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C++ program to print the maximum possible time using six of nine given single digits

The Objective is to form the maximum possible time (in HH:MM:SS and 12 hour format) using any six of nine given single digits (not necessarily distinct).
Submitted by Abhishek Jain, on August 05, 2017

Given a set of nine single digits ((not necessarily distinct) say 0,0,1,3,4,6,7,8,9. It is possible to form many distinct times in a 12 hour time format HH:MM:SS, such as 10:36:40 or 01:39:46 by using each of the digits only once. The objective is to find the maximum possible valid time (00:00:01 to 12:00:00) that can be formed using some six of the nine digits exactly once. In this case, it is 10:49:38.

Example:

1) Set={0,1,3,4,2,1,5,8,0}
It will print-:  11:58:43

2) Set={0,9,6,9,7,8,9,6,3}
It will print-: Impossible Operation!

In this Program, I’m using the concept of COUNT ARRAY. To understand about count array please go through the link: C++ program to find the frequency of a character in a string using Count Array.

Consider the program:

#include<iostream>
using namespace std;

//Applying the concept of count Array
int count[10]={0};        

//This function will return the maximum value from count array upto the given index n
int MAX(int n)            
{ 
	int i;
	for(i=n;i>=0;i--)
	{  
		if(count[i]!=0)
		{
			count[i]--;
			return i;
		}
	}
	return -1;
}

//main program
int main()
{ 
	int x,i,y=0;
	char A[8];
	for(i=0;i<9;i++)
	{
		cin>>x;
		if(x>=0 && x<=9)
			count[x]++;
		else
		{
			cout<<"Wrong Input!Please Enter Again"<<endl;
			i--; 
		}
	}

	if(count[2]>=1 && count[1]>=1 && count[0]>=4)
		cout<<"12:00:00"<<endl;
	else
	{
		//All works for char array A[]
		for(i=0;i<8 && y!=1;i++)
		{
			if(i%3==2)
			{
				A[i]=':';
			}
			else if(i%3==0 && i>0)
			{
				//Adding '0' to convert the value into its character format
				A[i]=MAX(5)+'0';
			}   
			else if(i%3==1)
			{
				if(A[0]==0+'0' || i>1)
				{
					A[i]=MAX(9)+'0';
				}
				else
				{
					A[i]=MAX(1)+'0';
				}
			}
			else if(i==0)
				A[i]=MAX(1)+'0';
			if(A[i]==-1+'0')
				y=1;
		}
		if(y==1)
		{ 
			cout<<"Impossible Operation!"<<endl;
		}
		else
		{
			for(i=0;i<8;i++)
				if(i%3==2)
					cout<<A[i];
				else
					cout<<A[i]-'0';
			cout<<endl;
		}
	}
	return 0;
}

Output

First run:
0 1 3 4 2 1 5 8 0 
11:58:43

Second run:
0 9 6 9 7 8 9 6 3 
Impossible Operation!