max_time

/*
given a set of nine single digits(not necesserily distinct) say 0,0,1,3,4,6,7,8,9 . it is possible to form many distinct times in 12 hour time format HH:MM:SS , such as 10:36:40 or 01:39:46 by use of digits  only once. the objective is to find maximum positive valid  time (00:00:01 to 12:00:00) . that can be formed using some six of nine digits exectly once. in this case it is 10:49:38.
example
(1) set={0,1,3,4,2,1,5,8,0}
    ans = 11:58:43
(2) set={0,9,6,9,7,8,9,6,3}
    ans = Impossible opretion!
*/
//9 digit to time

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
int sec,min,hh,ten=0,top=9;
class time
{
    int i;
    public:
    int getdata()
    {

            cout<<"\nEnter the value: ";
            cin>>i;
            return i;

    }
    int getans(int num[],int no)
    {
        int i,k,com,max;
        max=com=0;
        for(i=0;i<top;i++)
        {
            for(k=0;k<top;k++)
            {
                if(i!=k)
                {
                    com=(num[i]*10)+num[k];
                    //cout<<" "<<com;
                }
                if(com>max && com<no)
                {
                    max=com;
                }
            }
        }
        return max;
    }
    void transfer(int used[],int no)
    {


            used[0]=no/10;
            used[1]=(no%10);

    }
    void cuts(int num[],int used[])
    {
        int j=0,k;

        for(i=0;i<top;i++)
        {
            if(num[i]==used[j]  && j<2)
            {
                if(i!=top-1)
                {
                    for(int k=i;k<top-1;k++)
                    {
                        num[k]=num[k+1];
                    }
                }
                  top--;
                  j++;
                  i--;
            }
        }

    }



};

int main()
{
    time t1;
    int num[9],i,k,used[3];
    clrscr();
    //insertion
    cout<<"\n Enter 9 digits: ";
    for(i=0;i<top;i++)
    {
        num[i]=t1.getdata();
    }
    // displayed
    cout<<"\n User's content\n (";
    for(i=0;i<top;i++)
    {
        cout<<" "<<num[i];

    }
     cout<<")\n";
    // hh
    hh = t1.getans(num,12);

    t1.transfer(used,hh);
    // cut digits from num array of hour
    t1.cuts(num,used);

    //min
    used[0]=0;
    used[1]=0;
    //get minute
    min=t1.getans(num,60);
    // cut digits from num array of minute
    t1.transfer(used,min);

    t1.cuts(num,used);

    //sec
    used[0]=0;
    used[1]=0;
    //get ans of second
    sec=t1.getans(num,60);
    // cut digits from num array of second
    t1.transfer(used,sec);

    // if h,r,m ==0 the operation is impossible
    if(hh==0 || min==0 || sec==0)
    {
        cout<<"\n operation is impossible!";
    }
    else
    {
        cout<<"\n hh: mm: ss  ";
        cout<<"\n "<<hh<<":"<<min<<":"<<sec;
    }
    getch();
    return 0;
}