Print all n-digit numbers whose sum of digits equals to given sum

Last Updated : 07 Jan, 2024

Given number of digits n, print all n-digit numbers whose sum of digits adds upto given sum. Solution should not consider leading 0’s as digits.
Examples:

Input:  N = 2, Sum = 3
Output:  12 21 30

Input:  N = 3, Sum = 6
Output:  105 114 123 132 141 150 204 
         213 222 231 240 303 312 321 
         330 402 411 420 501 510 600 

Input:  N = 4, Sum = 3
Output:  1002 1011 1020 1101 1110 1200
         2001 2010 2100 3000

A simple solution would be to generate all N-digit numbers and print numbers that have sum of their digits equal to given sum. The complexity of this solution would be exponential.
A better solution is to generate only those N-digit numbers that satisfy the given constraints. The idea is to use recursion. We basically fill all digits from 0 to 9 into current position and maintain sum of digits so far. We then recurse for remaining sum and number of digits left. We handle leading 0’s separately as they are not counted as digits.
Below is a simple recursive implementation of above idea –

C++

// A C++ recursive program to print all n-digit
// numbers whose sum of digits equals to given sum
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to print all n-digit numbers
// whose sum of digits equals to given sum
 
// n, sum --> value of inputs
// out --> output array
// index --> index of next digit to be filled in
//           output array
void findNDigitNumsUtil(int n, int sum, char* out,
                        int index)
{
    // Base case
    if (index > n || sum < 0)
        return;
 
    // If number becomes N-digit
    if (index == n)
    {
        // if sum of its digits is equal to given sum,
        // print it
        if(sum == 0)
        {
            out[index] = '\0';
            cout << out << " ";
        }
        return;
    }
 
    // Traverse through every digit. Note that
    // here we're considering leading 0's as digits
    for (int i = 0; i <= 9; i++)
    {
        // append current digit to number
        out[index] = i + '0';
 
        // recurse for next digit with reduced sum
        findNDigitNumsUtil(n, sum - i, out, index + 1);
    }
}
 
// This is mainly a wrapper over findNDigitNumsUtil.
// It explicitly handles leading digit
void findNDigitNums(int n, int sum)
{
    // output array to store N-digit numbers
    char out[n + 1];
 
    // fill 1st position by every digit from 1 to 9 and
    // calls findNDigitNumsUtil() for remaining positions
    for (int i = 1; i <= 9; i++)
    {
        out[0] = i + '0';
        findNDigitNumsUtil(n, sum - i, out, 1);
    }
}
 
// Driver program
int main()
{
    int n = 2, sum = 3;
 
    findNDigitNums(n, sum);
 
    return 0;
}

Java

// Java recursive program to print all n-digit
// numbers whose sum of digits equals to given sum
import java.io.*;
 
class GFG 
{
    // Recursive function to print all n-digit numbers
    // whose sum of digits equals to given sum
  
    // n, sum --> value of inputs
    // out --> output array
    // index --> index of next digit to be 
    // filled in output array
    static void findNDigitNumsUtil(int n, int sum, char out[],
                                   int index)
    {
        // Base case
        if (index > n || sum < 0)
            return;
  
        // If number becomes N-digit
        if (index == n)
        {
            // if sum of its digits is equal to given sum,
            // print it
            if(sum == 0)
            {
                out[index] = '\0'   ;
                System.out.print(out);
                System.out.print(" ");
            }
            return;
        }
  
        // Traverse through every digit. Note that
        // here we're considering leading 0's as digits
        for (int i = 0; i <= 9; i++)
        {
            // append current digit to number
            out[index] = (char)(i + '0');
  
            // recurse for next digit with reduced sum
            findNDigitNumsUtil(n, sum - i, out, index + 1);
        }
    }
     
    // This is mainly a wrapper over findNDigitNumsUtil.
    // It explicitly handles leading digit
    static void findNDigitNums(int n, int sum)
    {
        // output array to store N-digit numbers
        char[] out = new char[n + 1];
  
