在通过从自然数中删除数字 K 形成的序列中找到第 N 项

给定整数N、K和一个无限的自然数序列，其中所有包含数字K (1<=K<=9) 的数字都被删除。任务是返回此序列的第 N 个数字。

例子：

Input: N = 12, K = 2
Output: 14
Explanation: The sequence generated for the above input would be like this: 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, up to infinity

Input: N = 10, K = 1
Output: 22

朴素方法：解决上述问题的基本方法是迭代到 N 并排除所有小于 N 的包含给定数字 K 的数字。最后，打印获得的第 N 个自然数。
时间复杂度：O(N)
辅助空间：O(1)

有效的方法：解决这个问题的有效方法是从第 N 个自然数中得到的灵感，在删除所有由数字 9 组成的数字之后。
如果 K 的值大于 8，则可以通过将K的值转换为以 9 为底的形式来解决给定的问题。可以遵循以下步骤：

• 将第 N 个自然数计算为以 9 为底的格式
• 以 9 为基数的大于等于 K 的每一位都加 1
• 下一个数字是想要的答案

以下是上述方法的代码：

C++

// C++ implementation for the above approach
 
#include 
using namespace std;
long long convertToBase9(long long n)
{
    long long ans = 0;
 
    // Denotes the digit place
    long long a = 1;
 
    // Method to convert any number
    // to binary equivalent
    while (n > 0) {
        ans += (a * (n % 9));
        a *= 10;
        n /= 9;
    }
    return ans;
}
 
long long getNthnumber(long long base9,
                       long long K)
{
    long long ans = 0;
 
    // denotes the current digits place
    long long a = 1;
    while (base9 > 0) {
        int cur = base9 % 10;
 
        // If current digit is >= K
        // increment its value by 1
        if (cur >= K) {
            ans += a * (cur + 1);
        }
 
        // Else add the digit as it is
        else {
            ans += a * cur;
        }
        base9 /= 10;
 
        // Move to the next digit
        a *= 10;
    }
    return ans;
}
 
// Driver code
int main()
{
    long long N = 10, K = 1;
    long long base9 = convertToBase9(N);
    cout << getNthnumber(base9, K);
    return 0;
}

Java

// Java implementation for the above approach
import java.io.*;
 
class GFG {
   
      static long convertToBase9(long n)
    {
        long ans = 0;
 
        // Denotes the digit place
        long a = 1;
 
        // Method to convert any number
        // to binary equivalent
        while (n > 0) {
            ans += (a * (n % 9));
            a *= 10;
            n /= 9;
        }
        return ans;
    }
 
    static long getNthnumber(long base9,
                          long K)
    {
        long ans = 0;
 
        // denotes the current digits place
        long a = 1;
        while (base9 > 0) {
            int cur = (int)(base9 % 10);
 
            // If current digit is >= K
            // increment its value by 1
            if (cur >= K) {
                ans += a * (cur + 1);
            }
 
            // Else add the digit as it is
            else {
                ans += a * cur;
            }
            base9 /= 10;
 
            // Move to the next digit
            a *= 10;
        }
        return ans;
    }
 
    // Driver code
    public static void main (String[] args) {
        long N = 10, K = 1;
        long base9 = convertToBase9(N);
        System.out.println(getNthnumber(base9, K));
    }
}
 
//  This code is contributed by Dharanendra L V.

Python3

# Python 3 implementation for the above approach
def convertToBase9(n):
    ans = 0
     
    # Denotes the digit place
    a = 1
 
    # Method to convert any number
    # to binary equivalent
    while(n > 0):
        ans += (a * (n % 9))
        a *= 10
        n //= 9
    return ans
 
def getNthnumber(base9, K):
    ans = 0
 
    # denotes the current digits place
    a = 1
    while (base9 > 0):
        cur = base9 % 10
 
        # If current digit is >= K
        # increment its value by 1
        if (cur >= K):
            ans += a * (cur + 1)
 
        # Else add the digit as it is
        else:
            ans += a * cur
        base9 //= 10
 
        # Move to the next digit
        a *= 10
    return ans
 
# Driver code
if __name__ == '__main__':
    N = 10
    K = 1
    base9 = convertToBase9(N)
    print(getNthnumber(base9, K))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# implementation for the above approach
using System;
 
class GFG {
   
      static long convertToBase9(long n)
    {
        long ans = 0;
 
        // Denotes the digit place
        long a = 1;
 
        // Method to convert any number
        // to binary equivalent
        while (n > 0) {
            ans += (a * (n % 9));
            a *= 10;
            n /= 9;
        }
        return ans;
    }
 
    static long getNthnumber(long base9,
                          long K)
    {
        long ans = 0;
 
        // denotes the current digits place
        long a = 1;
        while (base9 > 0) {
            int cur = (int)(base9 % 10);
 
            // If current digit is >= K
            // increment its value by 1
            if (cur >= K) {
                ans += a * (cur + 1);
            }
 
            // Else add the digit as it is
            else {
                ans += a * cur;
            }
            base9 /= 10;
 
            // Move to the next digit
            a *= 10;
        }
        return ans;
    }
 
    // Driver code
    public static void Main (String[] args) {
        long N = 10, K = 1;
        long base9 = convertToBase9(N);
        Console.Write(getNthnumber(base9, K));
    }
}
 
//  This code is contributed by shivanisinghss2110

Javascript

// Javascript implementation for the above approach
function convertToBase9(n)
{
  let ans = 0;
 
  // Denotes the digit place
  let a = 1;
 
  // Method to convert any number
  // to binary equivalent
  while (n > 0) {
    ans += a * (n % 9);
    a *= 10;
    n = Math.floor(n / 9);
  }
  return ans;
}
 
function getNthnumber(base9, K) {
  let ans = 0;
 
  // denotes the current digits place
  let a = 1;
  while (base9 > 0) {
    let cur = base9 % 10;
 
    // If current digit is >= K
    // increment its value by 1
    if (cur >= K) {
      ans += a * (cur + 1);
    }
 
    // Else add the digit as it is
    else {
      ans += a * cur;
    }
    base9 = Math.floor(base9 / 10);
 
    // Move to the next digit
    a *= 10;
  }
  return ans;
}
 
// Driver code
let N = 10,
  K = 1;
let base9 = convertToBase9(N);
document.write(getNthnumber(base9, K));
 
// This code is contributed by gfgking.

输出：

22

时间复杂度： O(log ₉ N)
辅助空间： O(1)