Want to print next biggest number - c++

Question

So Here is the question:
Consider a class named Job that has deadline as a data member and relevant getter/setter
method(s). Assume you have to schedule two most earliest jobs on the basis of their deadlines. That is,
if there are three jobs in the system with deadlines (deadline1, deadline2, and deadline3, respectively)
then the system should report the top two earliest jobs (with smallest deadline value). You might need
to find the deadline with smallest and second most smallest value.
Here is my code:
#include<iostream>
using namespace std;
class job
{
private:
int Deadline;
public:
static int i;
void setDeadline(int a);
int getDeadline();
};
void job::setDeadline(int a)
{
Deadline = a;
cout << "Job " << i << " Has Deadline " << Deadline << endl;
}
int job::getDeadline()
{
return Deadline;
}
int job::i = 1;
int main()
{
job job1, job2, job3, job4, job5, job6, job7, job8, job9, job10, count;
job1.setDeadline(5);
count.i++;
job2.setDeadline(3);
count.i++;
job3.setDeadline(6);
count.i++;
job4.setDeadline(12);
count.i++;
job5.setDeadline(31);
count.i++;
job6.setDeadline(20);
count.i++;
job7.setDeadline(19);
count.i++;
job8.setDeadline(2);
count.i++;
job9.setDeadline(8);
count.i++;
job10.setDeadline(7);
int array[10] = { job1.getDeadline(), job2.getDeadline(),job3.getDeadline(),job4.getDeadline(),job5.getDeadline(),job6.getDeadline(),job7.getDeadline(),job8.getDeadline(),job9.getDeadline(),job10.getDeadline() };
int temp = array[0], store = 0, first = 0, second = 0;
for (int i = 0; i <= 9; i++)
{
if (temp > array[i])
{
temp = array[i];
}
}
for (int i = 0; i <= 9; i++)
{
if (temp == array[i])
{
first = i + 1;
break;
}
}
temp = array[0];
for (int i = 0; i <= 9; i++)
{
if (temp > array[i])
{
temp = array[i];
}
}
for (int i = 0; i <= 9; i++)
{
if (temp == array[i])
{
second = i + 1;
}
}
cout << "\nJob " << first << " and Job " << second << " are earliest";
return 0;
}
the problem I am facing is that both times it print the first smallest value. I want to print first 2 smallest value. How can I do that?

Answer 1

When you look for the first value, you go through the array and find the job with the lowest deadline and fill "first" with the job id (by the way, if you want to relate the "job id" to the index, personally I would make the variables zero-based, so job0, job1 and so on up to job9)
When you look for the second value, you do the exact same operations so the job you find is again the job with the lowest deadline and use that info to fill "second".
So, as you do the exact same thing both times, you are getting the exact same value for both. When you search for the second value you should take "first" into account so that you ignore the job that has already been used for "first" and not update "temp" in that case (this is a little bit easier if you name the jobs starting with zero because you don't have to constantly add one to the index).

Answer 2

the problem I am facing is that both times it print the first smallest value. I want to print first 2 smallest value. How can I do that?
With your solution: By creating another temp array with the first element removed you already discovered.
I wrote you as a minimal executable example using a vector of pairs since you didn't mention restrictions or anything - that is the most elegant solution I am able to come up with:
#include <iostream>
#include <vector>
#include <string>
bool sortVecBySec(const std::pair<std::string, int> &a,
const std::pair<std::string, int> &b)
{
return (a.second < b.second);
}
int main()
{
//Driver function to sort the vector elements by
//the second element of pairs
std::vector<std::pair <std::string,int>> v;
std::string job[10];
int deadline[10] = {5,3,6,12,31,20,19,2,8,7};
int n = sizeof(job)/sizeof(job[0]);
//Propagate string array
for(int i{}; i < 10;++i){
job[i] = "Job " + (std::to_string(i+1));
}
//Enter values in vector of pairs
for(int i{}; i<n; ++i){
v.push_back(make_pair(job[i],deadline[i]));
}
//Printing the vector of pairs
std::cout << "Vector of pairs as it is before\n";
for(int i{}; i< n; i++){
std::cout << v[i].first << " = " <<v[i].second << '\n';
}
//Using sort() function to sort by 2nd element of pairs
sort(v.begin(), v.end(),sortVecBySec);
//Printing the vector of pairs
std::cout << "\nVector of pairs from shortest to longest\n";
for(int i{}; i< 2; i++){
std::cout << v[i].first << " = " <<v[i].second << '\n';
}
std::cin.get();
return 0;
}
If you want to print all jobs, just use the n variable in the last print section instead of 2.
I think that should help you and shouldn't be to hard to parse it in a class.

Answer 3

First, you can save an extra round by saving the index right there, when you locate the earliest deadline
for (int i = 0; i <= 9; i++)
{
if (temp > array[i])
{
temp = array[i];
first = i + 1;
}
}
When searching for the next higher deadline value, you must take first into account. Both for the start value, and later when comparing with other values
temp = first > 1 ? array[0] : array[1];
for (int i = 0; i <= 9; i++)
{
if (temp > array[i] && array[i] > array[first])
{
temp = array[i];
second = i + 1;
}
}
Be aware, that this does not work properly, when you have multiple equal values.
For this case, compare the index values instead, e.g.
if (temp > array[i] && i != first)
{
temp = array[i];
second = i + 1;
}