The algorithm S(A, n, i) selects all the j-th smallest elements (with j <= i) fr

Question

The algorithm S(A, n, i) selects all the j-th smallest elements (with j <= i) from an array A of n elements, by using linearselect to select each of the j-th smallest elements (with j <= i). One could also implement S alternatively as T (A, n, i), which first sort A (on average-case and on worst- case, the sorting takes time O(n log n) using mergesort) and then select the first i elements. Compare the average-case time complexities of the two algorithms; i.e., For the average-case complexities, under what conditions (on the choices for i), S is better than T or vice versa.

Explanation / Answer

Design an algorithm to search a given value in an array.

Solution:

Assumptions:

Let the array A is unordered.

Algorithm Name: LinearSearchArray(A, N, LB, UB, X, R)

Input : A is any numeric array with N number of elements, we have to search X in the given array A. LB & UB are the Lower Bound & Upper Bound of the array segment where we are going to search for the element X.

Output: R will hold -1 if X is not found in the array A, otherwise R will hold the index of X in the given array A.

Step 1: R = -1

Step 2: i =LB

Step 3: WHILE ( i < UB )

          IF ( A[i] = X ) Then

             R = i

             Return

          EndIf

          i = i + 1

        EndWhile

Step 4: Return

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