Hi I need help finding the time complexity The word case scenario for the following for loop. ```java/* Worst case scenario: what is the value of count for an array of size 'n' */public static boolean linS(int[] A, int st) { int count = 0; for( int i=0; i<10; i++ ) { count++; System.out.println(count); } int idx = 0; while( idx

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Database System Concepts

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ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

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Question

Hi I need help finding the time complexity The word case scenario for the following for loop.

```java
/* Worst case scenario:
what is the value of count for an array of size 'n' */
public static boolean linS(int[] A, int st) {
int count = 0;
for( int i=0; i<10; i++ ) {
count++;
System.out.println(count);
}
int idx = 0;
while( idx<A.length ) {
count++;
if( A[idx] == st ) {
break;
}
idx++;
}
return idx < A.length;
}
/* How would we describe the runtime of the above code? */
```

**Explanation:**

This Java code defines a method `linS` that performs a linear search on an array `A` to find a specific integer `st`. It first executes a simple loop 10 times, incrementing and printing a `count` variable each time. It then performs a linear search on the array `A`:

1. **Initialize `count` to 0**: This variable tracks the number of operations.

2. **First loop (constant time loop)**:
- Runs 10 times regardless of the size of `A`.
- Increments the `count` and prints it.

3. **Second loop (linear search on array)**:
- Iterates through each element of `A` from the beginning.
- Increments `count` by 1 for each iteration.
- Breaks the loop if `st` is found.

**Runtime Analysis:**

- **Best Case:** O(1) if `st` is found immediately.
- **Worst Case:** O(n) as the loop iterates through each element of the array `A` once until `st` is found or the end is reached.
- The first loop executes 10 times, so its runtime is constant, contributing a fixed amount of operations.

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