由数字和为 N 的非重复数字组成的最小正数

给定一个正整数N ，任务是找到由不同数字组成的最小正数，其数字之和等于N 。如果不存在这样的数字，则打印“-1” 。

例子：

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

Input: N = 46
Output: -1

方法：这个想法基于以下观察：

• 如果 N < 10：要求的答案是N本身。
• 如果 N > 45：要求答案为-1 ，因为可以使用非重复数字组成的最小数字为 123456789，其数字总和为 45。
• 否则：可以根据以下模式获得答案。

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.

请按照以下步骤解决问题：

• 如果给定的数字小于 10，则打印数字本身。
• 如果给定的数字大于 45，则打印“-1” 。
• 否则，请执行以下步骤：
  • 初始化一个字符串res以存储所需的答案并初始化一个变量，比如digit ，为9 。
  • 迭代一个循环直到N ≤ digit并执行以下步骤：
    • 在res开始处按数字。
    • 按数字递减N。
    • 将数字减1 。
  • 如果发现N的值大于0 ，则压入res开头的字符。
  • 完成以上步骤后，打印res的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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

 
// 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);
 

输出：

289

时间复杂度： O(N)
辅助空间： O(N)