The pseudocode provided is an implementation of Bubble Sort. Let’s analyze the number of swap operations in the case where the array is initially sorted in descending order:
Step 1: Understanding the Loop Execution
The outer loop runs from \( i = 0 \) to \( n-2 \), where \( n = 30 \). Therefore, the outer loop executes \( 29 \) times.
The inner loop runs from \( j = 0 \) to \( n-i-2 \), meaning the number of iterations decreases as \( i \) increases.
Step 2: Calculating Total Comparisons
The number of comparisons for each value of \( i \) is:
- For \( i = 0 \), the inner loop runs \( n-1 \) times.
- For \( i = 1 \), the inner loop runs \( n-2 \) times.
- For \( i = 2 \), the inner loop runs \( n-3 \) times.
- …
- For \( i = n-2 \), the inner loop runs 1 time.
Thus, the total number of comparisons is the sum of the first \( n-1 \) integers:
\[ \text{Total comparisons} = (n-1) + (n-2) + \ldots + 1 = \frac{n(n-1)}{2} \] For \( n = 30 \):
\[ \text{Total comparisons} = \frac{30 \times 29}{2} = 435 \] Step 3: Number of Swaps
In Bubble Sort, a swap occurs every time a comparison finds that \( A[j] > A[j+1] \). Since the array is initially sorted in descending order, every comparison will result in a swap.
Thus, the number of swaps is equal to the total number of comparisons, which is 435.
Final Answer: The number of swap operations performed is 435.
Consider the following code:
main() {
int x = 126, y = 105;
{
if (x > y)
x = x - y;
else
y = y - x;
}
while (x != y)
printf("%d", x);
}
Consider the following code:
int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);