Given three positive integers **n**, **k** and **x**. The task is to count the number of different array that can be formed of size n such that each element is between 1 to k and two consecutive element are different. Also, the first and last elements of each array should be 1 and x respectively.

Examples:

Input : n = 4, k = 3, x = 2 Output : 3

The idea is to use Dynamic Programming and combinatorics to solve the problem.

First of all, notice that the answer is same for all x from 2 to k. It can easily be proved. This will be useful later on.

Let the state f(i) denote the number of ways to fill the range [1, i] of array A such that A_{1} = 1 and A_{i} ≠ 1.

Therefore, if x ≠ 1, the answer to the problem is f(n)/(k – 1), because f(n) is the number of way where A_{n} is filled with a number from 2 to k, and the answer are equal for all such values A_{n}, so the answer for an individual value is f(n)/(k – 1).

Otherwise, if x = 1, the answer is f(n – 1), because A_{n – 1} ≠ 1, and the only number we can fill A_{n} with is x = 1.

Now, the main problem is how to calculate f(i). Consider all numbers that A_{i – 1} can be. We know that it must lie in [1, k].

- If A
_{i – 1}≠ 1, then there are (k – 2)f(i – 1) ways to fill in the rest of the array, because A_{i}cannot be 1 or A_{i – 1}(so we multiply with (k – 2)), and for the range [1, i – 1], there are, recursively, f(i – 1) ways. - If A
_{i – 1}= 1, then there are (k – 1)f(i – 2) ways to fill in the rest of the array, because A_{i – 1}= 1 means A_{i – 2}≠ 1 which means there are f(i – 2)ways to fill in the range [1, i – 2] and the only value that A_{i}cannot be 1, so we have (k – 1) choices for A_{i}.

By combining the above, we get

f(i) = (k - 1) * f(i - 2) + (k - 2) * f(i - 1)

This will help us to use dynamic programming using f(i).

Below is the implementation of this approach:

## C++

// CPP Program to find count of arrays. #include <bits/stdc++.h> #define MAXN 109 using namespace std; // Return the number of arrays with given constartints. int countarray(int n, int k, int x) { int dp[MAXN] = { 0 }; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for (int i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]); } // Driven Program int main() { int n = 4, k = 3, x = 2; cout << countarray(n, k, x) << endl; return 0; }

## Java

// Java program to find count of arrays. import java.util.*; class Counting { static int MAXN = 109; public static int countarray(int n, int k, int x) { int[] dp = new int[109]; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for (int i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]); } // driver code public static void main(String[] args) { int n = 4, k = 3, x = 2; System.out.println(countarray(n, k, x)); } } // This code is contributed by rishabh_jain

## Python3

# Python3 code to find count of arrays. # Return the number of lists with # given constraints. def countarray( n , k , x ): dp = list() # Initalising dp[0] and dp[1] dp.append(0) dp.append(1) # Computing f(i) for each 2 <= i <= n. i = 2 while i < n: dp.append( (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]) i = i + 1 return ( (k - 1) * dp[n - 2] if x == 1 else dp[n - 1]) # Driven code n = 4 k = 3 x = 2 print(countarray(n, k, x)) # This code is contributed by "Sharad_Bhardwaj".

Output:

3

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