1. Array Mystery Consider the following method: CSE 142...

1. Array Mystery Consider the following method: CSE 142 Section Handout #9 Sample Final Exam public static void mystery(int[] arr) { for (int i = 1; i < arr.length; i++) { if (arr[i - 1] < arr[i]) { arr[i] = arr[i] / 2; } } } Indicate in the right-hand column what values would be stored in the array after the method mystery executes if the array in the left-hand column is passed as its parameter. Original Contents of Array int[] a1 = {4, 7}; mystery(a1); int[] a2 = {3, 16, 5, 10}; mystery(a2); int[] a3 = {5, 6, 5, 4}; mystery(a3); int[] a4 = {3, 8, 13, 6, 12}; mystery(a4); 2. Reference Semantics Mystery Final Contents of Array Write the output of the following program below, as it would appear on the console.

Transcript

00:01 Hello student here the array is given and the elements are given and array is unsorted.

00:07 So total these element we have to search the elements are 25 30 50 75 and 92 and we have to find out how many comparison linear search is doing and after that we have to apply any bubble sort selection sort type of algorithm and we have to arrange the array in the sorted array and after that we have to apply the binary search algorithm so here i have written all the steps which i have described here and now i will go for the linear search linear search as we know with the help of for loop and with the help of if statement this will be performed with the help of for loop the elements will be stored from 0 up to 22 so for finding the 25 element i have to do total 19 comparison it will be at the position 18th, but total 19 comparison as array is starting with 0 so here the element is printed found and their comparison after this i have to find the 30 which is not present in the list so total 23 comparison we are doing again for the 50 element is found so total 20 comparison we are doing as it is present on the 19th position and for the 75 element is at the second position so total 2 comparison we are doing and again for 92 no element is there so for the 92 we are doing 23 comparisons so in this way the linear search work and here i have written the algorithm for the bubble sort and the bubble sort we know very well it works on bringing the last element at the last position which is the largest one so second largest one at the second last position and it is on the third last position so in this way bubble sort work and now again i have written the element but in the sorted because i applied the bubble sort and after applying the bubble sort the element in the sorted array will be like this and after this i have to apply the binary search as we know very well the binary search here we have to first find the mid element then we have to break the list into left and right part the left part to the right to the left of mid and the right area will be the right of the mid and here mid element is calculated high plus low divided by 2 high means index low means also index in our case high is 22 and low is 0 so 22 plus 0 divided by 2 11 so we will find the 11th element in this given array.

02:58 So 11th element in this given array is a 11 which is equal to 44 so i write the mid here this is the mid and to the left of the mid to the right of the mid to the left of the mid this list is to the right of the mid this list is now we are doing the search for the 25 element so 25 element is present to the left of the mid list so i will go towards the left of the mid list...

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