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
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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
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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.
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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
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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)
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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
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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)
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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
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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)
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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
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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.
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13. Question
&
Answers
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