Count of arrays having consecutive element with different values

3

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".


Output:

3

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