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

Given two integers N and K, the task is to find the total count of N-digit number such that the sum of every K consecutive digits of the number is equal.

Examples:

Input: N = 2, K = 1
Output: 9
Explanation: 
The numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99 with sum of every 1 consecutive digits equal to 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.

Input: N = 3, K = 2
Output: 90

Naive Approach: Iterate for all possible N-digit numbers and calculate the sum of every K consecutive digits of the number. If all the sums are equal then include this is the count else check for the next number.



Below is the implementation of the above approach:

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
  
class GFG {
  
    // Function to count the number of
    // N-digit numbers such that sum of
    // every k consecutive digits are equal
    static int countDigitSum(int N, int K)
    {
        // Range of numbers
        int l = (int)Math.pow(10, N - 1),
            r = (int)Math.pow(10, N) - 1;
        int count = 0;
  
        for (int i = l; i <= r; i++) {
            int num = i;
  
            // Extract digits of
            // the number
            int digits[] = new int[N];
  
            for (int j = N - 1;
                 j >= 0; j--) {
  
                digits[j] = num % 10;
                num /= 10;
            }
            int sum = 0, flag = 0;
  
            // Store the sum of
            // first K digits
            for (int j = 0; j < K; j++)
                sum += digits[j];
  
            // Check for every
            // k-consective digits
            for (int j = 1;
                 j < N - K + 1; j++) {
  
                int curr_sum = 0;
  
                for (int m = j;
                     m < j + K; m++) {
  
                    curr_sum += digits[m];
                }
  
                // If sum is not equal
                // then break the loop
                if (sum != curr_sum) {
                    flag = 1;
                    break;
                }
            }
  
            // Increment the count if it
            // satisfy the given condition
            if (flag == 0) {
                count++;
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void
        main(String[] args)
    {
        // Given N and K
        int N = 2, K = 1;
  
        // Function call
        System.out.print(countDigitSum(N, K));
    }
}

chevron_right






                        filter_none











Output:


9





Time Complexity: O(10N * N * K)
Auxiliary Space: O(N)


Efficient Approach: To optimize the above naive approach the idea is to use the Sliding window technique to check if the sum of K-Consecutive digits of the number are equal or not. Below are the steps:


    

  1. Get the range of numbers i.e., 10N-1  to 10N.
    2. 

  2. For each number in the above range,  consider a window of length K and find the sum of each digit. Store this sum as S.
    3. 

  3. Find the sum of next K digits using the sliding window by including the next K digits in the sum and remove the previous K digits from the sum.
    4. 

  4. If the sum obtained is equal to the above sum S then check for next K digits.
    5. 

  5. Otherwise, repeat the above step for the next numbers.
    6. 



Below is the implementation of the above approach:




Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program for the above approach 


class GFG { 


  


    // Function to count the number of 


    // N-digit numbers such that sum of 


    // every k consecutive digits are equal 


    static int countDigitSum(int N, int K) 


    { 


        // Range of numbers 


        int l = (int)Math.pow(10, N - 1), 


            r = (int)Math.pow(10, N) - 1; 


        int count = 0; 


        for (int i = l; i <= r; i++) { 


            int num = i; 


  


            // Extract digits of the number 


            int digits[] = new int[N]; 


            for (int j = N - 1; j >= 0; j--) { 


                digits[j] = num % 10; 


                num /= 10; 


            } 


            int sum = 0, flag = 0; 


  


            // Store the sum of 


            // first K digits 


            for (int j = 0; j < K; j++) 


                sum += digits[j]; 


  


            // Check for every 


            // k-consective digits 


            // using sliding window 


            for (int j = K; j < N; j++) { 


  


                if (sum - digits[j - K] 


                        + digits[j] 


                    != sum) { 


                    flag = 1; 


                    break; 


                } 


            } 


            if (flag == 0) { 


                count++; 


            } 


        } 


        return count; 


    } 


  


    // Driver Code 


    public static void main(String[] args) 


    { 


        // Given integer N and K 


        int N = 2, K = 1; 


        System.out.print(countDigitSum(N, K)); 


    } 


}



















                        chevron_right







                        filter_none











Output:


9





Time Complexity: O(10N *N)
Auxiliary Space: O(N)


competitive-programming-img





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
jrishabh99
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.