Smallest positive number made up of non-repeating digits whose sum of digits is N

Given a positive integer N, the task is to find the smallest positive number made up of distinct digits having sum of its digits equal to N. If no such number exists, print “-1”.

Examples:

Input: N = 11
Output: 29
Explanation: The sum of the digits = 2 + 9 = 11 ( = N).

Input: N = 46
Output: -1

Approach: The idea is based on the following observations:

• If N < 10: Required answer is N itself.
• If N > 45: Required answer is -1 as the smallest number that can be made using non-repeating digits is 123456789, whose sum of digits is 45.
• Otherwise: The answer can be obtained based on the following pattern.

Consider the following examples, to understand the pattern,

For N = 10: The output is 19.
For N = 11: The output is 29.
For N = 12: The output is 39.
For N = 13: The output is 49.
…
For N = 21: The output is 489.
For N = 22: The output is 589.
For N = 23: The output is 689.

On observing the above results, it is clear that the answer contains digits in decreasing order having a difference of 1 starting from one’s digit until the digit sum is less than N.

Follow the steps below to solve the problem:

• If the given number is less than 10, then print the number itself.
• If the given number is greater than 45, then print “-1”.
• Otherwise, perform the following steps:
  • Initialize a string res to store the required answer and initialize a variable, say digit, as 9.
  • Iterate a loop until N ? digit and perform the following steps:
    • Push digit at the starting of res.
    • Decrement N by digit.
    • Decrement digit by 1.
  • If the value of N is found to be greater than 0, then push the character at the starting of res.
  • After completing the above steps, print the value of res as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find smallest positive
// number made up of non-repeating digits
// whose sum of its digits is equal to n
void result(int n)
{
    if (n > 45) {

        // No such number exists
        cout << -1;
        return;
    }

    // Stores the required answer
    string res;

    // Store the digit at unit's place
    int digit = 9;

    // Iterate until n > digit
    while (n > digit) {

        // Push digit at the start of res
        res = char('0' + digit) + res;

        // Decrement n by digit
        n -= digit;

        // Decrement digit by 1
        digit -= 1;
    }

    if (n > 0) {

        // Push the remaining number
        // as the starting digit
        res = char('0' + n) + res;
    }

    // Print the required number
    cout << res;
}

// Driver Code
int main()
{
    int N = 19;

    // Function Call
    result(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG
{

// Function to find smallest positive
// number made up of non-repeating digits
// whose sum of its digits is equal to n
static void result(int n)
{
    if (n > 45)
    {

        // No such number exists
        System.out.print(-1);
        return;
    }

    // Stores the required answer
    String res="";

    // Store the digit at unit's place
    int digit = 9;

    // Iterate until n > digit
    while (n > digit)
    {

        // Push digit at the start of res
        res = (char)('0' + digit) + res;

        // Decrement n by digit
        n -= digit;

        // Decrement digit by 1
        digit -= 1;
    }

    if (n > 0)
    {

        // Push the remaining number
        // as the starting digit
        res = (char)('0' + n) + res;
    }

    // Print the required number
    System.out.print(res);
}

// Driver Code
public static void main(String[] args)
{
    int N = 19;

    // Function Call
    result(N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find smallest positive
# number made up of non-repeating digits
# whose sum of its digits is equal to n
def result(n):

    if (n > 45):

        # No such number exists
        print(-1, end = "")
        return

    # Stores the required answer
    res = ""

    # Store the digit at unit's place
    digit = 9

    # Iterate until n > digit
    while (n > digit):

        # Push digit at the start of res
        res = str(digit) + res

        # Decrement n by digit
        n -= digit

        # Decrement digit by 1
        digit -= 1

    if (n > 0):

        # Push the remaining number
        # as the starting digit
        res = str(n) + res

    # Print the required number
    print(res)

# Driver Code
if __name__ == '__main__':

    N = 19

    # Function Call
    result(N)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
class GFG
{

// Function to find smallest positive
// number made up of non-repeating digits
// whose sum of its digits is equal to n
static void result(int n)
{
    if (n > 45)
    {

        // No such number exists
        Console.Write(-1);
        return;
    }

    // Stores the required answer
    string res = "";

    // Store the digit at unit's place
    int digit = 9;

    // Iterate until n > digit
    while (n > digit)
    {

        // Push digit at the start of res
        res = (char)('0' + digit) + res;

        // Decrement n by digit
        n -= digit;

        // Decrement digit by 1
        digit -= 1;
    }

    if (n > 0)
    {

        // Push the remaining number
        // as the starting digit
        res = (char)('0' + n) + res;
    }

    // Print the required number
    Console.Write(res);
}

// Driver Code
public static void Main()
{
    int N = 19;

    // Function Call
    result(N);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript program for the above approach

// Function to find smallest positive
// number made up of non-repeating digits
// whose sum of its digits is equal to n
function result(n)
{
    if (n > 45) {

        // No such number exists
        document.write(-1);
    }

    // Stores the required answer
    var res = "";

    // Store the digit at unit's place
    var digit = 9;

    // Iterate until n > digit
    while (n > digit) {

        // Push digit at the start of res
        res = String.fromCharCode(48 + digit) + res;

        // Decrement n by digit
        n -= digit;

        // Decrement digit by 1
        digit -= 1;
    }

    if (n > 0) {

        // Push the remaining number
        // as the starting digit
        res = String.fromCharCode(48 + n) + res;
    }

    // Print the required number
    document.write(res);
}

// Driver Code
var N = 19;

// Function Call
result(N);

</script>

Output:

289

Time Complexity: O(N)
Auxiliary Space: O(N)