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- Find the complexity of the function used to find the kth smallest integer in an unordered array of integers: int selectkth(int a[], int k, int n) {int i, j, mini, tmp; for (i = 0; i < k; i++) { mini = i; for (j = i+1; j < n; j++) if (a[j]<a[mini]) mini = j; tmp = a[i]; a[i] = a[mini]; a[mini] = tmp; }return a[k-1];}Write a function that gets an array of ints of length n, and a number k, and returns the longest subsequence of consecutive k’s in the array. For example - On input ([1, 1, 3, 3, 3, 1, 5, 1], 1), the function returns 3 - On input ([1, 3, 3, 2, 3, 3, 3, 3], 3), the function returns 4 - On input ([3, 5, 2, 5, 2, 5, 5], 3), the function returns 2 - On input ([1, 2, 3, 3, 3, 3], 5), the function returns 0 // returns the length of the longest subsequence // of consecutive k’s in the array int longest_seq(const int* ar, int n, int k); test for the function: void test_q1() { inta1[] = {1, 1, 3, 3, 3, 1, 5, 1}; if (longest_seq(a1, 8, 1) == 2) printf("Q1-1 ok\n"); else printf("Q1-1 ERROR\n"); inta2[] = {1,3,3,2,3,3,3,3}; if (longest_seq(a2, 8, 3) == 4) printf("Q1-2 ok\n"); else printf("Q1-2 ERROR\n"); }Modify the Partition function so that after running it, any input array A is partitioned into three parts: the left part consisting of all elements < pivot, the middle part consisting of all elements = pivot, and the right part consisting of the rest Write down the pseudo-code for this modified version of the Partition function. (must be an in-place algorithm with Θ(n) time complexity.) Hint: Define three indices: i as the end of the left part, k as the end of the middle part, and j denoting the current index which is used in the for-loop.
- Write a function that gets an array of points and sorts the array using qsort(). Given two points a=(ax,ay) and b=(bx,by) we compare them as follows: 1) if (ax)2+(ay)2 < (bx)2+(by)2, then a should come before b in the sorted array. 2) if (ax)2+(ay)2 = (bx)2+(by)2, then we compare the points by the x coordinate. Remark: For a point a=(ax,ay) the quantity ((ax)2+(ay)2)½ is the distance of a from the (0,0). That is, we sort the points according to their distance to (0,0), and if for points at the same distance, then we sort them according to the first coordinate. You will need to implement the comparison function, and apply qsort() on the array with this comparison function. For example: - Input: [(3,2), (7,1), (1,1), (3,4), (5,0), (7,1)] - Expected output: [(1,1), (3,2), (3,4), (5,0), (7,1), (7,1) ] Explanation: (1,1) is first because 12+12=2 is the smallest (3,2) is next because 32+22 = 13 is the second smallest Both (3,4) and (5,0) have the same sum of squares, 25, but 3<5 so…Given an array of integers, write a function to find the maximum sum of any contiguous subarray (i.e., a subarray with consecutive elements) in the array. For example, in the array [1, -2, 3, 4, -1, 2, 1, -5, 4], the maximum sum of a contiguous subarray is 6 (from index 2 to index 5).Given an array A = [10, 7, 4, 2, 1], and target = 7, return the index of the target if found, else return -1. 1. Can this problem be solved in O(logN)? 2. If so, write the implementation. Mention time and space complexity 3. If this problem cannot be solved using O(logN), what is the solution that you suggest - write the code and also mention time and space complexity.
- Modify the quicksort function so that it calls insertion sort to sort any sublistwhose size is less than 50 items. Compare the performance of this version withthat of the original one, using data sets of 50, 500, and 5,000 items. Then adjust thethreshold for using the insertion sort to determine an optimal setting. Use python’s time module to calculate the duration of the original quicksort version and the modified version. Do this for 3 different data sets of 50, 500, and 5000 items. These datasets are not going to be provided, so you have to come up with them. You can use python’s random module to help come up with the random item. Experiment with a different threshold value for the size of the sublist that indicates a switch to insertion sort, and report which value was optimal. Use this template: import timedef original_quicksort(input_list):sorted_list = []#TODO: Your work here# Return sorted_listreturn sorted_list def modified_quicksort(input_list):sorted_list = []#TODO: Your…For a string P of length m, define a function ∆P :{1,...,m}→{1,...,m} as follows: ∆P[q] is the length of the shortest actual suffix of Pq which is also a prefix of P. If there is no actual suffix of Pq which is also a prefix of P then ∆P[q]=0. Give an efficient algorithm for calculating ∆P. (The algorithm should output the entire array ∆P .)Given a non-empty list items of positive integers in strictly ascending order, find and return thelongest arithmetic progression whose all values exist somewhere in that sequence. Return the answer as a tuple (start, stride, n) of the values that define the progression. To ensure unique results to facilitate automated testing, if there exist several progressions of the same length, this function should return the one with the lowest start. If several progressions of equal length emanate from the lowest start, return the progression with the smallest stride. INTRUCTIONS: Input a comment or a solution on how you solve the problem, write a psuodocode and flow chart about th problem
- Let A and B be two arrays of length n, each containing a random permutation of the numbers from 1 to n. An inversion between the two permutations A and B is a pair of values (x, y) where the index of x is less than the index of y in array A, but the index of x is more than the index of y in array B. Design an algorithm which counts the total number of inversions between A and B that runs in O(n log n) time.For code. an array of meeting time intervals consisting ofstart and end times [[s1,e1],[s2,e2],...] (si < ei),determine if a person could attend all meetings. For example,Given [[0, 30],[5, 10],[15, 20]],return false.""".Find the function that best represents the temporal order of the following algorithms in the worst case.Count the operations of addition (+) and multiplication (×). Suppose that the array K contains n values.polyEval(K: whole array, x: whole): whole y ← K[n − 1] for i ← n − 2, n − 3 .. 0 y ← y × x + K[i] end for return y end polyEval justify your answer please