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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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". 

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 

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. 
?> 

Output :

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

Recommended Posts: