2
if A[middle] == p
return middle;
else if A[middle] > p
return BinarySearch(A, p, low, middle-1)
else
return BinarySearch(A,p,middle+1,high)
10](https://image.slidesharecdn.com/lecture3-datastructuresalgorithms-sortingtechniques-130521015559-phpapp01/85/Lecture-3-data-structures-algorithms-sorting-techniques-http-techieme-in-10-320.jpg)
![Special Case Substring Searching:
Basic Idea: In a character string search a substring and return the index of first occurrence.
Naive Solution: Start from the first index of both the strings, compare the characters, if
character matches, compare the next character and so on till the substring exhausts.
Return the start index of the substring in the main string.
Algorithm:
SubStringSearch (S, sb)
j=0;
match = false;
while i < S.length or j < sb.length
if S[i] == sb[j]
match = true
i++, j++
else
match = false
j=0, i++
if match == true and j = sb.length
return i-sb.length
else
return -1
11](https://image.slidesharecdn.com/lecture3-datastructuresalgorithms-sortingtechniques-130521015559-phpapp01/85/Lecture-3-data-structures-algorithms-sorting-techniques-http-techieme-in-11-320.jpg)
Lecture 3 data structures & algorithms - sorting techniques - http://techieme.in
- 1. Description: A detailed discussion about algorithms and their measures, and understanding sorting. Duration: 90 minutes Starts at: Saturday 12th May 2013, 11:00AM -by Dharmendra Prasad 1
- 2. Table of Contents 1. Continuing the Sorting Algorithms. 1. Quick Sort (in place sorting algorithm) 2. Searching Algorithms 1. Binary search 3. Some real world problem scenarios. 1. Substring search 2
- 3. Algorithm: It’s a divide and conquer algorithm. Step1(Divide): Partition a given array into 2 sub arrays around a pivot ‘x’ such that the elements in lower sub array <= x <= elements in upper sub array. Step2(Conquer):Recursively sort 2 sub arrays. Step3(Combine):Do nothing as it is in place sorting. 3
- 4. Partition(A, p, q) //A[p , q] X ← A[p] i ← p for j ← p+1 to q do if A[j] <= x then i ← i+1 exchange A[i] ↔ A[j] exchange A[p] ↔ A[i]; return i 4 A p qi j
- 5. 5 Example: 6 10 13 5 8 3 2 11 X = 6, i = 0, j = 1 6 5 13 10 8 3 2 11 6 5 3 10 8 13 2 11 6 5 3 2 8 13 10 11 Swap pivot with i 2 5 3 6 8 13 10 11
- 6. Algorithm: QuickSort(A, p, q) if p < q then r <- Partition(A, p, q) QuickSort( A, p, r-1) QuickSort( A, r+1, q) Initial Call : QuickSort( A, 1, n) 6
- 7. Order Statistics: Problem Statement: Given an array of numbers, find the kth smallest number. Naïve Solution: Sort the array and return the element at index k. Case1: if k = 1, we are referring to the minimum number in the array. Case2: if k = length of the array, we are referring to the maximum number in the array. Case3: when k lies between 1 and n where n is the length of the array 7
- 8. Algorithm: OrderStat(A,p,q,k) // means kth smallest number in A between index p and q if p==q return A[p] r <- Partition(A,p,q) i <- r – p + 1; if k == i return A[r] if k < i return OrderStat(A,p,r-1,k) else return OrderStat(A,r+1,q,k-i) 8
- 9. Searching: Basic Idea: In an array A[a1,a2,a3,a4,…,an] find the index k such that p = A[k] Naïve Solution: Traverse through the array in a loop and compare each element with the given number. If the number matches, return the index of the number else return null. Algorithm: Search (A[1.. n], p) for i<- 1 to n do if A[i] == p return i return null 9
- 10. Searching: Basic Idea: In an array A[a1,a2,a3,a4,…,an] find the index k such that p = A[k] Binary Search Solution: Only if the array is sorted. Divide it into two halves, check the element at the center, if it is less than what we are searching, look into the upper half else look into the lower half. Repeat till you find the number or the array exhausts. Algorithm: BinarySearch (A, p, low,high) middle = (low+high)/2 if A[middle] == p return middle; else if A[middle] > p return BinarySearch(A, p, low, middle-1) else return BinarySearch(A,p,middle+1,high) 10
- 11. Special Case Substring Searching: Basic Idea: In a character string search a substring and return the index of first occurrence. Naive Solution: Start from the first index of both the strings, compare the characters, if character matches, compare the next character and so on till the substring exhausts. Return the start index of the substring in the main string. Algorithm: SubStringSearch (S, sb) j=0; match = false; while i < S.length or j < sb.length if S[i] == sb[j] match = true i++, j++ else match = false j=0, i++ if match == true and j = sb.length return i-sb.length else return -1 11
- 12. Special Case Substring Searching: Basic Idea: In a character string search a substring and return the index of first occurrence. Better Solution: Boyre Moore algorithm is used to effectively search substring in a given string. 12