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Please solve my following problem!
suppose we have a code like;
struct team
{
    int runs[15];
    char *name[15];
} ind,pak,aus;

// now first murge these int arrays in a single array
int d[45];
for(int i=0; i<15; i++)
{
    d[i]=pak.runs[i];
    d[15+i]=ind.runs[i];
    d[30+i]=aus.runs[i];
}


Now , I sort them using any method,
finally I get sorted values but I need the names adjacent to these runs.
like;
pak.runs[5] is placed at d[14] after sorting , then how can i display the corresponding name at pak.name[5] after sorting?
I hope you understand my question...
Please help me!
Posted
Updated 15-Mar-11 23:20pm
v3
Comments
JF2015 16-Mar-11 1:40am    
Edited for code formatting.
Emilio Garavaglia 16-Mar-11 4:21am    
More code formatting

As a quick solution,
Copy all 3x15 names into single array in the above for loop.
While sorting d[45] array, exchange the name[] values according to the sorting index(i do not know which algorithm you are using for sorting but you will exchange d[] values through indexes, at the same time you exchange the values).I am not considering any performance issues here.
 
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Comments
Check Solution from santosh for the code implementation of this solution
Dalek Dave 16-Mar-11 4:45am    
That would work.
How about using STL instead? You can use std::list<std::string> class, which has the method sort.

—SA
 
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Dalek Dave 16-Mar-11 4:45am    
Ah, Sage Advice!
Sergey Alexandrovich Kryukov 16-Mar-11 4:50am    
Thank you, Dalek,
--SA
Nuri Ismail 16-Mar-11 5:07am    
Good advice, my 5.

Personally I'd use std::multimap and already gave an example to the OP with this approach.
In this case the container does everything automatically behind the scene. :)
Sergey Alexandrovich Kryukov 16-Mar-11 5:11am    
I see. (Already voted :-)
I did not bother to dig into the code, offered the simplest container which would work better than OP's programming from scratch (which still would be good to learn, of course).
--SA
This may help you.
Before sorting of array you should fill name array like,

char *dn[45];

for(int j=0; j<15; j++)
{
dn[j]=pak.name[j];
dn[15+j]=ind.name[j];
dn[30+j]=aus.name[j];
}


now your sorting logic may differ but with linear sorting with do same,

void selectionSort(int arr[],char namearray[], int size)
{
     int indexOfMin, pass, j;

     for (pass = 0; pass < size - 1; pass++)
     {
           indexOfMin = pass;

           for (j = pass + 1; j < size; j++)
                 if (arr[j] < arr[indexOfMin])
                       indexOfMin = j;

           swap(arr[pass], arr[indexOfMin]);
           swapName(namearray[pass],namearray[indexOfMin]);
     }
}// end selectionSort()

void swap(int& x, int& y)
{
     int temp;
     temp = x;
     x = y;
     y = temp;
}// end swap()
void swapName(char& x, char& y)
{
     int temp;
     temp = x;
     x = y;
     y = temp;
}// end swapNames
 
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v3
Comments
Dalek Dave 16-Mar-11 4:44am    
Good Answer. 5!
[no name] 16-Mar-11 4:59am    
Thanks dalek,
Nuri Ismail 16-Mar-11 5:07am    
Good advice indeed! 5 :)
[no name] 16-Mar-11 6:08am    
Thanks nuri
mbue 16-Mar-11 14:01pm    
brr, bubble sort ;)
You already have your C-style solution.
I want to give you a simple C++ - style alternative.

Here is the code:
C++
#include <iostream>
#include <string>
#include <map>

int main()
{
	// Your data structure:
	static const int runs_size = 15;
	struct team
	{
		int runs[runs_size];
		std::string name;
	} ind, pak, aus;


	// Now we will fill some data. 
	// The next block has nothing to do with the solution.
	// It is just an example initialization for testing.
	
	// ====================== Example initialization ==========================
	// First clear evrything
	memset(&ind, 0, sizeof(ind));
	memset(&pak, 0, sizeof(pak));
	memset(&aus, 0, sizeof(aus));

	// Fill names:
	ind.name = "India";
	pak.name = "Pakistan";
	aus.name = "Australia";

	// Fill some runs:
	ind.runs[0] = 10; ind.runs[5] = 20; ind.runs[10] = 25;
	pak.runs[1] = 17; pak.runs[2] = 18; pak.runs[12] = 23;
	aus.runs[4] = 10; aus.runs[5] = 19; aus.runs[6] = 24;
	// =======================================================================


	// ======================== Here the important part begins ===========================
	typedef std::multimap<int, const team*, std::greater<int> > mmap_runs_teams_t;
	typedef std::pair<int, const team*> mmap_pair_t;

	mmap_runs_teams_t mmap_runs_teams;
	for (int i = 0; i < runs_size; i++)
	{
		mmap_runs_teams.insert(mmap_pair_t(ind.runs[i], &ind));
		mmap_runs_teams.insert(mmap_pair_t(pak.runs[i], &pak));
		mmap_runs_teams.insert(mmap_pair_t(aus.runs[i], &aus));
	}
	// ===== And here the important part ends - That's it! This is your solution! ;) =====


	// Now check the results:
	std::cout << "Best runs: " << std::endl;
	for (mmap_runs_teams_t::iterator it = mmap_runs_teams.begin(); it != mmap_runs_teams.end(); it++)
	{
		std::cout << "\nRun: " << it->first << "\tTeam: " << it->second->name;
	}

	std::cin.get();
	return 0;
}


The output of this program:
Best runs:
Run: 25 Team: India
Run: 24 Team: Australia
Run: 23 Team: Pakistan
Run: 20 Team: India
Run: 19 Team: Australia
Run: 18 Team: Pakistan
Run: 17 Team: Pakistan
Run: 10 Team: India
Run: 10 Team: Australia
........... all zero runs to the end..........



From the code above you can see that the actual solution is less then 10 lines of code! :)
This is possible with the Standard C++ Library. I used std::multimap[^] as a container.

Intentionally I will not explain how this solution works.
As an exercise, you can check out the documentation for std::multimap.
This way you can understand how this code works.

If you have some questions after reading the documentation, please feel free to ask here. :)
 
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v3
Comments
Sergey Alexandrovich Kryukov 16-Mar-11 5:08am    
Remarkable effort, my 5.
--SA
Nuri Ismail 16-Mar-11 5:10am    
Thank you very much! :)
[no name] 16-Mar-11 6:10am    
Woh Nice effort
my 5+
Nuri Ismail 16-Mar-11 6:21am    
Thank you, Santosh! :)

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