# 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 [Last updated : February 26, 2023]

## Printing the maximum possible time using six of nine given single digits

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

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

```#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!
```