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 $\in [1, 1000000000]$

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 $10^9$ .

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 <cmath> 
#include <cstdlib> 
#include <iostream> 
#include <vector> 
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

<?php 
// PHP program to find x such that  
// X + sumOfDigits(X) = N  
  
// Computing the sum of digits of x  
function 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.  
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 $O(100*len)$ where len is the number of digits in the number max(len) = 9. Thus the complexity can almost be said to be $O(len)$

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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