Count of N digit Numbers whose sum of every K consecutive digits is equal | Set 2

Given two integers N and K, the task is to find the count of all possible N-digit numbers having sum of every K consecutive digits of the number are equal.
Examples: 
 

Input: N = 2, K=1 
Output: 9 
Explanation: 
All two digit numbers satisfying the required conditions are {11, 22, 33, 44, 55, 66, 77, 88, 99}
Input: N = 3, K = 2 
Output: 90 
 

Naive and Sliding Window Approach: Refer to Count of N digit Numbers whose sum of every K consecutive digits is equal for the simplest approach and Sliding Window technique based approach.
Logarithmic Approach: 
For the sum of K-consecutive elements to be always equal, the entire number is governed by its first K digits. 
 

• The i^th digit will be equal to the (i-k)^th digit for the number to satisfy the condition such that the sum of every K consecutive digit is the same.

Illustration: 
N = 5 and K = 2 
If the first two digits are 1 and 2, then the number has to be 12121 so that sum of every 2 consecutive digits is 3. 
Observe that the first 2 digits i.e. the first K digits repeats itself. 
 

Therefore, to solve the problem, the task is now to find out the total count of K-digit numbers which is equal to 10^K - 10^(K-1). Therefore, print the calculated value as the answer.
Below is the implementation of the above approach:
 

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// N-digit numbers such that sum of
// every K consecutive digits are equal
void count(int n, int k)
{
    long count = (long)(pow(10, k) - pow(10, k - 1));

    // Print the answer
    cout << (count);
}

// Driver Code
int main()
{
    int n = 2, k = 1;
    count(n, k);
}

// This code is contributed by Ritik Bansal

// Java Program to implement
// the above approach
class GFG {

    // Function to count the number of
    // N-digit numbers such that sum of
    // every K consecutive digits are equal
    public static void count(int n, int k)
    {
        long count
            = (long)(Math.pow(10, k)
                     - Math.pow(10, k - 1));

        // Print the answer
        System.out.print(count);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 2, k = 1;
        count(n, k);
    }
}

# Python3 program to implement
# the above approach

# Function to count the number of
# N-digit numbers such that sum of
# every K consecutive digits are equal
def count(n, k):
    
    count = (pow(10, k) - pow(10, k - 1));

    # Print the answer
    print(count);
    
# Driver Code
if __name__ == '__main__':
    
    n = 2;
    k = 1;
    
    count(n, k);
    
# This code is contributed by 29AjayKumar 

// C# Program to implement
// the above approach
using System;
class GFG{
  
  // Function to count the number of
  // N-digit numbers such that sum of
  // every K consecutive digits are equal
  public static void count(int n, int k)
  {
    long count = (long)(Math.Pow(10, k) - 
                        Math.Pow(10, k - 1));

    // Print the answer
    Console.Write(count);
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int n = 2, k = 1;
    count(n, k);
  }
}

// This code is contributed by Rohit_ranjan 

<script>
// Javascript Program to implement
// the above approach

// Function to count the number of
// N-digit numbers such that sum of
// every K consecutive digits are equal
function count(n, k)
{
    let count = Math.pow(10, k) - Math.pow(10, k - 1);

    // Print the answer
    document.write(count);
}

// Driver Code
let n = 2, k = 1;
count(n, k);

// This code is contributed by souravmahato348.
</script>

9

Time Complexity: O(log K) 
Auxiliary Space: O(1)