Let array A= {20, 3, 10, 9, 17, 12, 8, 13, 2, 5, 11} to find the median using Lomuto partition the array after the first iteration is
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Let array A= {20, 3, 10, 9, 17, 12, 8, 13, 2, 5, 11} to find the median using Lomuto partition the array after the first iteration is
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- // function to generate all subsequences of given arrayfunction generate_subseq(arr): // length of arr n = len(arr) // total subsequences for array of length n m = 2**n // to store all subsequences seqs = [] // running a loop for m times for i in range(1, m): // creating an array of zeros of length n a = [0]*n num = i // to use as an index for 'a' j = n-1 // run this loop till num > 0 while num > 0: if num is odd: a[j] = 1 // divide num by 2 num = num/2 // subtract 1 from 'j' j -= 1 // to store current subsequence seq = [] // iterating for n times for i in range(n): // add ith index value to 'seq' if a[i] == 1: seq.append(arr[i]) // add 'seq' to 'seqs' seqs.append(seq) // return seqs return seqs // given listsS1 = ['B','C','D','A','A','C','D']S2 = ['A','C','D','B','A','C']…Given an array, find the next greater element for each element in the array, ifavailable. If not available, print the element itself. The next greater element y for anelement x in the array is the first element that is greater than x and occurs on its rightside. The next greater element of the right most element in an array is the elementitself.Example: Given A = [ 6 8 4 3 9] the next greater element listB = [8 9 9 9 9]You are given two arrays, one shorter (with all separate items) and one longer (with no unique elements). Find the shortest subarray in the larger array that contains all of the shorter array's items. The things can be shown in any order.EXAMPLEInput: {1, 5, 9}{7, 5, 9, 0, 2, 1, 3, 5. 7, 9. 1, 1, 5, 8, 8, 9, 7}Results: [7, 10] (the underlined portion above)
- Write a program to ask the user for the number of arrays he wants to maintain and then populate eacharray by taking elements from the user. Merge all these arrays in another array and finally remove duplicateelements from the merged array.Write method findZero, as started below. findZero should return the index of the first element of array A that contains the value zero, starting from position 0 up through position pos. If no element of A from position 0 to position pos contains the value zero, then findZero should return –1. For example: Array (A) Position (pos) Value returned by findZero(A, pos) 1 0 2 5 6 0 -1 1 0 2 5 6 1 1 1 0 2 5 6 2 1 1 0 2 0 6 2 1 1 0 2 0 6 4 1 1 2 3 4 5 4 -1 Complete method findZero below. Assume that it is called only with values that satisfy its assumptions as stated below. // Description: Returns smallest index k // such that (0 <= k <= pos) and (A[k] == 0) // Assumptions: 0 <= pos < A.length public int findZero(int[] A, int pos) {Let’s try this calculation, assuming we have an array of size 10. If rear is currently 5, it will be set to 6%10, or 6. If rear is currently 9, it will be set to 10%10or 0. Try this calculation using various situations to see that it works no matterhow big the array is
- # the method take another circular array and returns a linear array containing the common elements between the two circular arrays. def intersection(self, c2): # To Do print("==========Test 10==========") lin_arr9 = [10, 20, 30, 40, 50, None, None, None] c10 = CircularArray(lin_arr9, 5, 5) c10.printFullLinear() # This should print: 40, 50, None, None, None, 10, 20, 30 lin_arr10 = [5, 40, 15, 25, 10, 20, 5, None, None, None, None, None] c11 = CircularArray(lin_arr10, 8, 7) c11.printFullLinear() # This should print: 10, 20, 5, None, None, None, None, None, 5, 40, 15, 25 output = c10.intersection(c11) print(output) # This should print: [10, 20, 40]Write method findZero, as started below. findZero should return the index of the first element of array A that contains the value zero, starting from position 0 up through position pos. If no element of A from position 0 to position pos contains the value zero, then findZero should return –1. For example:(in java) Array (A) Position (pos) Value returned by findZero(A, pos) 1 0 2 5 6 0 -1 1 0 2 5 6 1 1 1 0 2 5 6 2 1 1 0 2 0 6 2 1 1 0 2 0 6 4 1 1 2 3 4 5 4 -1 Complete method findZero below. Assume that it is called only with values that satisfy its assumptions as stated below. // Description: Returns smallest index k // such that (0 <= k <= pos) and (A[k] == 0) // Assumptions: 0 <= pos < A.length public int findZero(int[] A, int pos) {Given a sorted array of n integers that has been rotated an unknownnumber of times, write code to find an element in the array. You may assume that the array wasoriginally sorted in increasing order.EXAMPLElnput:find5in{15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14}Output: 8 (the index of 5 in the array)
- Create an array of 10 elements. Initialize it as follows. INITIAL array : 58 24 13 15 63 9 8 81 1 a 78 Now split it into middle and store the elements in two different arrays. E.g.-After spliting : 58 24 13 15 b 63 9 8 81 1 c 78 Now create another array “d” of size 5. Fill this array by adding elements of array “b” and array “c”. 67 32 94 16 d 141Write a program in ¢++ that asks the user to enter 10 integers of an array . The program must return the position pointer of an element that is equal to the square of another element.and then print the array example:- array={2.6,8,4,5} ------- 1 is squar of 2 position=3: In an array of integers, a "peak" is an element which is greater than or equal tothe adjacent integers and a "valley" is an element which is less than or equal to the adjacent integers. For example, in the array {5, 8, 6, 2, 3, 4, 6}, {8, 6} are peaks and {5, 2} are valleys. Given an arrayof integers, sort the array into an alternating sequence of peaks and valleys.EXAMPLEInput: {5, 3, 1, 2, 3}Output: {5, 1, 3, 2, 3}