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Find smallest number with given number of digits and sum of digits

How to find the smallest number with given digit sum s and number of digits d?

Examples :

Input  : s = 9, d = 2
Output : 18
There are many other possible numbers 
like 45, 54, 90, etc with sum of digits
as 9 and number of digits as 2. The 
smallest of them is 18.

Input  : s = 20, d = 3
Output : 299



A Simple Solution is to consider all m digit numbers and keep track of minimum number with digit sum as s. A close upper bound on time complexity of this solution is O(10m).

There is a Greedy approach to solve the problem. The idea is to one by one fill all digits from rightmost to leftmost (or from least significant digit to most significant).
We initially deduct 1 from sum s so that we have smallest digit at the end. After deducting 1, we apply greedy approach. We compare remaining sum with 9, if remaining sum is more than 9, we put 9 at the current position, else we put the remaining sum. Since we fill digits from right to left, we put the highest digits on the right side. Below is implementation of the idea.

C++

// C++ program to find the smallest number that can be
// formed from given sum of digits and number of digits.
#include <iostream>
using namespace std;
  
// Prints the smallest possible number with digit sum 's'
// and 'm' number of digits.
void findSmallest(int m, int s)
{
    // If sum of digits is 0, then a number is possible
    // only if number of digits is 1.
    if (s == 0)
    {
        (m == 1)? cout << "Smallest number is " << 0
                : cout << "Not possible";
        return ;
    }
  
    // Sum greater than the maximum possible sum.
    if (s > 9*m)
    {
        cout << "Not possible";
        return ;
    }
  
    // Create an array to store digits of result
    int res[m];
  
    // deduct sum by one to account for cases later
    // (There must be 1 left for the most significant
    //  digit)
    s -= 1;
  
    // Fill last m-1 digits (from right to left)
    for (int i=m-1; i>0; i--)
    {
        // If sum is still greater than 9,
        // digit must be 9.
        if (s > 9)
        {
            res[i] = 9;
            s -= 9;
        }
        else
        {
            res[i] = s;
            s = 0;
        }
    }
  
    // Whatever is left should be the most significant
    // digit.
    res[0] = s + 1;  // The initially subtracted 1 is
                     // incorporated here.
  
    cout << "Smallest number is ";
    for (int i=0; i<m; i++)
        cout << res[i];
}
  
// Driver code
int main()
{
    int s = 9, m = 2;
    findSmallest(m, s);
    return 0;
}

Java

// Java program to find the smallest number that can be
// formed from given sum of digits and number of digits
  
class GFG 
{
    // Function to print the smallest possible number with digit sum 's'
    // and 'm' number of digits
    static void findSmallest(int m, int s)
    {
        // If sum of digits is 0, then a number is possible
        // only if number of digits is 1
        if (s == 0)
        {
            System.out.print(m == 1 ? "Smallest number is 0" : "Not possible");
              
            return ;
        }
   
        // Sum greater than the maximum possible sum
        if (s > 9*m)
        {
            System.out.println("Not possible");
            return ;
        }
   
        // Create an array to store digits of result
        int[] res = new int[m];
   
        // deduct sum by one to account for cases later
        // (There must be 1 left for the most significant
        //  digit)
        s -= 1;
   
        // Fill last m-1 digits (from right to left)
        for (int i=m-1; i>0; i--)
        {
            // If sum is still greater than 9,
            // digit must be 9
            if (s > 9)
            {
                res[i] = 9;
                s -= 9;
            }
            else
            {
                res[i] = s;
                s = 0;
            }
        }
   
        // Whatever is left should be the most significant
        // digit
        res[0] = s + 1// The initially subtracted 1 is
                        // incorporated here
   
        System.out.print("Smallest number is ");
        for (int i=0; i<m; i++)
            System.out.print(res[i]);
    }
      
    // driver program
    public static void main (String[] args) 
    {
        int s = 9, m = 2;
        findSmallest(m, s);
    }
}
  
// Contributed by Pramod Kumar

Python3

# Prints the smallest possible
# number with digit sum 's'
# and 'm' number of digits.
  
def findSmallest(m,s):
  
    # If sum of digits is 0,
    # then a number is possible
    # only if number of digits is 1.
    if (s == 0):
          
        if(m == 1) :
              print("Smallest number is 0"
        else
              print("Not possible")
        return 
   
