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Q: a. Give another implementation where update takes O(n) time while isSorted takes 0(1) time.
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- II. Read each problem carefully and present an algorithm with the required running-time to solve each problem. 2. Let A be a sorted array of integers. You want to implement two functions: update (i, x) - which takes as input an index i in A and sets A[i] = x. For example, calling update (3,10) will set A[3] = 10. Note that array A might not be completely sorted after an update. is Sorted (i) - returns true if the subarray A[0... i] is sorted, otherwise it returns false. A straightforward implementation of the two operations using only the given array can be done as follows: update operation can directly change the value of A[i] to x. is Sorted will scan A[0] up to A[i] while checking if elements are non-decreasing. With the above implementations, update takes 0(1) time but is Sorted takes O(n) time. a. Give another implementation where update takes O(n) time while isSorted takes O(1) time. b. Describe how to use a balanced BST to implement both operations in O(log n) time. Discuss why…Please help me with this: using js create an array of 30 random numbers that range between 1and 100. And yet again, write a function that will receive a number from the userand determine if that number exists in the array or not. But this time, start bySORTING your input list. After a sort, the list in problem 1 is as follows:[2, 2, 3, 5, 12, 14, 14, 15, 23, 36, 39, 41, 44, 44, 45, 48,49, 50, 52, 52, 59, 71, 81, 82, 88, 89, 89, 93, 96, 97] Approach: Implement a method called findC(x, A, i, j), where x is the number we arelooking for in array A, the first index of the array is i and the last index is j. We wantto determine whether x exists in A anywhere between index i and index j. Your firstcall to this method will therefore look like this: findC(x, A, 0, A.length-1). In the body of your function, compare x with the item that is in the middle of thearray, as you did before. As before, call the middle of index of the array mid. But thistime, if x<=a[mid], recursively call your…create an array of 30 random numbers that range between 1and 100. And yet again, write a function that will receive a number from the userand determine if that number exists in the array or not. But this time, start bySORTING your input list. After a sort, the list in problem 1 is as follows:[2, 2, 3, 5, 12, 14, 14, 15, 23, 36, 39, 41, 44, 44, 45, 48,49, 50, 52, 52, 59, 71, 81, 82, 88, 89, 89, 93, 96, 97] Approach: Implement a method called findC(x, A, i, j), where x is the number we arelooking for in array A, the first index of the array is i and the last index is j. We wantto determine whether x exists in A anywhere between index i and index j. Your firstcall to this method will therefore look like this: findC(x, A, 0, A.length-1). In the body of your function, compare x with the item that is in the middle of thearray, as you did before. As before, call the middle of index of the array mid. But thistime, if x<=a[mid], recursively call your function to search ONLY the first half of…
- Write a Python function def isSubArray(A,B) which takes two arrays and returns True if the first array is a (contiguous) subarray of the second array, otherwise it returns False. You may solve this problem using recursion or iteration or a mixture of recursion and iteration. For an array to be a subarray of another, it must occur entirely within the other one without other elements in between. For example: [31,7,25] is a subarray of [10,20,26,31,7,25,40,9] [26,31,25,40] is not a subarray of [10,20,26,31,7,25,40,9] A good way of solving this problem is to make use of an auxiliary function that takes two arrays and returns True if the contents of the first array occur at the front of the second array, otherwise it returns False. Then, A is a subarray of B if it occurs at the front of B, or at the front of B[1:], or at the front of B[2:], etc. Note you should not use A == B for arrays.Implement a function GetMinimumCutSegments(int[] arr, int k) that takes in an array arr of positive integers and an integer k, and returns the minimum number of contiguous segments of the array that must be concatenated to form an array of size k. The function should return -1 if it is not possible to create an array of size k by concatenating contiguous segments of the input array. Example: Input: [1, 2, 3, 4], 6 Output: 2 Explanation: The minimum number of contiguous segments that must be concatenated to form an array of size 6 is 2. The two segments that can be concatenated are [1, 2] and [3, 4]. Input: [1, 2, 3, 4], 7 Output: -1 Explanation: It is not possible to create an array of size 7 by concatenating contiguous segments of the input array. Constraints: The input array arr will have at most length 100. The integer k will be at least 1 and at most 10^6. Can you write a C# function that solves this problem? public int GetMinimumCutSegments(int[] arr, int k) { // Your code…Consider an array of size n. Write a function that takes n values as input for the array, swaps every 2 adjacent elements in the array and prints the updated array. For example, if the initial array is [1,2,4,8,9,10] then the final output shall be [2,1,8,4,10,9].