        // fill 1st position by every digit from 1 to 9 and
        // calls findNDigitNumsUtil() for remaining positions
        for (int i = 1; i <= 9; i++)
        {
            out[0] = (char)(i + '0');
            findNDigitNumsUtil(n, sum - i, out, 1);
        }
    }
     
    // driver program to test above function
    public static void main (String[] args) 
    {
             int n = 2, sum = 3;
             findNDigitNums(n, sum);
    }
}
 
// This code is contributed by Pramod Kumar

Python 3

# Python 3 recursive program to print 
# all n-digit numbers whose sum of 
# digits equals to given sum
 
# Recursive function to print all 
# n-digit numbers whose sum of 
# digits equals to given sum
 
# n, sum --> value of inputs
# out --> output array
# index --> index of next digit to be 
#            filled in output array
def findNDigitNumsUtil(n, sum, out,index):
 
    # Base case
    if (index > n or sum < 0):
        return
 
    f = ""
     
    # If number becomes N-digit
    if (index == n):
     
        # if sum of its digits is equal
        # to given sum, print it
        if(sum == 0):
            out[index] = "\0"
             
            for i in out:
                f = f + i
            print(f, end = " ")
         
        return
 
    # Traverse through every digit. Note 
    # that here we're considering leading
    # 0's as digits
    for i in range(10):
         
        # append current digit to number
        out[index] = chr(i + ord('0'))
 
        # recurse for next digit with reduced sum
        findNDigitNumsUtil(n, sum - i, 
                           out, index + 1)
 
# This is mainly a wrapper over findNDigitNumsUtil.
# It explicitly handles leading digit
def findNDigitNums( n, sum):
 
    # output array to store N-digit numbers
    out = [False] * (n + 1)
 
    # fill 1st position by every digit 
    # from 1 to 9 and calls findNDigitNumsUtil() 
    # for remaining positions
    for i in range(1, 10):
        out[0] = chr(i + ord('0'))
        findNDigitNumsUtil(n, sum - i, out, 1)
 
# Driver Code
if __name__ == "__main__":
    n = 2
    sum = 3
 
    findNDigitNums(n, sum)
 
# This code is contributed 
# by ChitraNayal

C#

// C# recursive program to print all n-digit
// numbers whose sum of digits equals to
// given sum
using System;
 
class GFG {
     
    // Recursive function to print all n-digit
    // numbers whose sum of digits equals to
    // given sum
 
    // n, sum --> value of inputs
    // out --> output array
    // index --> index of next digit to be 
    // filled in output array
    static void findNDigitNumsUtil(int n, int sum,
                             char []ou, int index)
    {
        // Base case
        if (index > n || sum < 0)
            return;
 
        // If number becomes N-digit
        if (index == n)
        {
            // if sum of its digits is equal to
            // given sum, print it
            if(sum == 0)
            {
                ou[index] = '\0';
                Console.Write(ou);
                Console.Write(" ");
            }
             
            return;
        }
 
        // Traverse through every digit. Note
        // that here we're considering leading
        // 0's as digits
        for (int i = 0; i <= 9; i++)
        {
            // append current digit to number
            ou[index] = (char)(i + '0');
 
            // recurse for next digit with
            // reduced sum
            findNDigitNumsUtil(n, sum - i, ou,
                                    index + 1);
         
        }
    }
 
    // This is mainly a wrapper over 
    // findNDigitNumsUtil. It explicitly
    // handles leading digit
    static void findNDigitNums(int n, int sum)
    {
         
        // output array to store N-digit
        // numbers
        char []ou = new char[n + 1];
 
        // fill 1st position by every digit
        // from 1 to 9 and calls 
        // findNDigitNumsUtil() for remaining 
        // positions
        for (int i = 1; i <= 9; i++)
        {
            ou[0] = (char)(i + '0');
            findNDigitNumsUtil(n, sum - i, ou, 1);
        }
    }
     
    // driver program to test above function
    public static void Main () 
    {
            int n = 2, sum = 3;
             
            findNDigitNums(n, sum);
    }
}
 
// This code is contributed by nitin mittal.

