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# 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:

## C++

 `// C++ program for``// 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``int` `countDigitSum(``int` `N, ``int` `K)``{``     ` `    ``// Range of numbers``    ``int` `l = (``int``)``pow``(10, N - 1),``        ``r = (``int``)``pow``(10, N) - 1;``    ``int` `count = 0;`` ` `    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ``int` `num = i;`` ` `        ``// Extract digits of the number``        ``int` `digits[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-consecutive 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``int` `main()``{``    ` `    ``// Given N and K``    ``int` `N = 2, K = 1;`` ` `    ``// Function call``    ``cout << countDigitSum(N, K);``    ``return` `0;``}` `// This code is contributed by target_2.`

## C

 `// C program for the above approach``#include ``#include ` `// Function to count the number of``// N-digit numbers such that sum of``// every k consecutive digits are equal``int` `countDigitSum(``int` `N, ``int` `K)``{``    ` `    ``// Range of numbers``    ``int` `l = (``int``)``pow``(10, N - 1),``        ``r = (``int``)``pow``(10, N) - 1;``    ``int` `count = 0;` `    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ``int` `num = i;` `        ``// Extract digits of the number``        ``int` `digits[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-consecutive 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``int` `main()``{``    ` `    ``// Given N and K``    ``int` `N = 2, K = 1;` `    ``// Function call``    ``printf``(``"%d"``, countDigitSum(N, K));``    ` `    ``return` `0;``}` `// This code is contributed by piyush3010`

## Java

 `// 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-consecutive 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));``    ``}``}`

## Python3

 `# Python3 program for the above approach` `# Function to count the number of``# N-digit numbers such that sum of``# every k consecutive digits are equal``def` `countDigitSum(N, K):``    ` `    ``# Range of numbers``    ``l ``=` `pow``(``10``, N ``-` `1``)``    ``r ``=` `pow``(``10``, N) ``-` `1``    ``count ``=` `0``    ` `    ``for` `i ``in` `range``(l, r ``+` `1``):``        ``num ``=` `i` `        ``# Extract digits of the number``        ``digits ``=` `[``0``] ``*` `N``        ` `        ``for` `j ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):``            ``digits[j] ``=` `num ``%` `10``            ``num ``/``/``=` `10``        ` `        ``sum` `=` `0``        ``flag ``=` `0` `        ``# Store the sum of first K digits``        ``for` `j ``in` `range``(``0``, K):``            ``sum` `+``=` `digits[j]` `        ``# Check for every``        ``# k-consecutive digits``        ``for` `j ``in` `range``(``1``, N ``-` `K ``+` `1``):``            ``curr_sum ``=` `0``            ` `            ``for` `m ``in` `range``(j, j ``+` `K):``                    ``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 ``+``=` `1``        ` `    ``return` `count` `# Driver code` `# Given N and K``N ``=` `2``K ``=` `1` `# Function call``print``(countDigitSum(N, K))` `# This code is contributed by sanjoy_62`

## C#

 `// C# program for 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``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-consecutive 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()``{``    ` `    ``// Given N and K``    ``int` `N = 2, K = 1;` `    ``// Function call``    ``Console.Write(countDigitSum(N, K));``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``

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. 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. Find the sum of the 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. If the sum obtained is equal to the above sum S then check for the next K digits.
5. Otherwise, repeat the above step for the next numbers.

Below is the implementation of the above approach:

## C++

 `// C++ program for 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``int` `countDigitSum(``int` `N, ``int` `K)``{``    ` `    ``// Range of numbers``    ``int` `l = (``int``)``pow``(10, N - 1),``        ``r = (``int``)``pow``(10, N) - 1;``    ` `    ``int` `count = 0;``    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ``int` `num = i;` `        ``// Extract digits of the number``        ``int` `digits[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-consecutive 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``int` `main()``{``    ` `    ``// Given integer N and K``    ``int` `N = 2, K = 1;``    ` `    ``cout<< countDigitSum(N, K)<

## C

 `// C program for the above approach``#include ``#include ` `// Function to count the number of``// N-digit numbers such that sum of``// every k consecutive digits are equal``int` `countDigitSum(``int` `N, ``int` `K)``{``    ` `    ``// Range of numbers``    ``int` `l = (``int``)``pow``(10, N - 1),``        ``r = (``int``)``pow``(10, N) - 1;``    ` `    ``int` `count = 0;``    ``for``(``int` `i = l; i <= r; i++)``    ``{``        ``int` `num = i;` `        ``// Extract digits of the number``        ``int` `digits[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-consecutive 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``int` `main()``{``    ` `    ``// Given integer N and K``    ``int` `N = 2, K = 1;``    ` `    ``printf``(``"%d"``, countDigitSum(N, K));``    ` `    ``return` `0;``}` `// This code is contributed by piyush3010`

## Java

 `// 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-consecutive 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));``    ``}``}` `/* This code is contributed by piyush3010 */`

## Python3

 `# Python3 program for the``# above approach` `# Function to count the``# number of N-digit numbers``# such that sum of every k``# consecutive digits are equal``def` `countDigitSum(N, K):``  ` `    ``# Range of numbers``    ``l ``=` `pow``(``10``, N ``-` `1``);``    ``r ``=` `pow``(``10``, N) ``-` `1``;``    ``count ``=` `0``;``    ` `    ``for` `i ``in` `range``(``1``, r ``+` `1``):``        ``num ``=` `i;` `        ``# Extract digits of``        ``# the number``        ``digits ``=` `[``0``] ``*` `(N);``        ` `        ``for` `j ``in` `range``(N ``-` `1``,``                       ``0``, ``-``1``):``            ``digits[j] ``=` `num ``%` `10``;``            ``num ``/``/``=` `10``;` `        ``sum` `=` `0``;``        ``flag ``=` `0``;` `        ``# Store the sum of``        ``# first K digits``        ``for` `j ``in` `range``(``0``, K):``            ``sum` `+``=` `digits[j];` `        ``# Check for every``        ``# k-consecutive digits``        ``# using sliding window``        ``for` `j ``in` `range``(K, N):``            ``if` `(``sum` `-` `digits[j ``-` `K] ``+``                ``digits[j] !``=` `sum``):``                ``flag ``=` `1``;``                ``break``;` `        ``if` `(flag ``=``=` `0``):``            ``count ``+``=` `1``;` `    ``return` `count;` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``  ` `    ``# Given integer N and K``    ``N ``=` `2``;``    ``K ``=` `1``;``    ``print``(countDigitSum(N, K));` `# This code is contributed by shikhasingrajput`

## C#

 `// C# program for 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``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-consecutive 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()``{``    ` `    ``// Given N and K``    ``int` `N = 2, K = 1;` `    ``// Function call``    ``Console.Write(countDigitSum(N, K));``}``}` `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`9`

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