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- Given two arrays A and B of equal size N, the task is to find a vector containing only those elements which belong to exactly one array - either A or B - but not both. That is, if some element x appears in both A and B it should not be included in the output vector. Complete the implementation of "getVector(vector<ll>A, vector<ll>B, int N)" function. Example: Input: 1 3 5 7 1 3 2 4 6 7 8 Output: 3 5 1 3 2 4 6 8 The driver codes are attached in the images below. DO NOT CHANGE THE PROVIDED DRIVER CODES AT ALL, LEAVE THEM EXACTLY THE SAME. ONLY ADD CODE TO THE "getVector(vector<ll>A, vector<ll>B, int N)" FUNCTIONWrite a function called genericSort() that takes in a numeric or integer vector, sorts it, and returns the indices of the sorted values. It should also print an error if the input is a character. For example, genericSort(c(1,3,7,5)) should return the vector (1,2,4,3). Programming in RFix the signed overflows in the following code so that it calculates the polygon area using the undefined behavior sanitizer. #include <cstdint>#include <vector>#include <iostream>using namespace std; uint64_t area(vector<pair<int, int>> &points) {int64_t total = 0;size_t n = points.size(); for (unsigned i = 0; i < n; i++) {unsigned j = (i + 1) % n;int x_i = points[i].first;int y_i = points[i].second;int x_j = points[j].first;int y_j = points[j].second;total += (x_i * (y_j - y_i) - y_i * (x_j - x_i));} total /= 2;return total;}
- What to input here to run properly? %Encode A1, b1 and x1 as the vector of unknowns. A1 = b1 = syms xv1 = %Check the size of A, set it as m1 and n1 [m1,n1] = %Augment A and b to form AM1 AM1 = %Solve the Reduced Rwo Echelon of AM1. RREFA1 = %Collect the last column and set as bnew1, set the remaining elements as Anew1 bnew1= Anew1 = %Check if Anew is an identity matrix, if it is, bnew is the solution if Anew1 =eye(m1,n1) Root1 = bnew1 else display("No Solution") end %Augment the matrix A1 with the identity Matrix of the same size, set the result as AMI1 AMI1 = %Find the reduced row echelon form of AMI1, set the result as RREFAI1 RREFAI1= %Collect the second half of the matrix as AInew1, set the remaining elements as AIold1 AInew1= AIold1 = %Check if AIold1 is an identity matrix, if it is, AInew1 is the inverse %Encode A2, b2 and xv2 as the vector of unknowns. A2 = b2 = syms xv2 = %Check the size of A2, set it as m2 and n2 [m2,n2] = %Augment…Write a function, slens, that borrows a Vector of Strings and returns aHashMap that maps each string to its length.fn slens(stuff: &Vec<String>) -> HashMap<String, usize> {...}fn main() {let stuff = vec![String::from("A"),String::from("fine"),String::from("mess"),];println!("{:?}", slens(&stuff));println!("{:?}", stuff);}/* Output:{"fine": 4, "mess": 4, "A": 1}["A", "fine", "mess"]*/6) While there is a built-in pop_back() method for vectors, there is no built-in pop_front method. Suppose a program needs a pop_front() method that will remove the first element from the vector. For example if the original vector is [1, 2, 3, 4, 5], then after passing in this vector to pop_front() the new resulting vector will be [2, 3, 4, 5]. Which of the options below is the correct implementation of pop_front()? Group of answer choices D-) void pop_front(vector<int> &v){ for(int i=0; i<(v.size()-1); i++)v[i+1] = v[i]; v.pop_back(); } C-) void pop_front(vector<int> &v){ for(int i=0; i<(v.size()-1); i++)v[i] = v[i+1]; v.pop_back(); } B-) void pop_front(vector<int> &v){ for(int i=0; i<v.size(); i++)v[i+1] = v[i]; v.pop_back(); } A-) void pop_front(vector<int> &v){ for(int i=0; i<v.size(); i++)v[i] = v[i+1]; v.pop_back(); } 7) Suppose a program has the following vector: [99, 23, 55, 71, 87, 64, 35, 42] The goal is to…
- In PYTHON Define a vector of integers with 5 elements Write the following function Given a vector and two indexes, returns a vector where the values at the provided indexes are swapped Example Input: my_vector = [40,51,62,73,84,95] Invokefunction:swap(my_vector,2,4) Output:[40,51,84,73,62,95]Which shows valid accesses for an array a and vector v, each with 10 elements? In other words, this question is asking what is the proper syntax for accessing the array and vector. a. array: a[7] vector: v.at(5) b. array: a.at(5) vector: v[9] c. array: a.at(5) vector: v[5] d. array: a[1] vector v.at[4] Coding is in C++.Create your own vector class which will test algorithms from the STL Derive class myVector from vector. myVector must implement the following methods: int seqSearch(T searchItem); int binarySearch(T searchItem); void bubbleSort(); void insertionSort(); Create a test program to create some vectors and test your methods above. Recall from your reading that binary search only works on a sorted list. Add a static member to the class to “remember” if the list is sorted ( i.e. binarySearch() should first sort the vector if it’s not sorted already). Use the template below as a starter for your assignment. All comments in bold represent code which you need to implement. #include <iostream> #include <string> #include <vector> using namespace std; template <class T> class myVector: public vector<T> { public: int seqSearch(T searchItem); int binarySearch(T searchItem); void bubbleSort(); void insertionSort(); }; template <class T>…
- Find the number of super values in vector<int>v where an array element is super if it is greater than all elements to its right. For ex in {4,27,1,20,17, 5} has 4 super values 27, 20,17,5 so its output should be 4. Code in C++Write a function that accepts a number, N, and a vector of numbers, V. The function will return two vectors which will make up any pairs of numbers in the vector that add together to be N. Do this with nested loops so the the inner loop will search the vector for the number N-V(n) == V(m). n and m are indices in the vector of numbers. Example A = [1,2,3,4,5,6, 7] Google(5, A) Return [1,2,3,4] and [4,3,2,1] being the pairs that sum to 5. Notice that each pair appears twice. Try to write code that does not do this.Implement a __setitem__ function that also supports negative indices. For example: W = L2(Node(10, Node(20, Node(30))))print W[ 10, 20, 30 ] W[1]=25print W[ 10, 25, 30 ] W[-1]=35print W[ 10, 25, 35 ] Complete the code: def L2(*args,**kwargs): class L2_class(L): def __getitem__(self, idx): <... YOUR CODE HERE ...> def __setitem__(self, idx, value): <... YOUR CODE HERE ...> return L2_class(*args,**kwargs) W = L2(Node(10, Node(20, Node(30))))print(W)W[1]=25print(W)W[-1]=35print(W)