What is the worst-case scenario for the number of comparisons needed in an insertion sort?

What is the worst-case scenario for the number of comparisons needed in an insertion sort?

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

Answer: (C)

In insertion sort, the algorithm processes each element of the array one by one, inserting the current element into its correct position among the already sorted elements. The worst-case scenario occurs when the array is sorted in reverse order, meaning every new element must be compared to all previously sorted elements before it can be correctly positioned. In this case, for the first element, there are no comparisons needed. For the second element, one comparison is needed. For the third element, two comparisons will be required, and so on. This results in a series of comparisons that can be expressed as: 1 + 2 + 3 + ... + (n - 1) The sum of the first (n - 1) natural numbers can be calculated using the formula \( \frac{n(n - 1)}{2} \). This simplifies to approximately \( \frac{n^2}{2} \) as n grows larger. Thus, in the worst case, the insertion sort requires on the order of \( n^2/2 \) comparisons, making that the correct answer. The other options do not reflect the nature of insertion sort's performance accurately; \( n \) would imply linear time, and \( n \log n \

In insertion sort, the algorithm processes each element of the array one by one, inserting the current element into its correct position among the already sorted elements. The worst-case scenario occurs when the array is sorted in reverse order, meaning every new element must be compared to all previously sorted elements before it can be correctly positioned.

In this case, for the first element, there are no comparisons needed. For the second element, one comparison is needed. For the third element, two comparisons will be required, and so on. This results in a series of comparisons that can be expressed as:

1 + 2 + 3 + ... + (n - 1)

The sum of the first (n - 1) natural numbers can be calculated using the formula ( \frac{n(n - 1)}{2} ). This simplifies to approximately ( \frac{n^2}{2} ) as n grows larger. Thus, in the worst case, the insertion sort requires on the order of ( n^2/2 ) comparisons, making that the correct answer.

The other options do not reflect the nature of insertion sort's performance accurately; ( n ) would imply linear time, and ( n \log n \