# Count of arrays having consecutive element with different values

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; ` `} ` |

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## 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 ` |

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## 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". ` |

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## C#

`// C# program to find count of arrays. ` `using` `System; ` ` ` `class` `GFG ` `{ ` `// 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() ` ` ` `{ ` ` ` `int` `n = 4, k = 3, x = 2; ` ` ` `Console.WriteLine(countarray(n, k, x)); ` ` ` `} ` `} ` ` ` ` ` `// This code is contributed by vt_m ` |

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## PHP

`<?php ` `// PHP Program to find ` `// count of arrays. ` ` ` `$MAXN` `= 109; ` ` ` `// Return the number of arrays ` `// with given constartints. ` `function` `countarray(` `$n` `, ` `$k` `, ` `$x` `) ` `{ ` ` ` `$dp` `= ` `array` `( 0 ); ` ` ` ` ` `// Initalising dp[0] and dp[1]. ` ` ` `$dp` `[0] = 0; ` ` ` `$dp` `[1] = 1; ` ` ` ` ` `// Computing f(i) for ` ` ` `// each 2 <= i <= n. ` ` ` `for` `( ` `$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` `[` `$n` `- 1]); ` `} ` ` ` `// Driven Code ` `$n` `= 4; ` `$k` `= 3; ` `$x` `= 2; ` `echo` `countarray(` `$n` `, ` `$k` `, ` `$x` `) ; ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

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**Output :**

3

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