The pre-processing step in the below pseudocode is: def median(L): S = merge_sort(L) n = len(L) mid = n/2 if n is odd then return S[mid] else return (S[mid-1] + S[mid]) / 2 S = merge_sort(L) mid = n/2 None n = len(L)
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- Add a loop over to this function that tracks how many times the sum functions is called. Also explain how the loop is working. int64_t sum(int64_t N, uint64_t A[]){ int64_t sum = 0; uint64_t indx = A[0]; for (int64_t i = 0; i < N; i++){ sum += A[indx]; indx = A[indx]; } return sum;}T/F: Selection Sort has both a best case and worst case runtime of O(n^2).Python|you can use modules if necessary order is a list of pairs in the form of 'X-Y' order = ['A-C', 'A-D', 'A-H', 'A-S', '2-C', '2-D', '2-H', '2-S', '3-C', '3-D', '3-H', '3-S', '4-C', '4-D', '4-H', '4-S', '5-C', '5-D', '5-H', '5-S', '6-C', '6-D', '6-H', '6-S', '7-C', '7-D', '7-H', '7-S', '8-C', '8-D', '8-H', '8-S', '9-C', '9-D', '9-H', '9-S', '10-C', '10-D', '10-H', '10-S', 'J-C', 'J-D', 'J-H', 'J-S', 'Q-C', 'Q-D', 'Q-H', 'Q-S', 'K-C', 'K-D', 'K-H', 'K-S'] Notice that there are 13 X's and each of them is paired with 4 values of Y for better understanding please take reference here Y = ['C','D','H','S']X = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K'] 1.) I have a code that prints only the third and the first pair from the list above? Meaning it prints an X with the third and first Y in the list (order). However, It prints with a bracket[] and is separated with quotes ' '{see picture for my output} both needs to be removed. Also it only needs to print the first…
- The following code doesn't correctly return the index values of the minimum contiguous subsequence. Please modify it so that it does, with comments if possible. def minSumRec(a, left, right): if left == right: return [a[left], left, right] center = int((left + right) / 2) min_left_sum = minSumRec(a, left, center) min_right_sum = minSumRec(a, center + 1, right) min_left_border_sum, left_border_sum = 1e9, 0 for i in range(center, left - 1, -1): left_border_sum += a[i] if left_border_sum < min_left_border_sum: min_left_border_sum = left_border_sum min_right_border_sum, right_border_sum = 1e9, 0 for i in range(center + 1, right + 1): right_border_sum += a[i] if right_border_sum < min_right_border_sum: min_right_border_sum = right_border_sum min_border_sum = min_right_border_sum + min_left_border_sum minSum = min(min_left_sum[0], min_right_sum[0], min_border_sum) if minSum == min_left_sum: return min_left_sum if minSum == min_right_sum: return min_right_sum return…nums = [9,-8,7,-6,3,2,-1,3,10,9] Use enumerate to iterate through the index and values of the nums list, and output the index where userNum is found. For example, if userNum is 3, your output should be: Number 3 found at index 4 Number 3 found at index 7The program written for binary search, calculates the midpoint of the span as mid: (Low + High)/2. The program works well if the number of elements in the list is small (about 32,000) but it behaves abnormally when the number of elements is large. This can be avoided by performing the calculation as: =
- Array implementation of List ADT Display the array elements from Left to Right and Right to left. Input: Enter the size of array MAX Enter number of elements N Enter the element to be inserted. Output: Display the Array For example: Test Input 1 2 10 W3 123 10 5 11 12 13 14 15 Result Forward: 1 2 3 Backward:3 2 1 Forward: 11 12 13 14 15 Backward:15 14 13 12 11A factor chainis a list where each previous element is a factor of the next consecutive element. The following is a factor chain: [3, 6, 18, 72] Example: is_factor_chain ([2, 4, 8, 9]) → False is_factor_chain ([2, 22, 66, 110]) → True WRITE IN PYTHON PLEASEl JO 100 4G 2:02 AM @ 52% assignment_3 v2 AI and ML Assignment 3 Write a python program to cluster candidates into 2/3 groups. The program performs dimensionality reduction, then clustering of candidates based on top 2 PCs. The results are plotted to visually show the performance of candidates. The program performs the following tasks: Read input "pca cluster.csv" file. Use Panda library. Example of the file: candi cand2 cand3 cand4 cand5 cad6 experience 10 9 7 3 education 10 10 8 10 health 4 6 8 9 9. interview 9 6 6 8 7 test 4 9. 10 communication skills 7. 6 9 10 10 writing skills team player leadership 10 9 6 6. 8 7 4 10 9 4 10 4 9 6 8 5 9 7 enthusiasm 7. 7 7 9 ready to start 5 6 6. 9 1 flexible in hours 10 1 9. 5 6. able to relocate 6 6 6. 10 4 references 9 9 7 9 4 10 • Apply PCA analysis to identify top 2 PCs. Eigen No. 0 Eigen value (33.50+0j) Contribution % (40.87+0j) Points on new scale [5 5-8 3 0-5] Eigen No. I Eigen value- (22.65+Oj) Contribution % (27.63+0j) Points on new…
- Write a java class named First_Last_Recursive_Merge_Sort that implements the recursive algorithm for Merge Sort. You can use the structure below for your implementation.The function sum_evens in python takes a list of integers and returns the sum of all the even integers in the list. For example: Test Result print(sum_evens([1, 5, 2, 5, 3, 5, 4])) 6 print(sum_evens([5, 5, -5, -5])) 0 print(sum_evens([16, 24, 30])) 70Language: Python 3 Autocomplete Ready O 1 v import ast 3. Hybrid Sort input() lst %3D 3 lst = ast.literal_eval(lst) 4 Insertion sort is a simple sorting algorithm that builds the final sorted array one item at a time. In each iteration, insertion sort inserts an element into an already sorted list (on left). The position where the item will be inserted is found through linear search. You decided to improve insertion sort by using binary search to find the position p where the new insertion should take place. 6 print(BinaryInsertionSort(lst)) Algorithm BinarylnsertionSort Input/output: takes an integer array a = {a[0], ..., a[n – 1]} of size n begin BinarylnsertionSort for i =1 to n val = a[i] p = BinarySearch(a, val, 0, i – 1) for j = i-1 to p a[j + 1]= a[i] j= j-1 end for a[p] = val i i+1 end for end BinarylnsertionSort Here, val = a[i] is the current value to be inserted at each step i into the already sorted part a[0], ..., ați – 1] of the array a. The binary search along that part…