    # Sum greater than the
    # maximum possible sum.
    if (s > 9*m):
      
        print("Not possible")
        return 
   
    # Create an array to
    # store digits of result
    res=[0 for i in range(m+1)]
   
    # deduct sum by one to
    # account for cases later
    # (There must be 1 left
    # for the most significant
    #  digit)
    s -= 1
   
    # Fill last m-1 digits
    # (from right to left)
    for i in range(m-1,0,-1):
      
        # If sum is still greater than 9,
        # digit must be 9.
        if (s > 9):
          
            res[i] = 9
            s -= 9
      
        else:
          
            res[i] = s
            s = 0
   
    # Whatever is left should
    # be the most significant
    # digit.
    # The initially subtracted 1 is
    # incorporated here.
    res[0] = s + 1 
                     
   
    print("Smallest number is ",end="")
    for i in range(m):
        print(res[i],end="")
  
  
s = 9
m = 2
findSmallest(m, s)
  
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find the smallest 
// number that can be formed from 
// given sum of digits and number 
// of digits
using System;
  
class GFG 
{
    // Function to print the smallest 
    // possible number with digit sum 's'
    // and 'm' number of digits
    static void findSmallest(int m, int s)
    {
        // If sum of digits is 0, 
        // then a number is possible
        // only if number of digits is 1
        if (s == 0)
        {
            Console.Write(m == 1 ? 
                          "Smallest number is 0"
                                  "Not possible");
              
            return ;
        }
  
        // Sum greater than the
        // maximum possible sum
        if (s > 9 * m)
        {
            Console.Write("Not possible");
            return ;
        }
  
        // Create an array to 
        // store digits of result
        int []res = new int[m];
  
        // deduct sum by one to account 
        // for cases later (There must be 
        // 1 left for the most significant
        // digit)
        s -= 1;
  
        // Fill last m-1 digits 
        // (from right to left)
        for (int i = m - 1; i > 0; i--)
        {
            // If sum is still greater 
            // than 9, digit must be 9
            if (s > 9)
            {
                res[i] = 9;
                s -= 9;
            }
            else
            {
                res[i] = s;
                s = 0;
            }
        }
  
        // Whatever is left should be 
        // the most significant digit
          
        // The initially subtracted 1 is
        // incorporated here
        res[0] = s + 1; 
  
        Console.Write("Smallest number is ");
        for (int i = 0; i < m; i++)
            Console.Write(res[i]);
    }
      
    // Driver Code
    public static void Main () 
    {
        int s = 9, m = 2;
        findSmallest(m, s);
    }
}
  
// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to find the smallest
// number that can be formed from 
// given sum of digits and number
// of digits.
  
// Prints the smallest possible 
// number with digit sum 's'
// and 'm' number of digits.
function findSmallest($m, $s)
{
    // If sum of digits is 0, then 
    // a number is possible only if
    // number of digits is 1.
    if ($s == 0)
    {
        if(($m == 1) == true)
          
        echo "Smallest number is " , 0;
        else
        echo "Not possible";
        return ;
    }
  
    // Sum greater than the
    // maximum possible sum.
    if ($s > 9 * $m)
    {
        echo "Not possible";
        return ;
    }
  
    // Create an array to store 
    // digits of result int res[m];
    // deduct sum by one to account 
    // for cases later (There must 
    // be 1 left for the most 
    // significant digit)
    $s -= 1;
  
    // Fill last m-1 digits
    // (from right to left)
    for ($i = $m - 1; $i > 0; $i--)
    {
        // If sum is still greater 
        // than 9, digit must be 9.
        if ($s > 9)
        {
            $res[$i] = 9;
            $s -= 9;
        }
        else
        {
            $res[$i] = $s;
            $s = 0;
        }
    }
  
    // Whatever is left should be 
    // the most significant digit.
      
    // The initially subtracted 1
    // is incorporated here.
    $res[0] = $s + 1; 
                        
    echo "Smallest number is ";
    for ($i = 0; $i < $m; $i++)
        echo $res[$i];
}
  
// Driver code
$s = 9; $m = 2;
findSmallest($m, $s);
  
// This code is contributed by ajit
?>


Output :

Smallest number is 18

Time Complexity of this solution is O(m).

We will soon be discussing approach to find the largest possible number with given sum of digits and number of digits.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



This article is attributed to GeeksforGeeks.org

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