Program to find the number from given holes

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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:

0, 4, 6, 9 has 1 holes each and 8 has 2 holes in it.
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:

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.
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.
Now, if value of rem variable is equal to 1 then first print 4 once and then print 8 quo number of times.
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:

Time Complexity: O(n)

Auxiliary Space: O(1)

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