# 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:
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 ith 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 10K – 10(K-1). Therefore, print the calculated value as the answer.
Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`9`

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

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