In this HackerRank Attending Workshops problem in c++ programming A student signed up for n workshops and wants to attend the maximum number of workshops where no two workshops overlap. You must do the following:
Implement 2 structures:
struct Workshop having the following members:
- The workshop’s start time.
- The workshop’s duration.
- The workshop’s end time.
struct Available_Workshops having the following members:
- An integer, n (the number of workshops the student signed up for).
- An array of type Workshop array having size n.
Implement 2 functions:
- Available_Workshops* initialize (int start_time[], int duration[], int n) Creates an Available_Workshops object and initializes its elements using the elements in the start_time[] and duration[] parameters (both are of size n). Here, start_time[i] and duration[i] are the respective start time and duration for the workshop. This function must return a pointer to an Available_Workshops object.
- int CalculateMaxWorkshops(Available_Workshops* ptr) Returns the maximum number of workshops the student can attend—without overlap. The next workshop cannot be attended until the previous workshop ends.
HackerRank Attending workshops problem solution in c++ programming.
#include<bits/stdc++.h>
using namespace std;
//Define the structs Workshops and Available_Workshops.
//Implement the functions initialize and CalculateMaxWorkshops
struct Workshops{
friend ostream &operator<<(ostream &os, const Workshops &obj);
int start_time;
int end_time;
int duration;
bool operator<(const Workshops &rhs){
return (this->end_time < rhs.end_time);
}
};
ostream &operator<<(ostream &os, const Workshops &obj){
os << obj.start_time << ": " << obj.end_time << ": "
<< obj.duration << "n";
return os;
}
struct Available_Workshops{
int n;
vector<Workshops> vec;
};
Available_Workshops* initialize(int start_time[], int duration[], int num)
{
Available_Workshops *avail = new Available_Workshops;
avail->n = num;
Workshops test;
for(int i{0}; i < num; i++){
test.start_time = start_time[i];
test.duration = duration[i];
test.end_time = start_time[i] + duration[i];
avail->vec.push_back(test);
}
sort(avail->vec.begin(), avail->vec.end());
return avail;
}
int CalculateMaxWorkshops(Available_Workshops *test){
int w_count = 1;
int test_end_time = test->vec.at(0).end_time;
for(int i{1}; i < test->n; i++){
if(test_end_time <= test->vec.at(i).start_time){
w_count++;
test_end_time = test->vec.at(i).end_time;
}
}
return w_count;
}
int main(int argc, char *argv[]) {
int n; // number of workshops
cin >> n;
// create arrays of unknown size n
int* start_time = new int[n];
int* duration = new int[n];
for(int i=0; i < n; i++){
cin >> start_time[i];
}
for(int i = 0; i < n; i++){
cin >> duration[i];
}
Available_Workshops * ptr;
ptr = initialize(start_time,duration, n);
cout << CalculateMaxWorkshops(ptr) << endl;
return 0;
}
Second solution
#include<bits/stdc++.h>
using namespace std;
typedef struct Workshops
{
int s,d,e;
}Workshops;
typedef struct Available_Workshops
{
int n;
Workshops *shops;
Available_Workshops(int t)
{
n = t;
shops = (Workshops*)malloc(t*sizeof(Workshops));
}
}Available_Workshops;
bool comp(const Workshops &a, const Workshops &b)
{
if(a.e < b.e)
return true;
return false;
}
Available_Workshops* initialize (int start_time[], int duration[] ,int N)
{
Available_Workshops* node = new Available_Workshops(N);
for(int i =0 ; i< node->n; i++)
{
(node->shops)[i].s = start_time[i];
(node->shops)[i].d = duration[i];
(node->shops)[i].e = start_time[i] + duration[i];
}
return node;
}
int CalculateMaxWorkshops (Available_Workshops* ptr)
{
sort(ptr->shops, ptr->shops + ptr->n, comp);
int ans = 1, t;
for(int i = 0 ; i < ptr->n; i++)
{
if(i == 0)
{
t = (ptr->shops[i]).e;
continue;
}
if((ptr->shops[i]).s>=t)
{
ans++;
t = (ptr->shops[i]).e;
}
}
return ans;
}
int main()
{
int n;
cin>>n;
int start_time[n],duration[n];
for(int i=0;i<n;i++)
{
cin>>start_time[i];
}
for(int i=0;i<n;i++)
{
cin>>duration[i];
}
Available_Workshops* ptr;
ptr=initialize(start_time,duration,n);
cout<<CalculateMaxWorkshops(ptr)<<endl;
return 0;
}