Javascript

<script>
 
// Javascript recursive program to print all n-digit
// numbers whose sum of digits equals to given sum
     
    // Recursive function to print all n-digit numbers
    // whose sum of digits equals to given sum
    
    // n, sum --> value of inputs
    // out --> output array
    // index --> index of next digit to be 
    // filled in output array
    function findNDigitNumsUtil(n, sum, out, index)
    {
     
        // Base case
        if (index > n || sum < 0)
            return;
    
        // If number becomes N-digit
        if (index == n)
        {
         
            // if sum of its digits is equal to given sum,
            // print it
            if(sum == 0)
            {
                out[index] = '\0';
                for(let i = 0; i < out.length; i++)
                 
                    document.write(out[i]);
                document.write(" ");
            }
            return;
        }
    
        // Traverse through every digit. Note that
        // here we're considering leading 0's as digits
        for (let i = 0; i <= 9; i++)
        {
            // append current digit to number
            out[index] = String.fromCharCode(i + '0'.charCodeAt(0));
    
            // recurse for next digit with reduced sum
            findNDigitNumsUtil(n, sum - i, out, index + 1);
        }
    }
     
    // This is mainly a wrapper over findNDigitNumsUtil.
    // It explicitly handles leading digit
    function findNDigitNums(n,sum)
    {
        // output array to store N-digit numbers
        let out = new Array(n+1);
        for(let i=0;i<n+1;i++)
        {
            out[i]=false;
        }
        // fill 1st position by every digit from 1 to 9 and
        // calls findNDigitNumsUtil() for remaining positions
        for (let i = 1; i <= 9; i++)
        {
            out[0] = String.fromCharCode(i + '0'.charCodeAt(0));
            findNDigitNumsUtil(n, sum - i, out, 1);
        }
    }
     
    // driver program to test above function
    let  n = 2, sum = 3;
    findNDigitNums(n, sum);
     
    // This code is contributed by avanitrachhadiya2155
</script>

PHP

<?php
// A PHP recursive program to print all 
// n-digit numbers whose sum of digits 
// equals to given sum
 
// Recursive function to print all n-digit
// numbers whose sum of digits equals to 
// given sum
 
// n, sum --> value of inputs
// out --> output array
// index --> index of next digit to be 
//             filled in output array
function findNDigitNumsUtil($n, $sum, $out,
                                    $index)
{
    // Base case
    if ($index > $n || $sum < 0)
        return;
 
    // If number becomes N-digit
    if ($index == $n)
    {
        // if sum of its digits is equal 
        // to given sum, print it
        if($sum == 0)
        {
            $out[$index] = '';
            foreach ($out as &$value)
            print($value);
            print(" ");
        }
        return;
    }
 
    // Traverse through every digit. Note 
    // that here we're considering leading
    // 0's as digits
    for ($i = 0; $i <= 9; $i++)
    {
        // append current digit to number
        $out[$index] = chr($i + ord('0'));
 
        // recurse for next digit with 
        // reduced sum
        findNDigitNumsUtil($n, $sum - $i, 
                           $out, $index + 1);
    }
}
 
// This is mainly a wrapper over findNDigitNumsUtil.
// It explicitly handles leading digit
function findNDigitNums($n, $sum)
{
    // output array to store N-digit numbers
    $out = array_fill(0, $n + 1, false);
 
    // fill 1st position by every digit from 
    // 1 to 9 and calls findNDigitNumsUtil() 
    // for remaining positions
    for ($i = 1; $i <= 9; $i++)
    {
        $out[0] = chr($i + ord('0'));
        findNDigitNumsUtil($n, $sum - $i, $out, 1);
    }
}
 
// Driver Code
$n = 2;
$sum = 3;
 
findNDigitNums($n, $sum);
 
// This code is contributed 
// by chandan_jnu
?>

Output:

12 21 30

Time Complexity: O(n*n!)

Auxiliary Space: O(n)

Smallest number whose product with N has sum of digits equal to that of N

Aditya Goel

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Article Tags :

Practice Tags :

Algorithms

Print all n-digit numbers whose sum of digits equals to given sum

C++

Java

Python 3

C#

Javascript

PHP

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