# Find a Number X whose sum with its digits is equal to N

Given a positive number N. We need to find number(s) such that sum of digits of those numbers to themselves is equal to N. If no such number is possible print -1. Here N

Examples:

Input : N = 21
Output : X = 15
Explanation : X + its digit sum
= 15 + 1 + 5
= 21

Input  : N = 5
Output : -1

Input : N = 100000001
Output : X = 99999937
X = 100000000


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 : (Naive Approach)
We have already discussed the approach here. The approach might not work for N as large as .

Method 2 : (Efficient)
It is a fact that for a number X < = 1000000000, the sum of digits never exceeds 100. Using this piece of information, we can iterate over all possibilities in the range 0 to 100 on both the sides of the number and check if the number X is eqaul to N – sum of digits of X. All the possibilities will be covered in this range.

## C++

 // CPP program to find x such that  // X + sumOfDigits(X) = N  #include  #include  #include  #include  using namespace std;     // Computing the sum of digits of x  int sumOfDigits(long int x)  {      int sum = 0;      while (x > 0) {          sum += x % 10;          x /= 10;      }      return sum;  }     // Checks for 100 numbers on both left  // and right side of the given number to  // find such numbers X such that X +   // sumOfDigits(X) = N and updates the answer  // vector accordingly  void compute(vector<long int>& answer, long int n)  {      // Checking for all possibilities of       // the answer      for (int i = 0; i <= 100; i++) {             // Evaluating the value on the left           // side of the given number          long int valueOnLeft = abs(n - i) +                        sumOfDigits(abs(n - i));             // Evaluating the value on the right          // side of the given number          long int valueOnRight = n + i + sumOfDigits(n + i);             // Checking the condition of equality           // on both sides of the given number N           // and updating the answer vector          if (valueOnLeft == n)              answer.push_back(abs(n - i));          if (valueOnRight == n)              answer.push_back(n + i);      }  }     // Driver Function  int main()  {          long int N = 100000001;         vector<long int> answer;      compute(answer, N);         // If no solution exists, print -1      if (answer.size() == 0)          cout << -1;      else {             // If one or more solutions are possible,          // printing them!          for (auto it = answer.begin(); it != answer.end(); ++it)              cout << "X = " << (*it) << endl;      }      return 0;  }

## Java

 // Java program to find x such that  // X + sumOfDigits(X) = N  import java.util.*;  import java.lang.*;  import java.io.*;     class GeeksforGeeks {         // Computing the sum of digits of x      static int sumOfDigits(long x)      {          int sum = 0;          while (x > 0) {              sum += (x % 10);              x /= 10;          }          return sum;      }         // Checks for 100 numbers on both left       // and right side of the given number to       // find such numbers X such that      // X + sumOfDigits(X) = N and prints solution.      static void compute(long n)      {          long answer[] = new long[100];          int pos = 0;             // Checking for all possibilities of the answer          // in the given range          for (int i = 0; i <= 100; i++) {                 // Evaluating the value on the left side of the              // given number              long valueOnLeft = Math.abs(n - i) +                                  sumOfDigits(Math.abs(n - i));                 // Evaluating the value on the right side of the              // given number              long valueOnRight = (n + i) + sumOfDigits(n + i);                 if (valueOnRight == n)                  answer[pos++] = (n + i);              if (valueOnLeft == n)                  answer[pos++] = Math.abs(n - i);          }             if (pos == 0)              System.out.print(-1);          else             for (int i = 0; i < pos; i++)                  System.out.println("X = " + answer[i]);      }      // Driver Function      public static void main(String[] args)      {          long N = 100000001;          compute(N);      }  }

## Python3

 # Python3 program to find x such that   # X + sumOfDigits(X) = N      # Computing the sum of digits of x   def sumOfDigits(x):          sum = 0;       while (x > 0):          sum += (x % 10);           x = int(x / 10);       return sum;      # Checks for 100 numbers on both left   # and right side of the given number   # to find such numbers X such that   # X + sumOfDigits(X) = N and prints   # solution.   def compute(n):          answer = [];       pos = 0;          # Checking for all possibilities       # of the answer in the given range       for i in range(101):             # Evaluating the value on the           # left side of the given number           valueOnLeft = (abs(n - i) +                         sumOfDigits(abs(n - i)));              # Evaluating the value on the right           # side of the given number           valueOnRight = (n + i) + sumOfDigits(n + i);              if (valueOnRight == n):               answer.append(n + i);           if (valueOnLeft == n):               answer.append(abs(n - i));          if (len(answer)== 0):           print(-1);       else:          for i in range(len(answer)):               print("X =", answer[i]);                 # Driver Code   N = 100000001;   compute(N);      # This code is contributed   # by mits

## C#

 // C# program to find x such that  // X + sumOfDigits(X) = N     using System;     public class GFG{      // Computing the sum of digits of x      static int sumOfDigits(long x)      {          int sum = 0;          while (x > 0) {              sum += (int)(x % 10);              x /= 10;          }          return sum;      }         // Checks for 100 numbers on both left       // and right side of the given number to       // find such numbers X such that      // X + sumOfDigits(X) = N and prints solution.      static void compute(long n)      {          long []answer = new long[100];          int pos = 0;             // Checking for all possibilities of the answer          // in the given range          for (int i = 0; i <= 100; i++) {                 // Evaluating the value on the left side of the              // given number              long valueOnLeft = Math.Abs(n - i) +                               sumOfDigits(Math.Abs(n - i));                 // Evaluating the value on the right side of the              // given number              long valueOnRight = (n + i) + sumOfDigits(n + i);                 if (valueOnRight == n)                  answer[pos++] = (n + i);              if (valueOnLeft == n)                  answer[pos++] = Math.Abs(n - i);          }             if (pos == 0)              Console.Write(-1);          else             for (int i = 0; i < pos; i++)                  Console.WriteLine("X = " + answer[i]);      }      // Driver Function                    static public void Main (){          long N = 100000001;          compute(N);      }  }

## PHP

  0)       {           $sum += ($x % 10);           $x = (int)$x / 10;       }       return $sum;  }    // Checks for 100 numbers on both left  // and right side of the given number  // to find such numbers X such that  // X + sumOfDigits(X) = N and prints  // solution.  function compute($n)   {       $answer = array(0);   $pos = 0;          // Checking for all possibilities       // of the answer in the given range       for ($i = 0; $i <= 100; $i++)   {     // Evaluating the value on the   // left side of the given number   $valueOnLeft = abs($n - $i) +                           sumOfDigits(abs($n - $i));              // Evaluating the value on the right           // side of the given number           $valueOnRight = ($n + $i) + sumOfDigits($n + $i);     if ($valueOnRight == $n)   $answer[$pos++] = ($n + $i);   if ($valueOnLeft == $n)   $answer[$pos++] =abs($n - $i);   }     if ($pos == 0)           echo (-1),"\n";       else         for ($i = 0; $i < $pos; $i++)               echo "X = ", $answer[$i], "\n";                 }      // Driver Code   $N = 100000001;  compute($N);      // This code is contributed   // by Sach_Code  ?>

Output:

X = 100000000
X = 99999937


The maximum complexity of this approach can be where len is the number of digits in the number max(len) = 9. Thus the complexity can almost be said to be

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Waba Laba Dub Dub

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