# 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++ 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 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`

Output:
```9
```

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

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