Solution: Largest Number with Given Number of Digits and Sum of Digits

int findSumOfDigits(int num) {
  int sum = 0;
  while (num != 0) {
    sum = sum + num % 10;
    num = num / 10;
  }
  return sum;
}
void findLargestNumber(int numberOfDigits, int sumOfDigits) {
  int max = 0;
  int startRange = pow(10, (numberOfDigits - 1));
  int endRange = pow(10, numberOfDigits);
  if (sumOfDigits == 0) {
    if (numberOfDigits == 1)
      cout << "Largest number is " << 0;
    else
      cout << "Largest number is Not possible";
    return;
  }
  // sumOfDigits is greater than the maximum possible sum. 
  if (sumOfDigits > 9 * numberOfDigits) {
    cout << "Largest number is Not possible";
    return;
  }
  while (startRange < endRange) {
    if (findSumOfDigits(startRange) == sumOfDigits) {
      if (max < startRange)
        max = startRange;
    }
    startRange++;
  }
  cout << "Largest number is " << max;
}
// Driver code 
int main() {
  int sumOfDigits = 20;
  int numberOfDigits = 3;
  cout << "If sum of digits is 20 and number of digits is 3 then ";
  findLargestNumber(numberOfDigits, sumOfDigits);
  cout << endl << endl;
  //Example 2
  sumOfDigits = 100;
  numberOfDigits = 2;
  cout << "If sum of digits is 100 and number of digits is 2 then ";
  findLargestNumber(numberOfDigits, sumOfDigits);
  return 0;
}