Input: marks > 9 Output: minimum mark of top 10% top = floor(n/10) for a = 0; . at+ if (0 0){ == for i = 1 ... top++ %3D for j = 0 ... n - i if marks [n-i] < marks [j] swap (marks[n-i], marks[j] return marks[n - top] } else { print("next number) for i = 1 .. top++ print(i) swap (marks[i], marks[i+1]) } print(a)

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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Solve the time Complexity of the above code snippet.

Please include details step by step. I don't need the answer alone.

I'll definitely show an upvote to your answer if it's correct and explicit.

Thanks for your help!

Input: marks[n], n > 9
Output: minimum mark of top 10%
1
2
3
4
top
floor(n/10)
%3D
for a = 0;
a++
...
6.
if (0 == 0){
1 ... top++
for j = 0 ... n - i
if marks [n-i] < marks [j]
swap (marks[n-i], marks[j])
7
for i
8
9.
10
11
return marks[n - top]
12
13
else {
print("next number)
for i = 1 ... top++
print(i)
swap (marks[i], marks[i+1])
14
15
16
17
18
}
19
20
print(a)
Transcribed Image Text:Input: marks[n], n > 9 Output: minimum mark of top 10% 1 2 3 4 top floor(n/10) %3D for a = 0; a++ ... 6. if (0 == 0){ 1 ... top++ for j = 0 ... n - i if marks [n-i] < marks [j] swap (marks[n-i], marks[j]) 7 for i 8 9. 10 11 return marks[n - top] 12 13 else { print("next number) for i = 1 ... top++ print(i) swap (marks[i], marks[i+1]) 14 15 16 17 18 } 19 20 print(a)
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