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Count of arrays having consecutive element with different values
  • Difficulty Level : Hard
  • Last Updated : 21 Apr, 2021

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++




// 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.
?>

Javascript




<script>
 
// Javascript program to find count of arrays.
let MAXN = 109;
   
function countarray(n, k, x)
{
    let dp = [];
 
    // Initalising dp[0] and dp[1].
    dp[0] = 0;
    dp[1] = 1;
 
    // Computing f(i) for each 2 <= i <= n.
    for(let 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
let n = 4, k = 3, x = 2;
 
document.write(countarray(n, k, x));
 
// This code is contributed by sanjoy_62
 
</script>

Output : 

3

 

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