class GFG { public static void main(String[] args) ( int i, n 8; for (i = 1; i <= n; i=i*2) { System.out.printf ("Hello World !!!\n");
class GFG { public static void main(String[] args) ( int i, n 8; for (i = 1; i <= n; i=i*2) { System.out.printf ("Hello World !!!\n");
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Question
![### 1.3. Describe the Worst Case Running Time of the Example Codes in "Big-Oh" Notation in Terms of the Variable n.
#### Examples:
**Example 1:**
```java
class GFG {
public static void main(String[] args) {
int i, n = 8;
for (i = 1; i <= n; i = i * 2) {
System.out.printf("Hello World !!!\n");
}
}
}
```
- **Analysis:**
- The loop starts with `i = 1` and continues as long as `i <= n`, with `i` being doubled (`i = i * 2`) in each iteration.
- The condition `i <= n` will be true for the values `1, 2, 4, 8` (in this case n=8).
- Therefore, the loop runs `log₂(n)` times.
- **Big-Oh Notation:** O(log n)
**Example 2:**
```java
/**
* Returns the sum of the integers in a given array.
*/
public static int example1(int[] arr) {
int n = arr.length, total = 0;
for (int j = 0; j < n; j++) // loop from 0 to n-1
total += arr[j];
return total;
}
```
- **Analysis:**
- The `for` loop iterates from `j = 0` to `j < n`, where `n` is the length of the array.
- Each element of the array is accessed once, resulting in the loop running `n` times.
- **Big-Oh Notation:** O(n)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ae76079-938b-4b5d-8ceb-1dfbc10f88ce%2Fe48df711-a981-4bad-908d-f68cf5b4df25%2Fr7u0xg_processed.png&w=3840&q=75)
Transcribed Image Text:### 1.3. Describe the Worst Case Running Time of the Example Codes in "Big-Oh" Notation in Terms of the Variable n.
#### Examples:
**Example 1:**
```java
class GFG {
public static void main(String[] args) {
int i, n = 8;
for (i = 1; i <= n; i = i * 2) {
System.out.printf("Hello World !!!\n");
}
}
}
```
- **Analysis:**
- The loop starts with `i = 1` and continues as long as `i <= n`, with `i` being doubled (`i = i * 2`) in each iteration.
- The condition `i <= n` will be true for the values `1, 2, 4, 8` (in this case n=8).
- Therefore, the loop runs `log₂(n)` times.
- **Big-Oh Notation:** O(log n)
**Example 2:**
```java
/**
* Returns the sum of the integers in a given array.
*/
public static int example1(int[] arr) {
int n = arr.length, total = 0;
for (int j = 0; j < n; j++) // loop from 0 to n-1
total += arr[j];
return total;
}
```
- **Analysis:**
- The `for` loop iterates from `j = 0` to `j < n`, where `n` is the length of the array.
- Each element of the array is accessed once, resulting in the loop running `n` times.
- **Big-Oh Notation:** O(n)
Expert Solution
Step 1: Introduction
Running time analysis of algorithms plays a pivotal role in computer science. The "big-Oh" notation, commonly known as O-notation, is used to describe the upper bound on the running time of an algorithm in terms of the size of its input. It gives us a high-level understanding of the time complexity of an algorithm, allowing us to compare its efficiency with other algorithms. The examples provided appear to be Java methods, and we're going to dive deep into understanding their time complexities.
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