# 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

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

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.

This article is contributed by Vaibhav Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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