Considering the following algorithm, analyze its best case, worst case and average case time complexity in terms of a polynomial of n and the asymptotic notation of ɵ. You need to show the steps of your analysis.
Considering the following algorithm, analyze its best case, worst case and average case time complexity in terms of a polynomial of n and the asymptotic notation of ɵ. You need to show the steps of your analysis.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
Considering the following
![Consider the algorithm for the sorting problem that sorts an array by counting,
for each of its elements, the number of smaller elements and then uses this
information to put the element in its appropriate position in the sorted array:
ALGORITHM Comparison Counting Sort(A[0..n-1])
//Sorts an array by comparison counting
//Input: Array A[0..n - 1] of orderable values
//Output: Array S[0..n-1] of A's elements sorted
// in nondecreasing order
0 ton - 1 do
for i
Count[i]<0
for i 0 to n - 2 do
for ji+ 1 to n - 1 do
if A[i]<A[j]
Count[j] Count[j]+1
else Count[i]Count[i] + 1
for i 0 ton - 1 do
←
S[Count[i]] A[i]
return S](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2724f57d-1ebd-4904-b3e9-94cc6d926852%2Fc43ed777-1d49-4de4-9ea7-b273dffb9c7f%2Fh87x2tc_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the algorithm for the sorting problem that sorts an array by counting,
for each of its elements, the number of smaller elements and then uses this
information to put the element in its appropriate position in the sorted array:
ALGORITHM Comparison Counting Sort(A[0..n-1])
//Sorts an array by comparison counting
//Input: Array A[0..n - 1] of orderable values
//Output: Array S[0..n-1] of A's elements sorted
// in nondecreasing order
0 ton - 1 do
for i
Count[i]<0
for i 0 to n - 2 do
for ji+ 1 to n - 1 do
if A[i]<A[j]
Count[j] Count[j]+1
else Count[i]Count[i] + 1
for i 0 ton - 1 do
←
S[Count[i]] A[i]
return S
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