Question

The average number of comparisons performed by linear search to find an item in an array of \(\mathrm{N}\) elements is ___________.

Short answer

Answer: The average number of comparisons performed by the linear search algorithm to find an item in an array of N elements is \(\frac{(N + 1)}{2}\).

Step-by-step solution

Understand the linear search algorithm

Linear search is an algorithm that finds a target value in a given array by checking each element until the target is found or the end of the array is reached. It performs comparisons sequentially to identify if an item is present in the array.

Find the possible number of comparisons for each element

The first element will be found with 1 comparison, the second with 2 comparisons, the third with 3 comparisons, and so on. The Nth element will be found with N comparisons. Thus, for each element in the array, we have a different number of comparisons that need to be made.

Calculate the average number of comparisons for the entire array

To find the average number of comparisons, we can add up the number of comparisons for each individual element and then divide by the total number of elements, which is N. This can be represented mathematically as: \(\text{Average number of comparisons} = \frac{1 + 2 + 3 + ... + (N-1) + N}{N}\)

Apply the formula for sum of first N natural numbers

We can simplify the above expression by using the formula to find the sum of first N natural numbers: \(\sum_{i=1}^N i = \frac{N (N + 1)}{2}\) Substituting this formula into the expression for the average number of comparisons, we get: \(\text{Average number of comparisons} = \frac{\frac{N (N + 1)}{2}}{N}\)

Simplify the expression

Now we can simplify the expression by canceling out the N from the numerator and denominator: \(\text{Average number of comparisons} = \frac{(N + 1)}{2}\) The average number of comparisons performed by linear search to find an item in an array of N elements is \(\frac{(N + 1)}{2}\).