Reference the algorithm in the attached image. (a) Write code to implement the algorithm in python. (b) Using the list L = 60, 35, 82, 112, 9, 50 as your input, print the output of the algorithm. (c) Modify the algorithm to a. print the array Count after each iteration of i in the nested for loop (second for loop). b. include code to count the number of times the if statement is executed. c. print the S array.
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Reference the
(a) Write code to implement the algorithm in python.
(b) Using the list L = 60, 35, 82, 112, 9, 50 as your input, print the output of the
algorithm.
(c) Modify the algorithm to
a. print the array Count after each iteration of i in the nested for loop
(second for loop).
b. include code to count the number of times the if statement is executed.
c. print the S array.
Step by step
Solved in 4 steps with 4 images
- ALGO1(A)// A is an integer array, an index that starts at 11): for i=1 to n-1 do2): minIndex = findSmallest(A,i)3): exchange A[i] with A[minIndex] The sub-routine find the smallest(A, i) in Line 2, and returns the index of the smallest element in thesub-array A[i:n]Suppose the array below is provided as input to ALGO1 2 5 6 7 3 8 1 4 Fill in the Blanks At the end of the first iteration of the for loop (i.e. with i=1)1) a) the element at index 1 is : b) the element at index 4 is : c) the element at index 7 is :Java Selection Sort but make it read the data 64, 25, 12, 22, 11 from a file not an array // Java program for implementation of Selection Sort class SelectionSort { void sort(int arr[]) { int n = arr.length; // One by one move boundary of unsorted subarray for (int i = 0; i < n-1; i++) { // Find the minimum element in unsorted array int min_idx = i; for (int j = i+1; j < n; j++) if (arr[j] < arr[min_idx]) min_idx = j; // Swap the found minimum element with the first // element int temp = arr[min_idx]; arr[min_idx] = arr[i]; arr[i] = temp; } } // Prints the array void printArray(int arr[]) { int n = arr.length; for (int i=0; i<n; ++i) System.out.print(arr[i]+" "); System.out.println(); } // Driver code to test above public…function 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…
- OGiven 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.int g(int nums[], int n) { if (n == 1) return 0; int val = g(nums, n-1); if (nums[n-1] > nums[val]) return val; return n-1; } Give a recurrence T(n) for the number of times the code nums[n-1] > nums[val] is executed when the array values are in ascending order. Then solve the recurrence231For the given array, find the kth largest element.** Note: Sorting is not allowed5Input format:Input:321564line 1 consist of array elements, separated by a spaceline 2 consist of k valueEg:output:Input:111output:
- Write the state/order of the elements of the following array after each pass of the outermost loop of the Selection sort algorithm [ 8, 5, -9, 14, 0, -1, -7, 3 ] for example, 1st pass: -9, 5, 8, 14, 0, -1, -7, 3 2nd pass: 3rd pass: 4th pass:Multiply left and right array sum. Basic Accuracy: 54.29% Submissions: 12630 Points: 1 Pitsy needs help in the given task by her teacher. The task is to divide a array into two sub array (left and right) containing n/2 elements each and do the sum of the subarrays and then multiply both the subarrays. Example 1: â€Bubble sort pseudocode: Bubblesort (A, n) // A is array, n items to sort For i = n-1 to 1 For j = 1 to i If A[j] < A[j+1] Swap (A[j], A[j+1]) (A) Determine the big-O running time (tight bound) of bubble sort in the left column. Show your derivation. This pseudocode assumes the array starts at index 1. Count comparisons as the critical operation. Your solution should count the exact number of comparisons as a function of n (number of items being sorted). In converting your formula for number of comparisons to a closed-form formula (as opposed to a summation formula), you do not need to show the derivation (such as writing the summation twice and dividing by two), though it is okay to do that. You do need to show the closed-form formula for exact number of comparisons, before simplifying it to a big-O representation. (B) Prove that f(n) = 1000n5 + 20000n2 + 32 is O(n6 ); Is this a tight upper bound? Why or why not?Bubble sort pseudocode: Bubblesort (A, n) // A is array, n items to sort For i = n-1 to 1 For j = 1 to i If A[j] < A[j+1] Swap (A[j], A[j+1]) (A) Determine the big-O running time (tight bound) of bubble sort in the left column. Show your derivation. This pseudocode assumes the array starts at index 1. Count comparisons as the critical operation. Your solution should count the exact number of comparisons as a function of n (number of items being sorted). In converting your formula for number of comparisons to a closed-form formula (as opposed to a summation formula), you do not need to show the derivation (such as writing the summation twice and dividing by two), though it is okay to do that. You do need to show the closed-form formula for exact number of comparisons, before simplifying it to a big-O representation. As a reference, study the stairs examples in the book. See figure 1.2, pg. 20. (B) Prove that f(n) = 1000n5 + 20000n2 + 32 is O(n6 ); Is this a tight upper bound? Why…An inversion in an array is a pair of elements that are "out of order," meaning that the element that occurs earlier in the array is bigger than the one that occurs later. What is the largest-possible total number of inversions an 8-element array can have? O 64 O 15 O None of the choices O 7 O 28 O 32sum++; 3. Consider the following algorithm for finding the distance between the two closest elements in an array of numbers. ALGORITHM MinDistance(A[0..n – 1]) //Input: Array A[0..n – 1] of numbers //Output: Minimum distance between two of its elements dminSEE MORE QUESTIONSRecommended textbooks for youEBK JAVA PROGRAMMINGComputer ScienceISBN:9781337671385Author:FARRELLPublisher:CENGAGE LEARNING - CONSIGNMENTProgramming Logic & Design ComprehensiveComputer ScienceISBN:9781337669405Author:FARRELLPublisher:CengageEBK JAVA PROGRAMMINGComputer ScienceISBN:9781337671385Author:FARRELLPublisher:CENGAGE LEARNING - CONSIGNMENTProgramming Logic & Design ComprehensiveComputer ScienceISBN:9781337669405Author:FARRELLPublisher:Cengage