Considering the following algorithm, analyze its best case, worst case and average case time complexity in terms of a polynomial of n and the asymptotic notation of ɵ. You need to show the steps of your analysis.
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- Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order. Example 1: Input: nums = [-4,-1,0,3,10] Output: [0,1,9,16,100] Explanation: After squaring, the array becomes [16,1,0,9,100]. After sorting, it becomes [0,1,9,16,100]. Example 2: Input: nums = [-7, -3,2,3,11] Output: [4.9,9,49,121]9.). integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], k = 1 Output: 1 Example 2: Input: nums = [1,2], k = 4 Output: -1 Example 3: Input: nums = [2,-1,2], k = 3 Output: 3.Consider the following algorithm, which may be called on an array that is not necessarily sorted. and returns the array in sorted form. GoofySort (A) : if (len (A) <= 3): SelectionSort(A) else: [len(A)/3] + GoofySort (A[1:2t]) GoofySort (A[t+1:n]) //sort the back 2/3rds + GoofySort (A[1:2t]) Let t + A[1:2t] //sort the front 2/3rds A[t+1:n] A[1:2t] //sort the front 2/3rds (a) Write a recurrence which captures the runtime of this algorithm. Your recurrence should be of the form T (n) =?. You can ignore any non-recursive operations (i.e. you only need to focus on operations in the last three lines). You do not need to handle base cases here. You can assumen is divisible by 3 for convenience. (b) Solve your recurrence using the substitution method. Show your work. You may assume a base case of T (1) = 1. State what the big-0 runtime is for this algorithm (hint: it should be a polynomial written in the form nº).
- 1. Consider the algorithm for the sorting problem that sorts an array by counting,for each of its elements, the number of smaller elements and then uses thisinformation to put the element in its appropriate position in the sorted array:ALGORITHMComparisonCountingSort(A[0..n − 1])//Sorts an array by comparison counting//Input: Array A[0..n//Output: Array S[0..n− 1] of orderable values− 1] of A’s elements sorted// in nondecreasing orderfor i ← 0 to nCount− 1 do[i]←0for i ← 0 to n − 2 dofor j ← i +1 to n − 1 doif A[i] < A[j ]Count[j ]← Count[j ] + 1else Count[i]← Count[i] + 1for i ←0 to n−1 doS[Count[i]]←A[i]return Sa. Apply this algorithm to sorting the list 60, 35, 81, 98, 14, 47.b. Is this algorithm stable?c. Is it in-place?Given a sorted array of positive integers. Your task is to rearrange the array elements alternatively i.e first element should be max value, second should be min value, third should be second max, fourth should be second min and so on.9).An array of integers nums sorted in ascending order, find the startingand ending position of a given target value. If the target is not found in thearray, return [-1, -1]. For example:Input: nums = [5,7,7,8,8,8,10], target = 8Output: [3,5]Input: nums = [5,7,7,8,8,8,10], target = 11Output: [-1,-1]. Please weite your Code.
- What is the time complexity for the following code/program? 1.5 A binary search works like this: in a sorted array, the search algorithm compares the target value to the middle element of the array. If they are not equal, the half in which the target cannot lie is eliminated and the search continues on the remaining half, again taking the middle element to compare to the target value and repeating this until the target value is found. If the search ends with the remaining half being empty, the target is not in the array. What is the complexity of binary search? Why?Write 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]) /2In programming language C Write a program to take input for two 1D array elements. Perform the sum of their elements and store the result in third array (1D). After this, arrange the elements of third array in ascending order and then implement binary search on the arranged array.
- 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++ languagefunction Sum(A,left,right) if left > right: return 0else if left = right: return A[left] mid = floor(N/2) lsum = Sum(A,left,mid) rsum = Sum(A,mid+1,right) return lsum + rsum function CreateB(A,N)B = new Array of length 1 B[0] = Sum(A,0,N-1) return B Building on the above, in a new scenario, given an array A of non-negative integers of length N, additionally a second array B is created; each element B[j] stores the value A[2*j]+A[2*j+1]. This works straightforwardly if N is even. If N is odd then the final element of B just stores A[N-1] as we can see in the figure below: (added in image) The second array B is now introducing redundancy, which allows us to detect if there has been a hardware failure: in our setup, such a failure will mean the values in the arrays are altered unintentionally. The hope is that if there is an error in A which changes the integer values then the sums in B are no longer correct and the algorithm says there has been an error; if there were an error in B…Given an array arr[] of N non-negative integers representing the height of blocks. If width of each block is 1, compute how much water can be trapped between the blocks during the rainy season. Example 1: Input: N = 6 arr[] = {3,0,0,2,0,4} Output: 10.