// A C++ recursive program to print all n-digit // numbers whose sum of digits equals to given sum #include    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] = '';  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 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] = '' ;  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 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 outindex): # 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] = '' 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# 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] = '';  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. 

<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] = '';  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(nsum)  {  // 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> 

 // 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 ?>