Program to find the number from given holes

Given a number H which represent the total number of holes. The task is to find the smallest number which has that many numbers of holes. 
NOTE: 

1. 0, 4, 6, 9 has 1 holes each and 8 has 2 holes in it.
2. The number should not contain leading zeros.

Examples:

Input: H = 1 
Output: 0
Input: H = 5 
Output: 488 
Explanation: 
Number which has 5 holes in it is 488. i.e (1 + 2 + 2) 

Refer: Count the number of holes in an integer

Approach: 

1. First of all, Check whether the number of holes given is 0 or 1, if 0 then print 1 and if 1 then print 0.
2. If number of holes given is more than 1 then divide the number of holes by 2 and store the remainder in ‘rem’ variable. and quotient in ‘quo’ variable.
3. Now, if value of rem variable is equal to 1 then first print 4 once and then print 8 quo number of times.
4. Else print 8 only quo number of times.

Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that will find out
// the number
void printNumber(int holes)
{

    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        cout << "1";

    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        cout << "0";

    // If number of holes
    // is more than 0 or 1.
    else {
        int rem = 0, quo = 0;

        rem = holes % 2;
        quo = holes / 2;

        // If number of holes is
        // odd
        if (rem == 1)
            cout << "4";

        for (int i = 0; i < quo; i++)
            cout << "8";
    }
}

// Driver code
int main()
{
    int holes = 3;

    // Calling Function
    printNumber(holes);

    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;

class GFG
{

// Function that will find out
// the number
static void printNumber(int holes)
{

    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        System.out.print("1");

    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        System.out.print("0");

    // If number of holes
    // is more than 0 or 1.
    else
    {
        int rem = 0, quo = 0;

        rem = holes % 2;
        quo = holes / 2;

        // If number of holes is
        // odd
        if (rem == 1)
            System.out.print("4");

        for (int i = 0; i < quo; i++)
                System.out.print("8");
    }
}

// Driver code
public static void main (String[] args)
{
    int holes = 3;

    // Calling Function
    printNumber(holes);
}
}

// This code is contributed by Sachin.

Python3

# Python3 implementation of
# the above approach

# Function that will find out
# the number
def printNumber(holes):

    # If number of holes
    # equal 0 then return 1
    if (holes == 0):
        print("1")

    # If number of holes
    # equal 0 then return 0
    elif (holes == 1):
        print("0", end = "")

    # If number of holes
    # is more than 0 or 1.
    else:
        rem = 0
        quo = 0

        rem = holes % 2
        quo = holes // 2

        # If number of holes is
        # odd
        if (rem == 1):
            print("4", end = "")

        for i in range(quo):
            print("8", end = "")

# Driver code
holes = 3

# Calling Function
printNumber(holes)

# This code is contributed by Mohit kumar

C#

// C# implementation of the above approach
using System;

class GFG
{

// Function that will find out
// the number
static void printNumber(int holes)
{

    // If number of holes
    // equal 0 then return 1
    if (holes == 0)
        Console.Write ("1");

    // If number of holes
    // equal 0 then return 0
    else if (holes == 1)
        Console.Write ("0");

    // If number of holes
    // is more than 0 or 1.
    else
    {
        int rem = 0, quo = 0;

        rem = holes % 2;
        quo = holes / 2;

        // If number of holes is
        // odd
        if (rem == 1)
        Console.Write ("4");

        for (int i = 0; i < quo; i++)
            Console.Write ("8");
    }
}

// Driver code
static public void Main ()
{
    int holes = 3;

    // Calling Function
    printNumber(holes);
}
}

// This code is contributed by jit_t

Javascript

<script>
    // Javascript implementation of the above approach

    // Function that will find out
    // the number
    function printNumber(holes)
    {

        // If number of holes
        // equal 0 then return 1
        if (holes == 0)
            document.write("1");

        // If number of holes
        // equal 0 then return 0
        else if (holes == 1)
            document.write("0");

        // If number of holes
        // is more than 0 or 1.
        else {
            let rem = 0, quo = 0;

            rem = holes % 2;
            quo = parseInt(holes / 2, 10);

            // If number of holes is
            // odd
            if (rem == 1)
                document.write("4");

            for (let i = 0; i < quo; i++)
                document.write("8");
        }
    }

    let holes = 3;

    // Calling Function
    printNumber(holes);

    // This code is contributed by divyeshrabadiya07.
</script>

Output:

48

Time Complexity: O(n)

Auxiliary Space: O(1)