The algorithm below searches for the maximum in an input array A. Assume A is a random sequence containing n distinguishable real numbers, what is the probability for line #4 to be executed exactly once? (The phrase "random sequence" here means the n numbers can appear in any order with equal chance.) mymax(A, n) { 1: max = A[1]; 2: for i = 2:n 3: if A[i] > max 4: max = A[i); 5: end; 6: end; 7: return max; }

JAVA PROGRAM ASAP *********** THIS PROGRAM MUST WORK IN HYPERGRADE AND PASS ALL THE TEST CASES. Perform algorithm time measurement for all three quadratic sorting algorithms for Best Case (already sorted), Average Case (randomly generated), and Worst Case (inverse sorted) inputs and compare them with O(N*logN) sorting algorithm using Arrays.sort() method. Use arrays sizes of 50K, 100K, and 200K to produce timing results. Ignore the message "There is an infinite loop!" that appears on HyperGrade after submitting your program. Unfortunately, HyperGrade is designed to limit the amount of time that your program is allowed to execute.

The PARTITION(A) procedure inputs an array A, and takes the fınal element x = A[r] as the pivot element. The output is an array where all elements to the left of x are less than x, and all elements to the right of x are greater than x. An example is provided on pg. 172 of the textbook. In this example, A = [2,8,7,1,3,5,6,4]. The pivot element is x=4 (highlighted in red). The end result of PARTITION is [2,1,3,4,7,5,6,8]. Notice that all elements to the left of 4 are less than 4, and all elements to the right of 4 are greater than 4. I strongly encourage you to go through this example, step by step. Let A = [1,2,7,8,9,6,3,4,5]. Determine PARTITION(A).

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: 5

15. Consider the problem of finding the first position in which an array b occurs as a subsequence of an array a. Write two nested loops: let result undefined for (let i = 0; i < a.length - b.length; i++) { for (let j = 0; j < b.length; j++) { if (a[i+j] = b[j]) . . . } } Complete with labeled break and continue statements.