How many comparisons does a bubble sort typically require for n elements?

How many comparisons does a bubble sort typically require for n elements?

• A. n log n
• B. n^2
• C. n^2/2
• D. log n

Answer: (C)

Bubble sort is a simple sorting algorithm that repeatedly steps through the list, compares adjacent elements, and swaps them if they are in the wrong order. The process is repeated until the list is sorted. To understand how many comparisons bubble sort typically makes, consider the following: In the worst-case scenario, where the list is sorted in reverse order, bubble sort will have to make comparisons and swaps across the entire array multiple times. For a list of n elements, during the first pass, it will require (n-1) comparisons, as it checks each adjacent pair up to the last element. During the second pass, it will require (n-2) comparisons, and this continues until the final pass where only one comparison is made. The total number of comparisons can be summed up as: (n - 1) + (n - 2) + ... + 1 = n(n - 1)/2 This sums up to about n²/2, which represents the average and worst-case scenario. Thus, the rationale behind this is that the algorithm will require approximately n² comparisons, especially evident as n becomes larger. Although (n²/2) appears to be a constant factor lower than (n²), it is the same

Bubble sort is a simple sorting algorithm that repeatedly steps through the list, compares adjacent elements, and swaps them if they are in the wrong order. The process is repeated until the list is sorted. To understand how many comparisons bubble sort typically makes, consider the following:

In the worst-case scenario, where the list is sorted in reverse order, bubble sort will have to make comparisons and swaps across the entire array multiple times. For a list of n elements, during the first pass, it will require (n-1) comparisons, as it checks each adjacent pair up to the last element. During the second pass, it will require (n-2) comparisons, and this continues until the final pass where only one comparison is made.

The total number of comparisons can be summed up as:

(n - 1) + (n - 2) + ... + 1 = n(n - 1)/2

This sums up to about n²/2, which represents the average and worst-case scenario. Thus, the rationale behind this is that the algorithm will require approximately n² comparisons, especially evident as n becomes larger. Although (n²/2) appears to be a constant factor lower than (n²), it is the same