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 A1 = 1 and Ai ≠ 1.
Therefore, if x ≠ 1, the answer to the problem is f(n)/(k – 1), because f(n) is the number of way where An is filled with a number from 2 to k, and the answer are equal for all such values An, so the answer for an individual value is f(n)/(k – 1).
Otherwise, if x = 1, the answer is f(n – 1), because An – 1 ≠ 1, and the only number we can fill An with is x = 1.
Now, the main problem is how to calculate f(i). Consider all numbers that Ai – 1 can be. We know that it must lie in [1, k].

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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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".

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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.
?>

chevron_right



Output :

3


My Personal Notes arrow_drop_up

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.



Improved By : vt_m