Explain step-by-step, showing all possible detail, how a binary search algorithm works on the array {1, 2, 3, 4, 5, 6, 7, 8} searching for the values of 5 and 12.
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Explain step-by-step, showing all possible detail, how a binary search
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- Given an unsorted array write the pseudo code to sort it and find a specificelement xa. Find (array, x)// first sort the array// search for element x using binary searchBinary Search algorithm uses the divide-and-conquer technique to find elements in O(logn) steps. Given the following sorted array that has 9 elements, what is the scope of the search (First index and last index of the array segment) in each iteration of the binary search algorithm while looking for the element "51"? Note: Numbers that show up in the first row are the array indexes 0. 1 3 4 6. 7 8. 9 12 19 21 30 32 37 44 51 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 5, Last = 8 3rd Iteration -> First = 5, Last = 6 %3D O 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 5, Last = 8 Brd Iteration -> First = 7, Last = 8 4th Iteration -> First = 8, Last = 8 O 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 0, Last = 3 3rd Iteration -> First = 2, Last = 3 4th Iteration -> First = 3, Last = 3 O1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 0, Last 3 Brd Iteration -> First = 2, Last = 3Pseudo Code shown in Figure Q1(a) is an algorithm for binary searching for an array with n number of elements. By applying this algorithm, show step by step approach on how to find number 11 in an array as depicted in Figure Q1(b). low + 0 high e n-1 while (low s high) do ix + (low + high) /2 if (t = Alix)) then return ix else if (t < A[ix]) then high e ix - 1 else low + ix + 1 return -1 Figure Ql(a) 2 4 6 9 11 12 | 25 [0] [1) [2] [3] (4) [5) [6] Figure Q1(b)
- Given an array A= {12, 11, 13, 5, 6}, sort it out using a technique illustrated in insertion sortWrite a program / Pseudo code / Algorithm that inserts the following numbers into two Separate arrays. L1 50 30 25 75 82 28 77 L2 50 40 25 75 80 21 37 30 After Insertion median should be calculated for each adjacent element from both array structures, after computing median print / Output that array M. if M is the median them M0 = (L1[0]+L2[0]) /2M1 = (L1[1]+L2[1]) /2Course: Data structure and algorithms: Topic: Algorithm Complexity: Please solve it o emergency basis: Question 2: Imagine that we want to keep track of friendships between n people. We can do this with an array of size nXn. Each row of the array represents the friends of an individual, with the columns indicating who has that individual as a friend. For example, if person j is a friend of person i, then we place a mark in column j of row i in the array. Likewise, we should also place a mark in column i of row j if we assume that friendship works both ways. What will be the space complexity of this problem.
- Write a program / Pseudo code / Algorithm that inserts the following numbers into two Separate arrays. L1 à 50 30 25 75 82 28 77 L2 à 50 40 25 75 80 21 37 30 After Insertion median should be calculated for each adjacent element from both array structures, after computing median print / Output that array M. if M is the median them M0 = (L1[0]+L2[0]) /2 M1 = (L1[1]+L2[1]) /2 ………….. please use c/c++ languageFollowing is the function for interpolation search. This searching algorithm estimates the position (index) of a key in array based on the elements in the first position and last position in the array, and the length of array. The array must be sorted in ascending order. Suppose array A contains the following 15 elements: A = [1, 3, 3, 10, 17, 22, 22, 22, 24, 25, 26, 27, 27, 28, 28] At first iteration, at which position (index) the element of 24 is estimated in array A? In which part of array (starting index and ending index) the searching should continue? How many iterations the searching are performed until the element of 24 is found? int InterpolationSearch(int x[], int key, int n) { int mid, min = 0, max = n-1; while(x[min] < key && x[max] > key) { mid = min + ((key-x[min])*(max-min)) / (x[max]-x[min]); if(x[mid] < key) min = mid + 1; else if(x[mid] > key) max = mid - 1; else return mid; } if…Consider the following array: A = {2, 5, 8, 17, 19, 34, 45, 67, 78, 87} What is the time complexity of the binary search algorithm when 98 is searched?
- : In searching an element in an array, linear search can be used, even though simple to implement, but not efficient, with only O(n) time complexity. Assuming the array is already in sorted order, modify the search function below, using a better algorithm, so the average time complexity for the search function is O(log n). include <iostream> using namespace std; int search(int al), int s, int v) { 1/ Modify below codes. for (int i = 0; i <s; i++) { if (a[i] = v) return i; return -1; int main() { int intArray:10] = { 5, 7, 8, 9, 10, 12, 13, 15, 20, 34); // Search for element '12' in 10-elements integer array. cout << search(intArray, 10, 12); // '5' will be printed out. // Search for element '35' in 10-elements integer array. cout << search(intArray, 10, 35); // '-1' will be printed out. // Index '-l' means that the element is not found. return 0;MATLAB PLEASE Write a general binary search code that can find a target with multiple appear- ances in an array. You can only use binary search algorithm. Use the following array to test your code to see if your code works: A=[2,4,0,1,1,2, 3, −3,4, −4,4,5,2,8,1,10,10,1]; and choose the test targets as 1. Your code must return the indices of the target number in the original un-sorted array. In your MATLAB code, no MATLAB built-in search or sort functions are allowed to be usedBinary Search algorithm uses the divide-and-conquer technique to find elements in O(logn) steps. Given the following sorted array that has 9 elements, what is the scope of the search (First index and last index of the array segment) in each iteration of the binary search algorithm while looking for the element "51"? Note: Numbers that show up in the first row are the array indexes 2 3 4 6 7 12 19 21 30 32 37 44 51 • 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 5, Last = 8 3rd Iteration -> First = 5, Last = 6 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 0, Last = 3 3rd Iteration -> First = 2, Last = 3 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 5, Last = 8 3rd Iteration -> First = 7, Last = 8 4th Iteration -> First = 8, Last = 8 1st Iteration -> First = 0, Last = 8 2nd Iteration -> First = 0, Last = 3 3rd Iteration -> First = 2, Last = 3 4th Iteration -> First = 3, Last = 3