def tri(n): if n == 1: return [1] prev= [0] + tri(n-1) for i in range(len(prev)-1): prev[i] return prev = prev[i] + prev[i+1] def quad (arr): n = len (arr) if n == 1: return arr[0] if n < 1: return -1 beg, mid, end = 0, n//2, n quad (arr[beg:mid]) qu = ad = quad(arr[mid:end]) return qu if qu › ad else a
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find the BIG O complexity of the two function do this by writing the big O complexity of each line in the snippet
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- in Python Minima in permutations. Write a program that takes an integer n from the command line, generates a random permutation, prints the permutation, and prints the number of left-to-right minima in the permutation (the number of times an element is the smallest seen so far). Then write a program that takes integers m and n from the command line, generates m random permutations of length n, and prints the average number of left-to-right minima in the permutations generated7) Implement counting sort over a list of keys that lie within the range (a, b),where a and b are small positive integers input by the user.8) Implement bucket sort with an array of lists data structure to sort i) a list ofstrings, ii) a list of floating point numbers and iii) a list of integers, which areuniformly distributed over a range, for an appropriate choice of the number ofbuckets.def find_root4(x, epsilon): ''' IN PYTHON Assume: x, epsilon are floating point numbers and epsilon > 0 Use bisection search to find the following root of x such that If x >=0, return y such that x - epsilon <= y ** 2 <= x + epsilon Else, return y such that x - epsilon <= y ** 7 <= x + epsilon Note: You must use bisection search to implement the function. ''' pass
- solve in java Integer arrays originalArr and filterArr are read from input, each containing two elements. If an element in originalArr is not a multiple of the corresponding element in filterArr, replace the element in originalArr with the corresponding element in filterArr times 2. Ex: If the input is: 16 9 8 4 then the output is: 16 8 Note: Given integers A and B, A is a multiple of B only if A % B == 0. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 import java.util.Scanner; public class ReplaceNumbers { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); finalintNUM_VALS=2; int[] originalArr=newint[NUM_VALS]; int[] filterArr=newint[NUM_VALS]; inti; for (i=0; i<NUM_VALS; ++i) { originalArr[i] =scnr.nextInt(); } for (i=0; i<NUM_VALS; ++i) { filterArr[i] =scnr.nextInt(); } /* Your code goes here */ for (i = 0; i < originalArr.length; ++i) { System.out.print(originalArr[i] + " "); } System.out.println();…Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Ridbit begins with an integer n. In one action, he can perform one of the accompanying tasks: partition n by one of its appropriate divisors, or take away 1 from n in case n is more prominent than 1. An appropriate divisor is a divisor of a number, barring itself. For instance, 1, 2, 4, 5, and 10 are appropriate divisors of 20, however 20 itself isn't. What is the base number of moves Ridbit is needed to make to decrease n to 1? Input The principal line contains a solitary integer t (1≤t≤1000) — the number of experiments. The main line of each experiment contains a solitary integer n (1≤n≤109). Output For each experiment, output the base number of moves needed to lessen n to 1.Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…Collapse positive integer intervals def collapse_intervals(items): This function is the inverse of the previous problem of expanding positive integer intervals. Given a nonempty list of positive integer items guaranteed to be in sorted ascending order, create and return the unique description string where every maximal sublist of consecutive integers has been condensed to the notation first-last. Such encoding doesn’t actually save any characters when first and last differ by only one. However, it is usually more important for the encoding to be uniform than to be pretty. As a general principle, uniform and consistent encoding of data allows the processing of that data to also be uniform in the tools down the line. If some maximal sublist consists of a single integer, it must be included in the result string all by itself without the minus sign separating it from the now redundant last number. Make sure that the string returned by your function does not contain any whitespace…In Java languge (Duplicate Elimination) Write a program that reads in a series of first names and eliminates duplicates by storing them in a Set. Allow the user to search for a first name. Add a name to set, use end to terminate input: Search a name, use end to terminate searching: Sample output as follows:
- package sorting; import java.util.Arrays; import java.util.Random; import sorting.Heap; public class ComparisonSorter { public static void insertionSort(int[] arr) { for(int i=1;i<arr.length;i++) { int j = i-1; int toPlace = arr[i]; while(j>=0&&arr[j]>toPlace) { arr[j+1] = arr[j]; j--; } arr[j+1] = toPlace; } } public static void mergeSort(int [] arr) { mergeSortRecurse(arr,0,arr.length-1); } private static void mergeSortRecurse(int[] arr, int start, int end) { if(start>=end) return; int mid = start + ((end-start)/2); mergeSortRecurse(arr,start,mid); mergeSortRecurse(arr,mid+1,end); merge(arr,start,mid,end); } private static void merge(int[] arr, int start, int mid, int end) { int leftSize = mid - start +1; int rightSize = end - mid; int[] left = new int[leftSize+1]; int[] right = new int[rightSize+1]; int leftIndex; int rightIndex; for(leftIndex = 0; leftIndex<leftSize;leftIndex++) left[leftIndex] = arr[start+leftIndex]; for(rightIndex = 0;…package sorting; import java.util.Arrays; import java.util.Random; import sorting.Heap; public class ComparisonSorter { public static void insertionSort(int[] arr) { for(int i=1;i<arr.length;i++) { int j = i-1; int toPlace = arr[i]; while(j>=0&&arr[j]>toPlace) { arr[j+1] = arr[j]; j--; } arr[j+1] = toPlace; } } public static void mergeSort(int [] arr) { mergeSortRecurse(arr,0,arr.length-1); } private static void mergeSortRecurse(int[] arr, int start, int end) { if(start>=end) return; int mid = start + ((end-start)/2); mergeSortRecurse(arr,start,mid); mergeSortRecurse(arr,mid+1,end); merge(arr,start,mid,end); } private static void merge(int[] arr, int start, int mid, int end) { int leftSize = mid - start +1; int rightSize = end - mid; int[] left = new int[leftSize+1]; int[] right = new int[rightSize+1]; int leftIndex; int rightIndex; for(leftIndex = 0; leftIndex<leftSize;leftIndex++) left[leftIndex] = arr[start+leftIndex]; for(rightIndex = 0;…Write a C++ or Java program that uses bit strings to find A ∪ B, A ∩ B, and A – B given subsets A and B of a universal set with 10 elements {0, 1, 2, 3, 4, 5, 6, 7, 8,9}. You have to use bit string Print to the screen set A, set B ( be sure to print out the name of the sets), as well as the set operation results(be sure to print out the name of the operations). The program requires that elements of subsets A and B are from user input. You can make the assumption that user input numbers are within the domain of the set {0, 1, 2, ..., 9} and there is no improper input.Additional requirements and reminders:• The use of STL, templates, and operator overloading is not permitted in any form.• Remember: You must use a bit string when representing a set. You may not use an array ofBoolean variables.• The elements of subset A and subset B must from user input. You may not hardcode the two subsets in your program.Hint:(1) It is a bit easier to program if we construct the bit string of a set…