- Write an algorithm to calculate the average of a list of numbers. The algorithm should be understandable by a person who does not know a computer programming language (not written in code). Do not use pre-defined functions. The algorithm should be able to be converted into a computer program on a line by line basis (or at least close to it). Example of an algorithm for a binary search: Find the midpoint of the sorted array / list by dividing its length by 2. Compare the midpoint to the value of interest. If the midpoint is larger than the value, perform binary search on right half of the array. If the midpoint is smaller than the value, perform binary search on left half of the array. Repeat these steps until the midpoint value is equal to the value of interest or we know the value is not in the array. PythonImplement the following function without using any additional data structure, and without sorting the input vectors. /* Given two vectors of integers, check if the two vectors contain same set of values: e.g., V1=[3,4,10,4,10,11] and V2=[3,3,4, 11, 10] stores same set int: {3, 4, 10, 11}. Note that duplicates are removed when considering set */ @param list1, list2: two vectors of integers@pre: list1, list2 have been initialized@post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */ bool SameSet (const vector<int> & list1, const vector<int> & list2) {//for each value in list1, check if it appears in list2, if not, return false //for each value in list2, check if the value appears in list1, if not return false //return true}implement QuickSort of ints that sorts the numbers in the non-decreasing order. Implement the rearrange function using QuickSort ( such that the pivot is set on the extreme left and the rearrangement is carried on on two pointers) using the O(n) time algorithmThe function gets as input an array, and index of the pivot.The function rearranges the array, and returns the index of the pivot after the rearrangement. int rearrange(int* A, int n, int pivot_index); Implement the QuickSort algorithm. - For n<=2 the algorithm just sorts the (small) array (smaller number first). - For n>=3 the algorithm uses the rearrange function with the pivot chosen to be the median of A[0], A[n/2], A[n-1]. void quick_sort(int* A, int n);
- Implement the following function, without using any data structure. /* Given two vectors of chars, check if the two vectors are permutations of each other, i.e., they contains same values, in same or different order.e.g., V1=[‘a’,’b’,’a’] and V2=[‘b’,’a’,’a’] stores same multi-set of data points: i.e., both contains two ‘a’, and one ‘b’. e.g., V3=[‘a’,’c’,’t’,’a’] and V4=[‘a’,’c’,’t’] are not same multi-set. V3 contains two ‘a’s, while V4 has only one ‘a’. Note: when considering multiset, the number of occurrences matters. @param list1, list2: two vectors of chars @pre: list1, list2 have been initialized @post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */ bool SameMultiSet (vector<char> list1, vector<char> list2)The insertion sort was discussed and the implementation was demonstrated in the sorting lecture. In this assignment, you are asked to re-implement the insertion sort (InsertionSort.java),with additional requirements. Particularly, you need to show:1. For each iteration, how a number from a unsorted region is placed in the correctionposion in the sorted region;2. How to make the whole array be sorted based on the previous step, and count the totalnumber of shifts during the whole insertion sort process.2 Details of the ProgramTo complete the whole implementation, you should write at least the following importantmethods:2.1 Part1: insertLast/**A method to make an almost sorted array into fully sorted.@param arr: an array of integers, with all the numbers are sortedexcepted the last one@param size: the number of elements in an array*/public static void insertLast(int[] arr, int size){// your work}To make it concrete, let’s use the example shown in Figure 1. In this example, the…01... ""Implementation of the Misra-Gries algorithm.Given a list of items and a value k, it returns the every item in the listthat appears at least n/k times, where n is the length of the array By default, k is set to 2, solving the majority problem. For the majority problem, this algorithm only guarantees that if there isan element that appears more than n/2 times, it will be outputed. If thereis no such element, any arbitrary element is returned by the algorithm.Therefore, we need to iterate through again at the end. But since we have filtredout the suspects, the memory complexity is significantly lower thanit would be to create counter for every element in the list. For example:Input misras_gries([1,4,4,4,5,4,4])Output {'4':5}Input misras_gries([0,0,0,1,1,1,1])Output {'1':4}Input misras_gries([0,0,0,0,1,1,1,2,2],3)Output {'0':4,'1':3}Input misras_gries([0,0,0,1,1,1]Output None"""..