In an unsorted array with N items, how many items must be examined to find a particular item on average case?Required to answer. Single line text.
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In an unsorted array with N items, how many items must be examined to find a particular item on average case?Required to answer. Single line text.
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- What will be the problem with the following algorithm for finding an item in an unsorted array? How can you solve it? int location = 0; bool moreToSearch = (location > length); found = false; while (moreToSearch && !found) { if(item == info[location]) { item = info[location]; } else { location++; moreToSearch = (location > length); } }Please keep the simulation neat and well organized. It is a request. Perform MergeSort on the given array. You need to show the whole simulation. A = [10,3,9,18,21,26,30,8,15,6,17].given an array of integer values , return true if 6 appears as either the first or last element in the array. firstLast6([1,2,36]) true firstLast([6,1,2,3]) true firstLast([13,6,1,2,3]) false
- A majority element is an element that makes up more than half of the items inan array. Given a positive integers array, find the majority element. If there is no majority element,return -1. Do this in O(N) time and 0(1) space.Input: 1 2 5 9 5 9 5 5 5Output: 5Hi, could you sort this array using quicksort with the first element being the pivot? Please show all steps. Thank you [6, 3, 9, 1, 8, 5, 7, 4, 2, 0]Let B be an array of size n ≥ 6 containing integers from 1 to n−5 inclusive, five of which are repeated. Write and test a method for finding the five integers in B that are repeated. What is the running time of this algorithm? Hint: Sort the array B then scan it from front to end looking for the repeated entries
- By using JavaScript find the sum of all the elements of a given array (before removing and after removing) and removes the last element from an array? [2, 3, 4, 6, 30, 23] Answer it as soon as possible to get the upvot.You are given two arrays, one shorter (with all separate items) and one longer (with no unique elements). Find the shortest subarray in the larger array that contains all of the shorter array's items. The things can be shown in any order.EXAMPLEInput: {1, 5, 9}{7, 5, 9, 0, 2, 1, 3, 5. 7, 9. 1, 1, 5, 8, 8, 9, 7}Results: [7, 10] (the underlined portion above)In the "Common Array Algorithms" section, you learned how to find the position of an element in an array. Suppose you want to find the position of the second match. You could remove the first match and start over, but that would be wasteful. Instead, adapt the algorithm from that section so that it starts the search at a given position, not zero. Complete the following code.
- # the method take another circular array and returns a linear array containing the common elements between the two circular arrays. def intersection(self, c2): # To Do print("==========Test 10==========") lin_arr9 = [10, 20, 30, 40, 50, None, None, None] c10 = CircularArray(lin_arr9, 5, 5) c10.printFullLinear() # This should print: 40, 50, None, None, None, 10, 20, 30 lin_arr10 = [5, 40, 15, 25, 10, 20, 5, None, None, None, None, None] c11 = CircularArray(lin_arr10, 8, 7) c11.printFullLinear() # This should print: 10, 20, 5, None, None, None, None, None, 5, 40, 15, 25 output = c10.intersection(c11) print(output) # This should print: [10, 20, 40]Let's say that B is an array with a size of n > 6 that has integers from 1 to n-5, inclusive, in it. There are exactly five repeats in B. Explain how to find the five numbers in B that are the same.Let A be an array with n = 2k − 1 elements, where k is some positive integer. Determine a formula (in terms of n) for the minimum possible number of total comparisons required by Quicksort, as well as a formula for the maximum possible number of total comparisons required by Quicksort.