Minimum count of digits required to obtain given Sum

Given an integer N, the task is to find the minimum number of digits required to generate a number having the sum of digits equal to N.

Examples:

Input: N = 18 
Output: 2 
Explanation: 
The number with smallest number of digits having sum of digits equal to 18 is 99.

Input: N = 28 
Output: 4 
Explanation: 
4-digit numbers like 8884, 6877, etc are the smallest in length having sum of digits equal to 28.

Approach: The problem can be solved by the following observations: 

1. Increment count by 9. Therefore, now count is equal to the number of 9's in the shortest number. Reduce N to N % 9
2. Now, if N exceeds 0, increment count by 1.
3. Finally, print count as the answer.

Below is the implementation of the above approach:

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// minimum count of digits
void mindigits(int n)
{
    // IF N is divisible by 9
    if (n % 9 == 0) {

        // Count of 9's is the answer
        cout << n / 9 << endl;
    }
    else {

        // If remainder is non-zero
        cout << (n / 9) + 1 << endl;
    }
}

// Driver Code
int main()
{
    int n1 = 24;
    int n2 = 14;
    mindigits(n1);
    mindigits(n2);
}

// Java program to implement
// the above approach
// required to make the given sum
import java.util.*;

class Main {

    // Function to print the minimum
    // count of digits
    static void mindigits(int n)
    {

        // IF N is divisible by 9
        if (n % 9 == 0) {

            // Count of 9's is the answer
            System.out.println(n / 9);
        }
        else {

            // If remainder is non-zero
            System.out.println((n / 9) + 1);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n1 = 24;
        int n2 = 18;
        mindigits(n1);
        mindigits(n2);
    }
}

# Python3 program to implement
# the above approach

# Function to print the minimum
# count of digits
def mindigits(n):
    
    # IF N is divisible by 9
    if (n % 9 == 0):

        # Count of 9's is the answer
        print(n // 9);
    else:

        # If remainder is non-zero
        print((n // 9) + 1);

# Driver Code
if __name__ == '__main__':
    
    n1 = 24;
    n2 = 18;
    
    mindigits(n1);
    mindigits(n2);

# This code is contributed by amal kumar choubey 

// C# program to implement
// the above approach
using System;

class GFG{

// Function to print the minimum
// count of digits
static void mindigits(int n)
{
    
    // IF N is divisible by 9
    if (n % 9 == 0)
    {
        
        // Count of 9's is the answer
        Console.WriteLine(n / 9);
    }
    else 
    {
        
        // If remainder is non-zero
        Console.WriteLine((n / 9) + 1);
    }
}

// Driver Code
public static void Main(String[] args)
{
    int n1 = 24;
    int n2 = 18;
    
    mindigits(n1);
    mindigits(n2);
}
}

// This code is contributed by 29AjayKumar

<script>

// JavaScript program to implement
// the above approach
// required to make the given sum

    // Function to print the minimum
    // count of digits
    function mindigits(n)
    {
 
        // IF N is divisible by 9
        if (n % 9 == 0) {
 
            // Count of 9's is the answer
            document.write(Math.floor(n / 9) + "<br/>");
        }
        else {
 
            // If remainder is non-zero
            document.write(Math.floor(n / 9) + 1 + "<br/>");
        }
    }

// Driver Code

        let n1 = 24;
        let n2 = 18;
        mindigits(n1);
        mindigits(n2);
           
</script>

3
2

Time Complexity: O(1)
Auxiliary Space: O(1)