Complete the following sorting program with bubble sort algorithm. #include using namespace std; template void Print (T& vec, int n, string s) { cout << s << ": [" << flush; for (int i=0; i

#include <iostream> using namespace std; #define SIZE 5   //creating the queue using array int A[SIZE]; int front = -1; int rear = -1;   //function to check if the queue is empty bool isempty() { if(front == -1 && rear == -1) return true; else return false; }   //function to enter elements in queue void enqueue ( int value ) { //if queue is full if ((rear + 1)%SIZE == front)    cout<<"Queue is full \n"; else {  //now the first element is inserted  if( front == -1)     front = 0; //inserting element at rear end rear = (rear+1)%SIZE;   A[rear] = value; } }   //function to remove elements from queue void dequeue ( ) { if( isempty() )  cout<<"Queue is empty\n"; else //only one element if( front == rear )  front = rear = -1; else  front = ( front + 1)%SIZE; }   //function to show the element at front void showfront() { if( isempty()) cout<<"Queue is empty\n"; else cout<<"element at front is:"<<A[front]; }   //function to display the queue void…

#include<bits/stdc++.h>using namespace std;bool isPalindrome(string &s){    int start=0;    int end=s.length()-1;    while(start<=end)    {        if(s[start]!=s[end])        {            return false;                    }        start++;        end--;    }    return true;}int main(){    string s;    cout<<"ENTER STRING:";    getline(cin,s);    int n=s.length();    bool flag=true;    for(int i=1;i<n;i++)    {        string lowerHalf=s.substr(0,i);        string upperHalf=s.substr(i,n-i);      if(isPalindrome(lowerHalf) && isPalindrome(upperHalf))      {          flag=false;          cout<<"String A is:"<<lowerHalf<<"\n";          cout<<"String B is:"<<upperHalf<<"\n";          break;      }    }    if(flag)    {        cout<<"NO\n";    }    return 0;} change to stdio.h string.h

Python Big-O Notation/Time Complexity Pls answer only if u know big-o Thanks!  Item #3.

void func(vector<string>& names){   sort(names.begin(), names.end());       //(a)   for(int i = names.size() - 1; i > 0; --i){.  //(b)     if(names[i] == names[i - 1]){        names[i] = move(names.back());        names.pop_back();     }   } } What are the reasons or consequences for the programmer’s choice to loop backwards in the line marked (b), and under what circumstances would the algorithm behave differently if that line were replaced with: for (int i = 1; i < names.size(); ++i) ?   What would change?

Use iterators to print each element of v separted by spaces. #include <iostream>#include <vector> using namespace std; int main(){vector<int> v = {1, 3, 5}; (.....) }

#include <iostream> #include <string> #include <cstdlib> using namespace std; template<class T> T func(T a) { cout<<a; return a; } template<class U> void func(U a) { cout<<a; } int main(int argc, char const *argv[]) { int a = 5; int b = func(a); return 0; }       Give output for this code.

task7.c Compare 2 strings function cmp will print out > if s1 is "bigger" than s2, < if s1 is "smaller than s2, = if s1 is "equal" to s2. */ #include <stdio.h> #include <string.h> void cmp(char *s1, char *s2){     return; } int main() {     char *string1 = "UNIX rules!";     char *string2 = "Windows drools!";     cmp(string1,string2);                 return 0; }     Start with task7.c, implement the cmp() function. The goal of this function is to compare two strings using the sum of the ASCII value of all the characters in each string. Output of your program should look like this:  UNIX rules! is smaller than Windows drools!

program Linked List: modify the following program to make a node containing data values of int, char, and string. #include <iostream> using namespace std; struct node { int data; struct Node *next; }; struct Node* head = nullptr;//or Null or 0; void insert(int new_data) { struct Node* new_node=(struct Node*) new(struct Node); new_mode->data=new_data; new_mode->next=head; head=new_node; } void display() { struct Node* ptr; ptr=head; while(ptr ! = NULL) {   cout<<ptr->data<<""; ptr=ptr->next; } } int main() { insert{2}; display{}; return0; }

Exercise 4 Write a function called chop that takes a list, modifies it by removing the first and last elements, and returns None. For example: >>> t = [1, 2, 3, 4] >>> chop(t) >>> t [2, 3]