Question 3 Analyze the time complexity of each block of code and give the big-O notation. 1) int n 87 for (int i 0; i < n; i++) System.out.println ("Hello World"); 2), int n 20; 0; i < n; i++) · i; j < n; j++) for (int i for (int j arr[i] [j] =i+j; 3) int n 20; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) %3D for (int k = 0; k < n; k++) C[i) []] C[1] [j] + A[i][k] + B[k] [j];
Question 3 Analyze the time complexity of each block of code and give the big-O notation. 1) int n 87 for (int i 0; i < n; i++) System.out.println ("Hello World"); 2), int n 20; 0; i < n; i++) · i; j < n; j++) for (int i for (int j arr[i] [j] =i+j; 3) int n 20; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) %3D for (int k = 0; k < n; k++) C[i) []] C[1] [j] + A[i][k] + B[k] [j];
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 3 Analyze the time complexity of each block of code and give the big-O notation.
1)
int n
8;
0; i < n; i++)
System.out.println ("Hello World");
for (int i =
2)
int n
= 203;
for (int i
= 0; i < n; i++)·
for (int j = i; j < n; j++)
i + j;
arr[i][j]
3)
int n
= 20;
for (int i = 0; i < n; i++)
for (int j
0; j<n; j++)
for (int k = 0; k < n; k++)
C[1] [j]
C[1][j] + A[i] [k] + B[k] []];
int n
20;
count
=03B
while (n > 1)
n =n / 2;
count++;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff16a53e-7a5b-4b93-b68f-c79dc179f09f%2F146bcafa-b9ab-45c0-96aa-10833f1c3dff%2Fh9tvup_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 3 Analyze the time complexity of each block of code and give the big-O notation.
1)
int n
8;
0; i < n; i++)
System.out.println ("Hello World");
for (int i =
2)
int n
= 203;
for (int i
= 0; i < n; i++)·
for (int j = i; j < n; j++)
i + j;
arr[i][j]
3)
int n
= 20;
for (int i = 0; i < n; i++)
for (int j
0; j<n; j++)
for (int k = 0; k < n; k++)
C[1] [j]
C[1][j] + A[i] [k] + B[k] []];
int n
20;
count
=03B
while (n > 1)
n =n / 2;
count++;
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