/** Returns true if and only if A[start..end] is sorted in ascending order. * Preconditon: A is not null, and 0 <= start <= end <= A.length. **/ private static boolean isSorted (int[] A, int start, int end) { // TODO return false;
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Question 66
Computer Science
in java
Step by step
Solved in 3 steps with 3 images
- Double pointers: Describe how this operation can be done in O(1) time if there are pointers in each node to both the previous and the next node.bool isSubsetOf(const IntSet& otherIntSet) const// Pre: (none)// Post: True is returned if all elements of the invoking IntSet// are also elements of otherIntSet, otherwise false is// returned.// By definition, true is returned if the invoking IntSet// is empty (i.e., an empty IntSet is always isSubsetOf// another IntSet, even if the other IntSet is also empty). bool isSubsetOf(const IntSet& otherIntSet) const; bool IntSet::isSubsetOf(const IntSet& otherIntSet) const{ cout << "isSubsetOf() is not implemented yet..." << endl; return false; // dummy value returned}C++ help please my function: bool palindrome(const int a[], int start, int end){ bool flag = true; for (int i = start; i <= end / 2 && a[i]!= 0; i++) { if (a[i] != a[end - i - 1]) { flag = false; cout << "Array Is Not Palindrome\n"; return false; } } if (flag == true) cout << "Palindrome"; return true;} the main: int a[]= {1,2,1,2,1};cout << "Array = [1,2,1,2,1}\n";cout << "Testing array palindrome using 3 parameters:\n";palindrome(a, 2, 3); please help my function is not doing what it is supposed to :( it returns true when its not supposed to
- Recursive Count Write a recursive function rc_count(L:List[int], x:int) -> int that takes a list L and an integer x. The function returns the number of occurrences of x in L. For example, rc_count([1,1,2,3], 1) returns 2 and rc_count([1,1,2,3], 4) returns 0. Your Answer: 1 # Put your answer here 2 SubmitRecursive Max! Write a recursive function rc_max(L:List[int]) -> int that takes a list L and returns an element with highest value in the list. For example, rc_max([1,2,3]) returns 3. Note that L is not necessarily sorted. You may assume that the input L is not empty. You cannot use the built-in max() Your Answer: 1 # Put your answer here 2 Submit# Initialize sets A and BA = { 'a', 'b', 'c', 'd', 'e', 'f' }B = { 'c', 'd', 'f', 'h', 'j', 'k' } print('A = ', sorted(A))print('B = ',sorted(B))print(' ') AsymdiffB = set()# put your code here to compute the symmetric difference of A and B# first add all the elements that are in A but not B# then add all the elements that are in B but not A print('Symmetric Difference of A and B = ', sorted(AsymdiffB))
- A/ find code optimization method to this code and then find type to this code 1- For(i=1;iCount dominators def count_dominators(items): An element of items is said to be a dominator if every element to its right (not just the one element that is immediately to its right) is strictly smaller than it. By this definition, the last item of the list is automatically a dominator. This function should count how many elements in items are dominators, and return that count. For example, dominators of [42, 7, 12, 9, 13, 5] would be the elements 42, 13 and 5. Before starting to write code for this function, you should consult the parable of "Shlemiel the painter" and think how this seemingly silly tale from a simpler time relates to today's computational problems performed on lists, strings and other sequences. This problem will be the first of many that you will encounter during and after this course to illustrate the important principle of using only one loop to achieve in a tiny fraction of time the same end result that Shlemiel achieves with two nested loops. Your workload…Double trouble def double_trouble(items, n): Suppose, if just for the sake of argument, that the following operation is repeated n times for the given list of items: remove the first element, and append that same element twice to the end of items. Which one of the items would be removed and copied in the last operation performed? Sure, this problem could be finger-quotes “solved” by actually performing that operation n times, but the point of this exercise is to come up with an analytical solution to compute the result much faster than actually going through that whole rigmarole. To gently nudge you towards thinking in symbolic and analytical solutions, the automated tester is designed so that anybody trying to brute force their way through this problem by performing all n operations one by one for real will run out of time and memory long before receiving the answer, as will the entire universe. To come up with this analytical solution, tabulate some small cases (you can implement the…Double trouble def double_trouble(items, n): Suppose, if just for the sake of argument, that the following operation is repeated n times for the given list of items: remove the first element, and append that same element twice to the end of items. Which one of the items would be removed and copied in the last operation performed?Sure, this problem could be finger-quotes “solved” by actually performing that operation n times, but the point of this exercise is to come up with an analytical solution to compute the result much faster than actually going through that whole rigmarole. To gently nudge you towards thinking in symbolic and analytical solutions, the automated tester is designed so that anybody trying to brute force their way through this problem by performing all n operations one by one for real will run out of time and memory long before receiving the answer, as will the entire universe.To come up with this analytical solution, tabulate some small cases (you can implement the…const int size =4; int r1[size]; int i; for (i=0;i<2;i++) r1[i]= 2 * i+3; r1[3]=2 * i+ i; r1[2]=r1[3]; for (i=0;i#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. 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