Kth largest N digit number divisible by M

Given three positive integers N, K, and M. The task is to find Kth largest N digit number divisible by M. 

Note: K will be such an integer that Kth largest N digit number divisible by M always exists.

Examples

Input: N = 2, K = 2, M = 2
Output: 96
Explanation: The 2nd largest 2 digit number divisible by 2 is 96. 

Input: N = 9, K = 6, M = 4
Output: 999999976

Approach: The problem is maths-based. Given three numbers N, K, and M. It is required to find the Kth largest N digit number divisible by M. To get the largest N digit divisible by M, at first it is required to find the largest N digit number(say P), which is N times 9. 
Now the largest N digit number divisible by M is (P - (P%M)).
Therefore, subtract (K-1) times M from this value to get the Kth largest value of N digit number which is divisible by M.
Given below is the conditions and mathematical expression to get Kth largest N digit number divisible by M.

Let P be the largest N digit number. 
Then the largest N digit number divisible by M is: (P - (P % M)). 
Now the Kth largest N digit number divisible by M is: [(P - (P % M)) - ((K - 1) * M)]

Below is the code according to the above formula.

// C++ program for above approach
#include <iostream>
using namespace std;

// Function to find Kth N
// digit number divisible by M
int findAnswer(int N, int K, int M)
{
    int i;
    long long int r = 0;

    // Loop to calculate the largest 
    // N digit number.
    for (i = 1; i <= N; i++) {
        r = r * 10 + 9;
    }

    // Kth largest N digit number 
    // divisible by M.
    long long int u = r - (r % M) 
        - M * (K - 1);

    return u;
}

// Driver Code
int main()
{
    int N = 9;
    int K = 6;
    int M = 4;

    cout << findAnswer(N, K, M);
    return 0;
}

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

class GFG{

  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    int i;
    int r = 0;

    // Loop to calculate the largest 
    // N digit number.
    for (i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }

    // Kth largest N digit number 
    // divisible by M.
    int u = r - (r % M) 
      - M * (K - 1);

    return u;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 9;
    int K = 6;
    int M = 4;

    System.out.print(findAnswer(N, K, M));
  }
}

// This code is contributed by 29AjayKumar 

# Python code for the above approach 

# Function to find Kth N
# digit number divisible by M
def findAnswer(N, K, M):
    i = None
    r = 0;

    # Loop to calculate the largest 
    # N digit number.
    for i in range(1, N + 1):
        r = r * 10 + 9;

    # Kth largest N digit number 
    # divisible by M.
    u = r - (r % M) - M * (K - 1);

    return u;

# Driver Code
N = 9;
K = 6;
M = 4;

print(findAnswer(N, K, M));

# This code is contributed by Saurabh Jaiswal

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

  // Function to find Kth N
  // digit number divisible by M
  static int findAnswer(int N, int K, int M)
  {
    long r = 0;

    // Loop to calculate the largest 
    // N digit number.
    for (int i = 1; i <= N; i++) {
      r = r * 10 + 9;
    }

    // Kth largest N digit number 
    // divisible by M.
    long u = r - (r % M) 
      - M * (K - 1);

    return (int)u;
  }

  // Driver Code
  public static void Main()
  {
    int N = 9;
    int K = 6;
    int M = 4;

    Console.Write(findAnswer(N, K, M));
  }
}

// This code is contributed by Samim Hossain Mondal.

    <script>
        // JavaScript code for the above approach 

        // Function to find Kth N
        // digit number divisible by M
        function findAnswer(N, K, M) {
            let i;
            let r = 0;

            // Loop to calculate the largest 
            // N digit number.
            for (i = 1; i <= N; i++) {
                r = r * 10 + 9;
            }

            // Kth largest N digit number 
            // divisible by M.
            let u = r - (r % M)
                - M * (K - 1);

            return u;
        }

        // Driver Code
        let N = 9;
        let K = 6;
        let M = 4;

        document.write(findAnswer(N, K, M));

         // This code is contributed by Potta Lokesh
    </script>

999999976

Time Complexity: O(MaxDigit), Where maxDigit is the largest N digit number.
Auxiliary Space: O(1)